Chapter 4: Motion in Two Dimensions PDF

60 134 Motion in Two Dimension E3 The motion of an object is called two dimensional, if two of the three co-ordinates are required to specify the position of the object in space changes w.r.t time. In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard t...

60 134 Motion in Two Dimension E3 The motion of an object is called two dimensional, if two of the three co-ordinates are required to specify the position of the object in space changes w.r.t time. In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard table, an insect crawling over the floor of a room, earth revolving around the sun etc. 1. Projectile motion 2. Circular ID Two special cases of motion in two dimension are motion D YG 3.1 Introduction. U PROJECTILE MOTION ST U A hunter aims his gun and fires a bullet directly towards a monkey sitting on a distant tree. If the monkey remains in his position, he will be safe but at the instant the bullet leaves the barrel of gun, if the monkey drops from the tree, the bullet will hit the monkey because the bullet will not follow the linear path. The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion. If the force acting on a particle is oblique with initial velocity then the motion of particle is called projectile motion. 3.2 Projectile. Motion in Two Dimension 135 A body which is in flight through the atmosphere but is not being propelled by any fuel is called projectile. Example: (i) A bomb released from an aeroplane in level flight (ii) A bullet fired from a gun 60 (iii) An arrow released from bow (iv) A Javelin thrown by an athlete 3.3 Assumptions of Projectile Motion. E3 (1) There is no resistance due to air. (2) The effect due to curvature of earth is negligible. (3) The effect due to rotation of earth is negligible. ID (4) For all points of the trajectory, the acceleration due to gravity ‘g’ is constant in magnitude and direction. U 3.4 Principles of Physical Independence of Motions. D YG (1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions. (2) The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical component. (3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component. (4) The horizontal motion is a uniform motion and the vertical motion is a uniformly U accelerated retarded motion. 3.5 Types of Projectile Motion. (2) Horizontal projectile motion ST (1) Oblique projectile motion (3) Projectile motion on an inclined plane X Y  X  X Y 3.6 Oblique Projectile. Y 136 Motion in Two Dimension In projectile motion, horizontal component of velocity (u cos), acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity (u sin ), momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point. 60 (1) Equation of trajectory : A projectile thrown with velocity u at an angle  with the horizontal. The velocity u can be resolved into two rectangular components. Y v cos  component along X–axis and u sin  component along Y–axis. y  (u sin  ) t  …. (i) u sin  u P y E3 For horizontal motion For vertical motion x x x = u cos  t  t  u cos   O u cos  1 2 gt 2 …. (ii) X ID   x  1  x2    g From equation (i) and (ii) y  u sin   2 2    u cos 2  u cos     1 gx 2 2 u 2 cos 2  U y  x tan   D YG This equation shows that the trajectory of projectile is parabolic because it is similar to equation of parabola y = ax – bx2 Note : Equation of oblique projectile also can be written as x  y  x tan  1   R  (where R = horizontal range = u 2 sin 2 ) g Sample problems based on trajectory The trajectory of a projectile is represented by y  3 x  gx 2 /2. The angle of projection is U Problem 1. (a) 30o comparing ST Solution : (c) By y  x tan   gx (b) 45o the coefficient (c) 60o of x in given (d) None of these equation with standard equation 2 2u cos 2  2 tan   3    60  Problem 2. The path followed by a body projected along y-axis is given as by y  3 x  (1 / 2)x 2 , if g = 10 m/s, then the initial velocity of projectile will be – (x and y are in m) (a) 3 10 m/s Solution : (b) By (b) 2 10 m/s comparing y  x tan   gx the coefficient 2 2u 2 cos 2 . (c) 10 3 m/s of x2 in given equation (d) 10 2 m/s with standard equation Motion in Two Dimension 137 g 2u cos  2 2  1 2 Substituting  = 60o we get u  2 10 m / sec. The equation of projectile is y  16 x  (a) 16 m 5x 2. The horizontal range is 4 (b) 8 m 60 Problem 3. (c) 3.2 m (d) 12.8 m x  Solution : (d) Standard equation of projectile motion y  x tan  1   R  x  5x 2  or y  16 x 1  64 / 5  4  E3 Given equation : y  16 x  64 =12.8 m. 5 By comparing above equations R  U ID  (2) Displacement of projectile ( r ) : Let the particle acquires a position P having the coordinates (x, y) just after time t from the instant of projection. The corresponding position  vector of the particle at time t is r as shown in the figure.  ….(i) r  xˆi  yˆj Y The horizontal distance covered during time t is given as ….(ii)  v D YG x  v x t  x  u cos  t The vertical velocity of the particle at time t is given as v y  (v 0 )y  gt, ….(iii) y P (x, y)   r   vy  O  vx X x vi Now the vertical displacement y is given as y  u sin  t  1 / 2 gt 2 ….(iv) U Putting the values of x and y from equation (ii) and equation (iv) in equation (i) we obtain the position vector at any time t as ST  1   r  (u cos  ) t ˆi   (u sin  ) t  gt 2  ˆj 2      r  (u t cos  )2   (u t sin  )  1 2 gt  2  2  ut sin   1 / 2 gt 2 gt sin   gt  and   tan 1 (y / x )  tan 1  r ut 1   (u t cos  ) u  2u   Note :  The angle of elevation  of the highest point of the     or   tan 1  2u sin   gt      2u cos    Y projectile and the angle of projection  are related to each other as 1 tan   tan  2 2 u   O H X R 138 Motion in Two Dimension Sample problems based on displacement A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 Newton’s in OF direction perpendicular to OE. The distance of the body from O after 4 seconds will be (a) 12 m (b) 28 m 60 Problem 4. (c) 20 m (d) Solution : (c) Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds  x  ut  3  4  12 m u O y E d 1 4 1F 1 y  ut  (a) t 2  0    t 2    4 2  16 m 22 2 m  2  d  20 m  x x 2  y 2  (12)2  (16) 2 ID So its distance from O is given by d  Problem 5. E3 and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds. F A body starts from the origin with an acceleration of 6 m/s2 along the x-axis and 8 m/s2 (a) U along the y-axis. Its distance from the origin after 4 seconds will be 56 m (b) 64 m 1 1 a x t 2   6  (4 ) 2  48 m 2 2 Displacement along Y- axis : y  u y t  1 1 a y t 2   8  (4 ) 2  64 m 2 2 (d) D YG Solution : (c) Displacement along X- axis : x  u x t  (c) 80 m Total distance from the origin  x 2  y 2  (48 ) 2  (64 ) 2  80 m U (3) Instantaneous velocity v : In projectile motion, vertical component of velocity changes but horizontal component of velocity remains always constant. ST Example : When a man jumps over the hurdle leaving behind its skateboard then vertical component of his velocity is changing, but not the horizontal component, which matches with the skateboard velocity. As a result, the skateboard stays underneath him, allowing him to land on it. Motion in Two Dimension 139 Let vi be the instantaneous velocity of projectile at time t direction of this velocity is along the tangent to the trajectory at point P.  v i  v x i  v y ˆj  v i  v x2  v y2  u 2 cos 2   (u sin   gt) 2 Direction of instantaneous velocity tan   vy vx  u sin   gt or u cos  60 v i  u 2  g 2 t 2  2u gt sin      tan 1 tan   gt  sec   u  E3 (4) Change in velocity : Initial velocity (at projection point) u i  u cos  ˆi  u sin  ˆj Final velocity (at highest point) u f  u cos  ˆi  0 ˆj ID (i) Change in velocity (Between projection point and highest point) u  u f  u i  u sin  ˆj When body reaches the ground after completing its motion then final velocity u f  u cos  ˆi  u sin  ˆj U (ii) Change in velocity (Between complete projectile motion) u  u f  u i  2u sin  ˆi Problem 6. D YG Sample problems based on velocity In a projectile motion, velocity at maximum height is (a) u cos  2 (b) u cos (c) u sin  2 (d) Solution : (b) In a projectile motion at maximum height body possess only horizontal component of velocity i.e. u cos. A body is thrown at angle 30o to the horizontal with the velocity of 30 m/s. After 1 sec, its U Problem 7. velocity will be (in m/s) (g = 10 m/s2) ST (a) 10 7 (b) 700 10 (c) 100 7 (d) Solution : (a) From the formula of instantaneous velocity v  u 2  g 2 t 2  2 u g t sin  v  (30) 2  (10) 2  1 2  2  30  10  1  sin 30 o  10 7 m / s Problem 8. A projectile is fired at 30o to the horizontal. The vertical component of its velocity is 80 ms– 1. Its time of flight is T. What will be the velocity of the projectile at t = T/2 (a) 80 ms–1 (b) 80 3 ms–1 (c) (80/ 3 ) ms–1 (d) Solution : (b) At half of the time of flight, the position of the projectile will be at the highest point of the parabola and at that position particle possess horizontal component of velocity only. 140 Motion in Two Dimension Given uvertical  u sin   80  u  sin 30 o  160 m / s u horizontal  u cos   160 cos 30 o  80 3 m / s.  Problem 9. 80 A particle is projected from point O with velocity u in a direction making an angle  with 60 the horizontal. At any instant its position is at point P at right angles to the initial direction of projection. Its velocity at point P is (b) u cot (c) u cosec (d) u sec 90o P E3 u tan v u  O ID (a) Solution : (b) Horizontal velocity at point ' O '  u cos  Horizontal velocity at point ' P '  v sin  remains constant throughout the motion P v sin 90o –  u u sin U In projectile motion horizontal component of velocity 90o v  O u cos D YG  v sin   u cos   v  u cot Problem 10. A particle P is projected with velocity u1 at an angle of 30o with the horizontal. Another particle Q is thrown vertically upwards with velocity u2 from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is u1  u 2 (b) u1  2u 2 (c) u1  ST U (a) (d) u2 2 u1  4u 2 u1 P u2 30o Q Solution : (b) Both particle collide at the highest point it means the vertical distance travelled by both the particle will be equal, i.e. the vertical component of velocity of both particle will be equal u 1 sin 30   u 2  u1  u2 2  u 1  2u 2 Problem 11. Two seconds after projection a projectile is travelling in a direction inclined at 30 o to the horizontal after one more sec, it is travelling horizontally, the magnitude and direction of its velocity are (a) 2 20 m /sec, 60 o [RPET 1999] (b) 20 3 m /sec, 60 o (c) 6 40 m/sec, 30 o (d) 40 6 m /sec, 30 o Motion in Two Dimension 141 Solution : (b) Let in 2 sec body reaches upto point A and after one more sec upto point B. Total time of ascent for a body is given 3 sec i.e. t   u sin   10  3  30 u sin  3 g v …..(i) A u …..(ii)  v sin 30  u sin   g  2 v sin 30 o  30  20 Using v  u  g t  As u sin   30   v  20 m / s. Substituting this value in equation (ii) u cos O E3 For vertical upward motion between point O and A o 60 Horizontal component of velocity remains always constant u cos   v cos 30  B 30o u cos  20 cos 30 o  10 3 …..(iii) ID From equation (i) and (iii) u  20 3 and   60  Problem 12. A body is projected up a smooth inclined plane (length = 20 2 m ) with velocity u from the point M as shown in the figure. The angle of inclination is 45 o and the top is connected to a U well of diameter 40 m. If the body just manages to cross the well, what is the value of v (a) 40 ms 1 45o D YG (b) 40 2 ms 1 (c) 20 ms 1 M 40 m (d) 20 2 ms 1 Solution : (d) At point N angle of projection of the body will be 45°. Let velocity of projection at this point is v. If the body just manages to cross the well then Range  Diameter of well v 2 sin 2  40 g U As  v 2  400   45  v N R v  20 m / s 45o u ST But we have to calculate the velocity (u) of the body at point M. For motion along the inclined plane (from M to N) M 40 m Final velocity (v) = 20 m/s, acceleration (a) = – g sin = – g sin 45o, distance of inclined plane (s) = 20 2 m (20 ) 2  u 2  2 g.20 2 [Using v2 = u2 + 2as] 2 u 2  20 2  400  u  20 2 m / s. Problem 13. A projectile is fired with velocity u making angle  with the horizontal. What is the change in velocity when it is at the highest point 142 Motion in Two Dimension u cos (a) (c) u sin (b) u (d) Solution : (c) Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes. Initially vertical component u sin  60 Finally it becomes zero. So change in velocity  u sin  (i) Change in momentum p  p f  p i   mu sin  ˆj (Between E3 (5) Change in momentum : Simply by the multiplication of mass in the above expression of velocity (Article-4). projection point and highest point) ID (ii) Change in momentum (For the complete projectile motion) p  p f  p i   2mu sin  ˆj (6) Angular momentum : Angular momentum of projectile at highest point of trajectory about the point of projection is given by Y U L  mvr  u 2 sin 2   Here r  H   2 g   u 2 sin 2  m u 3 cos  sin 2  L  m u cos   2g 2g D YG  P = mv u  r X O Sample problems based on momentum and angular momentum Problem 14. A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30o with the horizontal. The change in momentum (in magnitude) of the body is U (a) 24.5 N–s (b) 49.0 N–s (c) 98.0 N–s (d) 50.0 N–s Solution : (b) Change in momentum between complete projectile motion = 2mu sin  2  0.5  98  sin 30  = ST 49 N–s. Problem 15. A particle of mass 100 g is fired with a velocity 20 m sec–1 making an angle of 30o with the horizontal. When it rises to the highest point of its path then the change in its momentum is (a) 3 kg m sec 1 (b) 1/2 kg m sec–1 (c) 2 kg m sec 1 (d) 1 kg m sec–1 Solution : (d) Horizontal momentum remains always constant  So change in vertical momentum ( p ) = Final vertical momentum – Initial vertical momentum  0  mu sin  | P |  0.1  20  sin 30 o  1 kg m / sec. Motion in Two Dimension 143 Problem 16. Two equal masses (m) are projected at the same angle () from two points separated by their range with equal velocities (v). The momentum at the point of their collision is (b) 2 mv cos (a) Zero (c) – 2 mv cos (d) None of these Solution : (a) Both masses will collide at the highest point of their trajectory with equal and opposite mv cos mv cos v v  E3  60 momentum. So net momentum of the system will be zero. Problem 17. A particle of mass m is projected with velocity v making an angle of 45o with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when (b) mv3/ (4 2 g) (a) Zero m u 3 cos  sin 2  mv 3 = 2g (4 2 g) (c) mv3/ ( 2 g) (d) mv2/2g [As  = 45o] U Solution : (b) L  ID the particle is at its maximum height is (where g = acceleration due to gravity) Problem 18. A body is projected from the ground with some angle to the horizontal. What happens to Solution : (b) D YG the angular momentum about the initial position in this motion (a) Decreases (b) Increases (c) Remains same (d) First increases and then decreases Problem 19. In case of a projectile, where is the angular momentum minimum (a) At the starting point At the highest point U (b) (c) On return to the ground (d) At some location other than those mentioned above ST Solution : (a) (7) Time of flight : The total time taken by the projectile to go up and come down to the same level from which it was projected is called time of flight. For vertical upward motion 0 = u sin  – gt  t = (u sin /g) Now as time taken to go up is equal to the time taken to come down so Time of flight T  2 t  2u sin  g 144 Motion in Two Dimension (i) Time of flight can also be expressed as : T  2.u y g (where uy is the vertical component of initial velocity). (ii) For complementary angles of projection  and 90o –  T1 T 2u sin  / g = tan   1  tan   T2 T2 2u sin( 90   ) / g (b) Multiplication of time of flight = T1 T 2  60 (a) Ratio of time of flight = 2u sin  2u cos  g g  T1 T2  2R g u sin   g(t1  t 2 ) 2 Y t ID or E3 (iii) If t1 is the time taken by projectile to rise upto point p and t2 is the time taken in 2u sin   time of flight falling from point p to ground level then t1  t 2  g and height of the point p is given by h  u sin  t1  1 2 gt1 2 P t 1 h 2 X O D YG U (t  t 2 ) 1 hg 1 t1  gt12 2 2 gt t by solving h  1 2 2 (iv) If B and C are at the same level on trajectory and the time difference between these two points is t1, similarly A and D are also at the same level and the time difference between these two positions is t2 then 8h g Y h ST B A t1 t2 C D O U t 22  t12  X Sample problems based on time of flight Problem 20. For a given velocity, a projectile has the same range R for two angles of projection if t1 and t2 are the times of flight in the two cases then (a) t1t 2  R 2 (b) t1t 2  R (c) t1t 2  Solution : (b) As we know for complementary angles t 1 t 2  1 R (d) t1t 2  1 R2 2R  t1 t 2  R. g Problem 21. A body is thrown with a velocity of 9.8 m/s making an angle of 30o with the horizontal. It will hit the ground after a time (a) 1.5 s (b) 1 s (c) 3 s (d) 2 s Motion in Two Dimension 145 Solution : (b) T  2u sin  2  9.8  sin 30 o  = 1sec g 9.8 Problem 22. Two particles are separated at a horizontal distance x as shown in figure. They are projected at the same time as shown in figure with different initial speed. The time after 60 which the horizontal distance between the particles become zero is u u/ 3 60o 30o A (b) x/u E3 (a) u /2 x B x (c) 2u/ x (d) u/ x ID Solution : (b) Let x 1 and x 2 are the horizontal distances travelled by particle A and B respectively in time t. u. cos 30   t …..(i) u ……(ii). cos 30 o  t  u cos 60 o  t  ut  x  ut  t  x / u 3 D YG x1  x 2  x 2  u cos 60 o  t and 3 U x1  Problem 23. A particle is projected from a point O with a velocity u in a direction making an angle  upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is (a) u sin  g (b) u cosec g (c) u tan  g (d) u sec  g Solution : (b) When body projected with initial velocity u by making angle  with the horizontal. Then U after time t, (at point P) it’s direction is perpendicular to u. Magnitude of velocity at point P is given by v  u cot . (from sample problem no. 9) ST For vertical motion : Initial velocity (at point O)  u sin  Final velocity (at point P)   v cos   u cot cos u sin 90o Time of flight (from point O to P) = t Applying first equation of motion v  u  g t O P u  u cos v cos (90 – ) v  u cot cos  u sin   g t  t   u sin   u cot  cos  u cosec u  sin 2   cos 2   g g sin  g Problem 24. A ball is projected upwards from the top of tower with a velocity 50 ms –1 making angle 30o with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground 146 Motion in Two Dimension (a) 2.33 sec (b) 5.33 sec (c) 6.33 sec (d) 9.33 sec Solution : (c) Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by So we can resolve the given velocity in vertical direction and can apply the above formula. Initial vertical component of E3 25  2  9.8  70  1  1   = 6.33 sec. 9.8  (25 ) 2  t  70 m velocity u sin  50 sin 30  25 m / s. 60 u u sin30o 30o u 2 gh  t  1  1  2  g  u  becomes Solution : (c) R  (b) 2 times u 2 sin 2 2u sin  and T  g g (c) 2 times (d) 1 / 2 times U (a) 4 times ID Problem 25. If for a given angle of projection, the horizontal range is doubled, the time of flight D YG  R  u 2 and T  u (If  and g are constant). In the given condition to make range double, velocity must be increased upto of previous value. So automatically time of flight will becomes 2 times that 2 times. Problem 26. A particle is thrown with velocity u at an angle  from the horizontal. Another particle is thrown with the same velocity at an angle  from the vertical. The ratio of times of flight of two particles will be (a) tan 2  : 1 (b) cot 2  : 1 (c) tan  : 1 (d) cot  : 1 U Solution : (c) For first particles angle of projection from the horizontal is . So T1  2u sin  g ST For second particle angle of projection from the vertical is . it mean from the horizontal is (90  ).  T2  T 2u sin (90   ) 2u cos  . So ratio of time of flight 1  tan . g T2 g Problem 27. The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = 10 ms–2). The time of flight will be decreased by (a) 0% T g 2u sin  1  2  Solution : (c) T  g T2 g1 (b) 1% (c) 9% g 10  11 g 10 g Fractional decrease in time of flight  T1  T2 1  T1 11 (d) 11% Motion in Two Dimension 147 Percentage decrease = 9% (8) Horizontal range : It is the horizontal distance travelled by a body during the time of flight. So by using second equation of motion R  u cos   T  u cos   (2u sin  / g)  u sin 2 g u 2  O (i) Range of projectile can also be expressed as : 2u x u y g Horizontal range X 2u sin  2 u cos  u sin  2u x u y   g g g ID R = u cos × T = u cos   R E3 R u sin 2 g 60 Y 2 (where ux and uy are the horizontal and vertical component of initial U velocity) D YG (ii) If angle of projection is changed from  to   = (90 – ) then range remains unchanged. 60o 30o U Blast u 2 sin 2 ' u 2 sin[ 2(90 o   )] u 2 sin 2   R g g g ST R'  So a projectile has same range at angles of projection  and (90 – ), though time of flight, maximum height and trajectories are different. These angles  and 90o –  are called complementary angles of projection and for complementary angles of projection ratio of range R1 u 2 sin 2 / g R1  2 1  1 o R 2 u sin [2 (90   )] / g R2 (iii) For angle of projection 1 = (45 – ) and 2 = (45 + ), range will be same and equal to u cos 2/g. 2 1 and 2 are also the complementary angles. Y u H 45o 148 Motion in Two Dimension (iv) Maximum range : For range to be maximum d  u 2 sin 2  dR 0   0 d   g d  60  cos 2 = 0 i.e. 2 = 90o   = 45o and Rmax = (u2/g) i.e., a projectile will have maximum range when it is projected at an angle of 45o to the horizontal and the maximum range will be (u2/g). When the range is maximum, the height H reached by the projectile E3 u 2 sin 2  u 2 sin 2 45 u 2 R max    2g 2g 4g 4 H i.e., if a person can throw a projectile to a maximum distance Rmax, The maximum height to which  .  ID  R max  4 it will rise is  (v) Relation between horizontal range and maximum height : R   R  4H cot  U R u 2 sin 2 / g  2  4 cot  H u sin 2  / 2 g  u 2 sin 2  u 2 sin 2 and H  g 2g (vi) If in case of projectile motion range R is n times the maximum height H u 2 sin 2 u 2 sin 2  n  tan   [4 / n] or   tan 1 [4 / n] g 2g D YG i.e. R = nH  The angle of projection is given by   tan 1 [4 / n] Note : If R = H then   tan 1 (4 ) or   76 o. If R = 4H then   tan 1 (1) or   45 o. U Sample problem based on horizontal range Problem 28. A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at ST an angle of 30o with the horizontal. How far from the throwing point will the ball be at 1 3 the height of 10 m from the ground (g = 10 m/s2, sin 30o = , cos 30o  ) 2 2 (a) 8.66 m (b) 5.20 m (c) 4.33 m (d) 2.60 m Solution : (a) Simply we have to calculate the range of projectile R u u sin 2 (10 ) sin(2  30 )  g 10 2 2 R  5 3  8.66 meter 30o 10 m 10 m Problem 29. Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long–jump event Motion in Two Dimension 149 (a) Speed before he jumps and his weight (b) The direction in which he leaps and the initial speed (c) The force with which he pushes the ground and his speed Solution : (b) Because range  (Velocity of projection ) 2  sin 2(Angle of project ion) g 60 (d) The direction in which he leaps and the weight Problem 30. For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms–2) (a) 5 : 4 Solution : (a) H  E3 [EAMCET (Med.) 2000] (b) 5 : 2 u 2 sin 2  2u sin and T  2g g (c) 5 : 1  H T2  (d) 10 : 1 g 10 5 u 2 sin 2  / 2 g    2 2 2 8 8 4 4 u sin  / g ID Problem 31. A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with which he throws the ball is (to the nearest integer) Solution : (c) R max  (b) 42 ms–1 u2 = 100 g (d) 35 ms–1 (when   45  )  31.62 m / s. D YG  u  1000 (c) 32 ms–1 U (a) 30 ms–1 Problem 32. If two bodies are projected at 30o and 60o respectively, with the same velocity, then [CBSE PMT 2000; JI (a) Their ranges are same same (b) (c) Their times of flight are same Their heights are (d) All of these Solution : (a) Because these are complementary angles. U Problem 33. Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank ST the paths according to initial horizontal velocity component, highest first (a) 1, 2, 3, 4 y 1 2 3 4 x 0 (b) 2, 3, 4, 1 (c) 3, 4, 1, 2 (d) 4, 3, 2, 1 Solution : (d) Range  horizontal component of velocity. Graph 4 shows maximum range, so football possess maximum horizontal velocity in this case. Problem 34. Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15o, 30o, 45o and 60o with the horizontal respectively. The body having shortest range is 150 Motion in Two Dimension (a) P (b) Q (c) R (d) S Solution : (a) Range of projectile will be minimum for that angle which is farthest from 45°. Problem 35. A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed Solution : (c) R  (b) 150 m u 2 sin 2 g  R  u 2 so (c) 200 m R2  u2  R1  u1 2 (d) 250 m 60 (a) 100 m  2u      R 2  4 R1 = 4  50 = 200 m  u   2 10 m/s2. Then the range is (a) 25 × 103 m (c) 50 × 102 m (d) 25 × 102 m u 2 sin 2 500 2 sin2  15   12500 m  12.5  10 3 m  g 10 ID Solution : (b) Range (R)  (b) 12.5 × 103 m E3 Problem 36. A bullet is fired from a canon with velocity 500 m/s. If the angle of projection is 15o and g = Problem 37. A projectile thrown with a speed v at an angle  has a range R on the surface of earth. For same v and , its range on the surface of moon will be u 2 sin 2 g R Moon g  Earth = 6 R Earth g Moon (c) R/36 (d) 36 R  R 1/ g D YG Solution : (b) R  (b) 6 R U (a) R/6 1    g Moon  6 g Earth    R Moon  6 R Earth  6 R Problem 38. A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of projection as origin, the co-ordinates of the point where the speed of the particle is U minimum are (a) (R, R)  R (b)  R,   2 R R (c)  ,  2 4  R (d)  R,   4 ST Solution : (c) For maximum horizontal Range   45  From R  4 H cot  = 4H [As  = 45o, for maximum Y range.] Speed of the particle will be minimum at the highest R/4 45o point of parabola. X R/2 So the co-ordinate of the highest point will be (R/2, R/4) Problem 39. The speed of a projectile at the highest point becomes horizontal range of the projectile will be 1 2 times its initial speed. The Motion in Two Dimension 151 (a) u2 g (b) u2 2g (c) u2 3g u Solution : (a) Velocity at the highest point is given by u cos   (d) (given) u2 4g   = 45o 2 u 2 sin 2 u 2 sin(2  45 o ) u 2   g g g 60 Horizontal range R  Problem 40. A large number of bullets are fired in all directions with same speed u. What is the (a)  u2 g E3 maximum area on the ground on which these bullets will spread u4 (b)  (c)  2 g2 u4 (d)  2 g2 u2 g2 Solution : (b) The maximum area will be equal to area of the circle with radius equal to the maximum Maximum area r   R max 2  2 ID range of projectile u2    g  2 4    u  g2  [As r  R max  u 2 / g for  = 45o] U Problem 41. A projectile is projected with initial velocity (6ˆi  8 ˆj) m / sec. If g = 10 ms–2, then horizontal range is (b) 9.6 metre D YG (a) 4.8 metre  (c) 19.2 metre (d) 14.0 metre  Solution : (b) Initial velocity  6ˆi  8 Jˆ m / s (given) Magnitude of velocity of projection u  u x2  u y2  6 2  8 2 = 10 m/s Angle of projection tan   uy ux  8 4 4 3   sin   and cos   6 3 5 5 U u 2 2 sin  cos  u 2 sin 2   Now horizontal range R  g g (10 ) 2  2  10 4 3  5 5  9.6 meter ST Problem 42. A projectile thrown with an initial speed u and angle of projection 15o to the horizontal has a range R. If the same projectile is thrown at an angle of 45o to the horizontal with speed 2u, its range will be (a) 12 R Solution : (c) R  (b) 3 R (c) 8 R (d) 4 R u 2 sin 2  R  u 2 sin 2 g R2  u2  R1  u 1    2  sin 2 2   sin 2 1   2u    R 2  R1    u   2  sin 90 o   sin 30 o     8 R1   Problem 43. The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is 152 Motion in Two Dimension (a) (b) u 2 / 3 g 3u 2 / 2 g (c) 3u 2 / 2 g (d) 3u 2 / g Solution : (a) If the velocity of projection is u then at the highest point body posses only u cos  u (given) 2 Now R     60  u 2 sin(2  60 ) 3 u2  2 g g 60 u cos   Problem 44. A projectile is thrown from a point in a horizontal place such that its horizontal and (a) 4.9 m Solution : (d) We know R  (b) 9.8 m 2u x u y g  E3 vertical velocity component are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is (c) 19.6 m 2  9.8  19.6  39.2 m 9.8 (d) 39.2 m ID Where u x  horizontal component of initial velocity, u y  vertical component of initial velocity. Problem 45. A particle is projected with a velocity v such that its range on the horizontal plane is twice U the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity) 4v 2 5g (b) 4g (c) 5v 2 D YG (a) [BHU 1984] v2 g (d) 4v 2 5g Solution : (a) We know R  4 H cot  2 H  4 H cot   cot   u 2. 2. sin . cos  Range  g As R  2H given  1 2 1 ; sin  ; cos   2 5 5 2u 2 2. 1 5 g  5  4u 2 5g U Problem 46. The range R of projectile is same when its maximum heights are h1 and h2. What is the relation between R and h1 and h2 ST (a) R  h1 h 2 (b) R  2 h1 h 2 (d) R  4 h1 h 2 (c) R  2 h1 h 2 Solution : (d) For equal ranges body should be projected with angle  or (90 o   ) from the horizontal. And for these angles : h1  u 2 cos 2  u 2 sin 2  and h 2  2g 2g by multiplication of both height : h1 h 2  u 2 sin 2 cos 2 4g2 1  16  u 2 sin 2    g      2  16 h1 h 2  R 2  R  4 h1 h 2 Problem 47. A grasshopper can jump maximum distance 1.6 m. It spends negligible time on the ground. How far can it go in 10 seconds Motion in Two Dimension 153 (a) 5 2 m (b) 10 2 m (c) 20 2 m (d) 40 2 m Solution : (c) Horizontal distance travelled by grasshopper will be maximum for   45  R max  u2  1.6 m g  u  4 m / s. 60 Horizontal component of velocity of grasshopper u cos   4 cos 45  2 2 m / s 1.6 m Total distance covered by it in10 sec. S  u cos   t  2 2  10  20 2 m Problem 48. A projectile is thrown with an initial velocity of v  aˆi  bˆj, if the range of projectile is (a) a = 2b (b) b = a vy vx (c) b = 2a  tan 1 b a From formula R  4 H cot   2 H  cot   b = 2a D YG U ST b a 1  tan   2 2 U From equation (i) and (ii)  tan   (d) b = 4a …(i) ID Solution : (c) Angle of projection   tan 1 E3 double the maximum height reached by it then …(ii) [As R = 2H given] 152 Motion in Two Dimension (9) Maximum height : It is the maximum height from the point of projection, a projectile can reach. So, by using v 2  u 2  2as Y 0  (u sin  ) 2  2 gH u y2 (where u y is the vertical component of initial velocity). 2g u2 2g (when sin2 = max = 1 i.e.,  = 90o) ID (ii) H max  X E3 (i) Maximum height can also be expressed as Max. height  O H 60 u u 2 sin 2  H 2g i.e., for maximum height body should be projected vertically upward. So it falls back to the point of projection after reaching the maximum height.  H1 sin 2  u 2 sin 2  / 2 g  2   tan 2  2 H 2 u sin 2 (90 o   )2 g cos  D YG Ratio of maximum height = U (iii) For complementary angles of projection  and 90o –  H1  tan 2  H2 Sample problem based on maximum height Problem 49. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same U effort, he throws the ball vertically upwards. The maximum height attained by the ball is (a) 100 m u2  100 m g (c) 60 m (d) 50 m (when   45  ) ST Solution : (d) R max  (b) 80 m u 2  100  10  1000 H max  u2 1000   50 metre. 2g 2  10 (when   90  ) Problem 50. A ball thrown by one player reaches the other in 2 sec. the maximum height attained by the ball above the point of projection will be about (a) 10 m Solution : (c) T  2u sin   2 sec g (b) 7.5 m (given) (c) 5 m u sin   10 (d) 2.5 m Motion in Two Dimension 153 Now H  u 2 sin 2  (10 ) 2   5 m. 2g 2  10 Problem 51. Two stones are projected with the same magnitude of velocity, but making different angles (a) Y (b) 2 Y 60 with horizontal. The angle of projection of one is /3 and its maximum height is Y, the maximum height attained by the other stone with as /6 angle of projection is (c) 3 Y (d) Y 3 Solution : (d) When two stones are projected with same velocity then for complementary angles  and Ratio of maximum heights : E3 (90o –  ) H1 H  Y  tan 2   tan 2  3  H 2  1  H2 3 3 3 maximum height reached by it will (a) Remain the same u 2 sin 2 2g (b) Be doubled (c) Be quadrupled (d) Be halved [As   constant] H  u 2 U Solution : (c) H  ID Problem 52. If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the If initial velocity of a projectile be doubled then H will becomes 4 times. D YG Problem 53. Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If Pankaj throws his ball vertically up and Sudhir at an angle , both of them stay in our view for the same period. The height attained by the two balls are in the ratio (a) 2 : 1 (c) 1 : cos (b) 1 : 1 Solution : (b) Time of flight for the ball thrown by Pankaj T1  U ST  Height of the ball thrown by Sudhir H 2   H1 u 2 / 2g  2 1 2 =1 H 2 u 2 cos  / 2 g 2u 2 sin(90 g o ) 2u 2 cos  g u2 90o –  2m  u1  u 2 cos  Short Trick : Maximum height H  T2 u 12 2g u 22   m 2u1 2u 2 cos   g g Height of the ball thrown by Pankaj H 1  2 u1 g u1 Time of flight for the ball thrown by Sudhir T2  According to problem T1  T2 (d) 1 : sec 2 sin (90 2g [As u1  u 2 cos  ] o )  u 22 2  T1  cos H   T  2  2 2g H  1  1 (As T1 = T2) H2 H 2 1 Problem 54. A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. If the gun can impart a velocity of 500 ms–1 to the bullet. At what height above the bird must he aim his gun in order to hit it (take g = 10 ms–2) 154 Motion in Two Dimension [CPMT 1996] (a) 20 cm (b) 10 cm (c) 50 cm (d) 100 cm Solution : (a) Time taken by bullet to travel a horizontal distance of 100 m is given by t  100 1  sec 500 5 60 In this time the bullet also moves downward due to gravity its vertical displacement 2 h 1 2 1 1 g t   10     1 / 5 m = 20 cm 2 2 5 So bullet should be fired aiming 20 cm above the bird to hit it. E3 Problem 55. The maximum horizontal range of a projectile is 400 m. The maximum height attained by it will be (a) 100 m (b) 200 m (c) 400 m Solution : (a) R max  400 m [when   45  ] (d) 800 m ID So from the Relation R  4 H cot   400  4 H cot 45   H  100 m. Problem 56. Two bodies are projected with the same velocity. If one is projected at an angle of 30 o and (a) 3 : 1 (b) 1 : 3 H1 sin 2  1 sin 2 30 o 1    2 o 2 H2 3 sin  2 sin 60 D YG Solution : (b) U the other at an angle of 60o to the horizontal, the ratio of the maximum heights reached is [EAMCET (Med.) 1995; Pb. PMT 2000; AIIMS 2001] (c) 1 : 2 (d) 2 : 1 Problem 57. If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained by it will be (a) 125 m (c) 100 m (d) 150 m u 2 sin 2  2u sin  (50 )2  10 sec  u sin   50 so H    125 m. 2g g 2  10 U Solution : (a) T  (b) 50 m [RPMT 1997; RPET 1998] ST Problem 58. A man can throw a stone 80 m. The maximum height to which he can raise the stone is (a) 10 m (b) 15 m (c) 30 m (d) 40 m Solution : (d) The problem is different from problem no. (54). In that problem for a given angle of projection range was given and we had find maximum height for that angle. But in this problem angle of projection can vary, R max  u2  80 m [for   45  ] g But height can be maximum when body projected vertically up H max  u 2 sin 2 90 o u2 1 u2    2g 2 g 2  g   = 40 m   Motion in Two Dimension 155 Problem 59. A ball is thrown at different angles with the same speed u and from the same points and it has same range in both the cases. If y1 and y2 be the heights attained in the two cases, then y1  y 2  u2 g 2u 2 g (b) (c) u2 2g (d) u2 4g 60 (a) Solution : (c) Same ranges can be obtained for complementary angles i.e.  and 90o –  u 2 sin 2  u 2 cos 2  and y 2  2g 2g  y1  y 2  u2 u 2 sin 2  u 2 cos 2    2g 2g 2g E3 y1  (10) Projectile passing through two different points on same height at time t1 and t2 : If the particle passes two points situated at equal height y at t  t1 and t  t 2 , then 1 2 gt 2 2 Y.....(ii)  Comparing equation (i) with equation (ii) y y X O D YG gt1  t 2  u sin   2 t = t2 t = t1 u U y  u sin   t 2  and 1 2 gt1.....(i) 2 ID (i) Height (y): y  u sin   t1  Substituting this value in equation (i) 1 t t  y  g 1 2  t1  gt12 2  2   y (ii) Time (t1 and t2): y  u sin  t  1 2 gt 2 U   2 gy u sin   1 1  t   u sin  g   2u sin  2y t  t 0 g g ST 2   2 gy u sin    t1  1  1   u sin  g    gt1 t 2 2 2    and      2           2 gy u sin    t2  1  1   u sin  g    2         (11) Motion of a projectile as observed from another projectile : Suppose two balls A and B are projected simultaneously from the origin, with initial velocities u1 and u2 at angle 1 and 2, respectively with the horizontal. The instantaneous positions of the two balls are given by Ball A : x1 = (u1 cos1)t y 1  (u1 sin  1 ) t  Y 1 2 gt 2 u1 y1 – y2 u2 1 2 O B X 156 Motion in Two Dimension Ball B : x2 = (u2 cos2)t y 2  (u 2 sin  2 ) t  1 2 gt 2 The position of the ball A with respect to ball B is given by x  x 1  x 2  (u1 cos  1  u 2 cos  2 ) t Now y  u1 sin  1  u 2 sin  2  x  u1 cos  1  u 2 cos  2 60 y  y 1  y 2  (u1 sin  1  u 2 sin  2 ) t    constant  E3 Thus motion of a projectile relative to another projectile is a straight line. (12) Energy of projectile : When a projectile moves upward its kinetic energy decreases, potential energy increases but the total energy always remain constant. ID If a body is projected with initial kinetic energy K(=1/2 mu2), with angle of projection  with the horizontal then at the highest point of trajectory  1 1 m (u cos  )2  mu 2 cos 2  2 2 K'  K cos  2 U (i) Kinetic energy  Y u cos K = Kcos2 u  K X 2 2   As H  u sin   2g  (iii) Total energy = Kinetic energy + Potential energy  1 1 mu 2 cos 2   mu 2 sin 2  2 2 D YG u 2 sin 2  1  mu 2 sin 2  2g 2 (ii) Potential energy  mgH  mg =     1 mu 2 = Energy at the point of projection. 2 U This is in accordance with the law of conservation of energy. Sample problems based on energy ST Problem 60. A projectile is projected with a kinetic energy K. Its range is R. It will have the minimum kinetic energy, after covering a horizontal distance equal to (a) 0.25 R (b) 0.5 R (c) 0.75 R (d) R Solution : (b) Projectile possess minimum kinetic energy at the highest point of the trajectory i.e. at a horizontal distance R / 2. Problem 61. A projectile is fired at 30o with momentum p. Neglecting friction, the change in kinetic energy when it returns to the ground will be (a) Zero (b) 30% (c) 60% (d) 100% Solution : (a) According to law of conservation of energy, projectile acquire same kinetic energy when it comes at same level. Problem 62. A particle is projected making angle 45 o with horizontal having kinetic energy K. The kinetic energy at highest point will be Motion in Two Dimension 157 (a) K (b) 2 K 2 (c) 2K (d) K Solution : (b) Kinetic energy at the highest point K '  K cos 2   K cos 2 45 o  K / 2 Problem 63. Two balls of same mass are projected one vertically upwards and the other at angle 60 o with the vertical. The ratio of their potential energy at the highest point is (b) 2 : 1 (c) 4 : 1 Solution : (c) Potential energy at the highest point is given by PE  1 mu 2 2 1 mu 2 sin 2  2 E3 For first ball   90   (PE)1  (d) 4 : 3 60 (a) 3 : 2 For second ball   (90 o  60 o )  30  from the horizontal  (PE) 2  PE I  4 : 1 (PE)II ID  1 1 mu 2 sin 2 30   mu 2 2 8 Problem 64. In the above problem, the kinetic energy at the highest point for the second ball is K. What is the kinetic energy for the first ball (b) 3 K (c) 2 K (d) Zero U (a) 4 K Solution : (d) KE at the highest point KE  1 mu 2 cos 2  2 D YG For first ball  = 90o  KE = 0 Problem 65. A ball is thrown at an angle  with the horizontal. Its initial kinetic energy is 100 J and it becomes 30 J at the highest point. The angle of projection is (a) 45o (b) 30o (d) cos 1 ( 3 / 10 ) (c) cos–1 (3/10) Solution : (d) KE at highest point K '  K cos 2  U 30  100 cos 2   cos 2    3  3     cos 1   10  10   ST 3.7 Horizontal Projectile. A body be projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time. (1) Trajectory of horizontal projectile : The horizontal displacement x is governed by the equation x = ut  t  x u …. (i) u O X y x Y P(x, y) 158 Motion in Two Dimension The vertical displacement y is governed by y  1 2 gt 2 …. (ii) By substituting the value of t in equation (ii) y  1 gx2 2 u2 60 (since initial vertical velocity is zero) Sample problems based on trajectory E3 Problem 66. An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth’s surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling [MP PET 1993] ID (a) On a parabolic path as seen by pilot in the plane (b) Vertically along a straight path as seen by an observer on the ground near the target (c) On a parabolic path as seen by an observer on the ground near the target U (d) On a zig-zag path as seen by pilot in the plane D YG Solution : (c) X Y U The path of the ball appears parabolic to a observer near the target because it is at rest. But to a Pilot the path appears straight line because the horizontal velocity of aeroplane and the ball are equal, so the relative horizontal displacement is zero. ST Problem 67. The barrel of a gun and the target are at the same height. As soon as the gun is fired, the target is also released. In which of the following cases, the bullet will not strike the target (a) Range of projectile is less than the initial distance between the gun and the target (b) Range of projectile is more than the initial distance between the gun and the target (c) Range of projectile is equal to the initial distance between the gun and target (d) Bullet will always strike the target Solution : (a) Condition for hitting of bullet with target initial distance between the gun and target  Range of projectile. Problem 68. A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h metre high and b mere wide, the ball will just hit the edge of nth step if n equals to Motion in Two Dimension 159 (a) hu 2 (b) gb 2 u28 gb 2 Solution : (c) By using equation of trajectory y  2hu 2 (c) gx 2 2u 2 (d) gb 2 for given condition 2u 2 g hb 2 u h 2u 2 n  2hu 2 gb b nh 2 60 nh  g(nb ) 2 E3 nb  (2) Displacement of Projectile (r ) : After time t, horizontal displacement x  ut and vertical 1 2 gt. 2  1 So, the position vector r  ut ˆi  gt 2 ˆj 2 2  gy  u   2    as t    D YG   tan 1   gt  and   tan 1    2u  U  gt  r  ut 1     2u  Therefore ID displacement y  2y g     (3) Instantaneous velocity : Throughout the motion, the horizontal component of the velocity is vx = u. The vertical component of velocity increases with time and is given by vy = 0 + g t = g t (From v = u + g t)   v  v x ˆi  v y ˆj = v  u ˆi  g t ˆj U So, Again  gt  v  u 2  gt  2  u 1    u  v  uˆi  2 gy ˆj i.e. v  u 2  2 gy  r P (x,y)  2 Direction of instantaneous velocity : tan   X  y vy ST i.e. x u O vx v Y vy vx   2 gy   vy     tan 1     u v  x      tan 1   gt   u   tan 1  Where  is the angle of instantaneous velocity from the horizontal. or 160 Motion in Two Dimension Sample problems based on velocity Problem 69. A body is projected horizontally from the top of a tower with initial velocity 18 ms–1. It hits (a) 9 ms–1 (b) 9 2 ms–1 (c) 18 ms–1 Solution : (c) When the body strikes the ground vx  vy u = 18 =1 18 2 ms–1 (d) 18 E3 tan 45 o  vy 60 the ground at angle 45o. What is the vertical component of velocity when it strikes the ground vx 45o vy ID v y  18 m / s. Problem 70. A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities (a) U when they reach the earth’s surface will be 2 gh  u 2 : u (b) 1 : 2 (c) 1 : 1 (d) D YG Solution : (c) For first particle : v 2  u 2  2 gh  v  u 2  2 gh For second particle : v  v x2  v y2  u 2   2 gh  2 2 gh  u 2 : 2 gh u u  u 2  2 gh h v So the ratio of velocities will be 1 : 1. vx v vy Problem 71. A staircase contains three steps each 10 cm high and 20 cm wide. What should be the U minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane ST u (a) 0.5 m/s h b (b) 1 m/s Solution : (c) Formula for this condition is given by n   3 2  10  u 2 10  20 2  u 2  200 cm / sec  2 m / sec (c) 2 m/s (d) 4 m/s 2hu 2 gb 2 where h = height of each step, b = width of step, u = horizontal velocity of projection, n = number of step. Motion in Two Dimension 161 (4) Time of flight : If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then 1 gT 2 (for vertical motion) 2 60 h0 2h g T  E3 Sample problems based on time of flight Problem 72. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first (b) Depends on their mass (c) The slower one (d) Both will reach simultaneously ID (a) The faster one Solution : (d) Problem 73. An aeroplane is flying at a height of 1960 m in horizontal direction with a velocity of 360 U km/hr. When it is vertically above the point. A on the ground, it drops a bomb. The bomb strikes a point B on the ground, then the time taken by the bomb to reach the ground is (a) 20 2 sec 2h  g (c) 10 2 sec (d) 10 sec 2  1960 = 20 sec 9.8 D YG Solution : (b) t  (b) 20 sec (5) Horizontal range : Let R is the horizontal distance travelled by the body u R  uT  h 2h g Range U R u 1 0 T 2 (for horizontal motion) 2 ST Sample problems based on horizontal range Problem 74. A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. (g = 9.8 m/s2) (a) 100 m 3 Solution : (b) S  u  t  u  (b) 500 m 3 (c) 200 m 3 (d) 400 m 3 500 5 2  490 2h  m  60   18 9.8 3 g Problem 75. A body is thrown horizontally with velocity 2 gh from the top of a tower of height h. It strikes the level ground through the foot of tower at a distance x from the tower. The value of x is (a) h (b) h 2 (c) 2 h (d) 2h 3 162 Motion in Two Dimension Solution : (c) x  u  2h 2h  2 gh  g g u   x  2h 2 gh h 60 x Problem 76. An aeroplane moving horizontally with a speed of 720 km/h drops a food packet, while flying at a height of 396.9 m. The time taken by a food packet to reach the ground and its horizontal range is (Take g = 9.8 m/sec2 ) Solution : (d) Time of descent t  (b) 5 sec and 500 m (c) 8 sec and 1500 m 2h  g 2  396. 9 9.8 u = 720 km/h  t  9 sec ID and horizontal distance S  u  t (d) 9 sec and 1800 m E3 (a) 3 sec and 2000 m  720  5  S    9  1800 m  18  s U (6) If projectiles A and B are projected horizontally with different initial velocity from same height and third particle C is dropped from same point then D YG (i) All three particles will take equal time to reach the ground. (ii) Their net velocity would be different but all three particle possess same vertical component of velocity. (iii) The trajectory of projectiles A and B will be straight line w.r.t. particle C. h C B ST U A (7) If various particles thrown with same initial velocity but indifferent direction then A u u A B u C D E u u h A E Motion in Two Dimension 163 (i) They strike the ground with same speed at different times irrespective of their initial direction of velocities. (ii) Time would be least for particle E which was thrown vertically downward. (iii) Time would be maximum for particle A which was thrown vertically upward. 60 3.8 Projectile Motion on an Inclined Plane. Let a particle be projected up with a speed u from an inclined plane which makes an angle  with the horizontal velocity of projection makes an angle  with the inclined plane. We have taken reference x-axis in the direction of plane. E3 Hence the component of initial velocity parallel and perpendicular to the plane are equal to u cos  and u sin  respectively i.e. u||  u cos  and u   u sin . Y g sin  The component of g along the plane is and  u perpendicular to the plane is g cos  as shown in the figure i.e. ID a||   g sin  and a   g cos . U Therefore the particle decelerates at a rate of g sin  as it moves from O to P. (1) Time of flight : We know for oblique projectile motion T  we can say T  2u  a   Time of flight on an inclined plane T  2u sin  g cos  U u 2 2a  ST  Maximum height on an inclined plane H  u 2 sin 2  2 g cos  (3) Horizontal range : For one dimensional motion s  ut  Horizontal range on an inclined plane R  u|| T  R  u cos  T  1 g sin  T 2 2  2u sin  R  u cos    g cos  By solving R   1  2u sin    g sin    2  g cos  2u 2 sin  cos(   ) g cos 2   g 2u sin  g (2) Maximum height : We know for oblique projectile motion H  or we can say H   O t =0 D YG or ax=–g sin     2 1 a|| T 2 2 1 2 at 2 u 2 sin 2  2g ay= g cos  P t =T X 164 Motion in Two Dimension (i) Maximum range occurs when      4 2 (ii) The maximum range along the inclined plane when the projectile is thrown upwards is given by u2 g (1  sin  ) 60 R max  (iii) The maximum range along the inclined plane when the projectile is thrown downwards is given by u2 g (1  sin  ) E3 R max  Sample problem based on inclined projectile Problem 77. For a given velocity of projection from a point on the inclined plane, the maximum range ID down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is (a) 30o (b) 45o (c) 60o (d) 90o u2 g(1  sin  ) U Solution : (a) Maximum range up the inclined plane (R max )up  Maximum range down the inclined plane (R max )down  u2 g(1  sin  ) u2 u2  3 g(1  sin  ) g(1  sin  ) D YG and according to problem : By solving  = 30o Problem 78. A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is  = 30o, and the angle of the barrel to the horizontal  = 60o. The initial velocity v of the shell is 21 m/sec. Then distance of point from the gun at which shell will fall (a) 10 m (b) 20 m (c) 30 m (d) 40 m o o o o Solution : (c) Here u = 21 m/sec,  = 30 ,  =  –  = 60 – 30 = 30 U Maximum range R  2u 2 sin  cos(   ) 2  (21 ) 2  sin 30 o cos 60 o   30 m g 9.8  cos 2 30 o cos 2  Problem 79. The maximum range of rifle bullet on the horizontal ground is 6 km its maximum range on an ST inclined of 30o will be (a) 1 km (b) 2 km (c) 4 km Solution : (c) Maximum range on horizontal plane R  2 u  6 km (given) g Maximum range on a inclined plane R max  Putting  = 30o R max  u2 g (1  sin 30 ) o  2  u 2 3  g u2 g (1  sin  )  2    6  4 km.  3  (d) 6 km 164 Motion in Two Dimension CIRCULAR MOTION Circular motion is another example of motion in two dimensions. To create circular motion in a body it must be given some initial velocity and a force must then  v2 60 act on the body which is always directed at right angles to instantaneous velocity.  v3 F Since this force is always at right angles to the displacement due to the initial velocity therefore no work is done by the force on the  F F E3 F particle. Hence, its kinetic energy and thus speed is unaffected. But due  v1  v4 to simultaneous action of the force and the velocity the particle follows 3.9 Variables of Circular Motion. ID resultant path, which in this case is a circle. Circular motion can be classified into two types – Uniform circular motion and non-uniform circular motion. (1) Displacement and distance : When particle moves in a circular path describing an D YG U angle  during time t (as shown in the figure) from the position A to the position B, we see that  the magnitude of the position vector r (that is equal to the radius of the circle) remains   constant. i.e., r1  r2  r and the direction of the position vector changes from time to time.  (i) Displacement : The change of position vector or the displacement r of the particle from position A to the position B is given by referring the figure.   v2 v1     r  r2  r1    r  r  r2  r1 r  r12  r22 B  r2  2r1 r2 cos  O  U Putting r1  r2  r we obtain ST  r  2r sin  r2 O  r   r1  2 (ii) Distance : The distanced covered by the particle during the time t is given as d = length of the arc AB = r  (iii) Ratio of distance and displacement : A B r  r 2  r 2  2r.r cos     r  2r 2 1  cos    2r 2  2 sin 2  2   r1 d r    cosec ( / 2) r 2r sin  / 2 2 A Motion in Two Dimension 165 Sample problems based on distance and displacement circle would be (a) r r (c) (b) 60 Problem 80. A particle is rotating in a circle of radius r. The distance traversed by it in completing half 2r Solution : (b) Distance travelled by particle = Semi-circumference = r. (d) E3 Problem 81. An athlete completes one round of a circular track of radius 10 m in 40 sec. The distance covered by him in 2 min 20 sec is (a) No. of revolution (n)  (b) 140 m (c) 110 m (d) Total time of mo tion 140 sec   3.5 Time period 40 sec ID Solution : (d) 70 m Distance covered by an athlete in revolution  n (2 r)  3.5 (2 r)  3.5  2  22  10 = 220 m. 7 [RPMT 1999; BHU 2000] (a) U Problem 82. A wheel covers a distance of 9.5 km in 2000 revolutions. The diameter of the wheel is 15 m D YG Solution : (c) Distance  n (2 r)  9.5  10 3  2000  (D) (b) 7.5 m  D (c) 1.5 m (d) 9.5  10 3  1.5 m. 2000   (2) Angular displacement () : The angle turned by a body moving on a circle from some reference line is called angular displacement. (i) Dimension = [M0L0T0] (as  = arc / radius). U (ii) Units = Radian or Degree. It is some times also specified in terms of fraction or multiple of revolution. (iii) 2 rad  360 o  1 Revolution ST (iv) Angular displacement is a axial vector quantity. Its direction depends upon the sense of rotation of the object and can be given by Right Hand Rule; which states that if the curvature of the fingers of right hand represents the sense of rotation of the object, then the thumb, held perpendicular to the curvature of the fingers, represents the direction of angular displacement vector. (v) Relation between linear displacement and angular displacement s  r or s  r Sample problem based on angular displacement S  O r 166 Motion in Two Dimension Problem 83. A flywheel rotates at a constant speed of 3000 rpm. The angle described by the shaft in radian in one second is (a) 2  (b) 30  (c) 100  (d) 3000  Solution : (c) Angular speed  3000 rpm  50 rps  50  2 rad / sec  100  rad / sec 60 i.e. angle described by the shaft in one second is 100  rad. Problem 84. A particle completes 1.5 revolutions in a circular path of radius 2 cm. The angular displacement of the particle will be – (in radian) (b) 3  (c) 2  (d)  E3 (a) 6  Solution : (b) 1 revolution mean the angular displacement of 2 rad 3  rad. 1.5 revolution means 1. 5  2 =  d dt D YG  angle traced  d  Lt  t 0 t time taken dt U (i) Angular velocity  = ID (3) Angular velocity () : Angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement. (ii) Dimension : [M0L0T–1] (iii) Units : Radians per second (rad.s–1) or Degree per second. (iv) Angular velocity is an axial vector. (v) Relation between angular velocity and linear velocity v    r U Its direction is the same as that of . For anticlockwise rotation of the point object on the circular path, the direction of , according to Right hand rule is along the axis of circular path directed upwards. For clockwise rotation of the point object on the circular path, the direction ST of  is along the axis of circular path directed downwards. Note :  It is important to note that nothing actually moves in the direction of the angular velocity vector . The direction of  simply represents that the rotational motion is taking place in a plane perpendicular to it. (vi) For uniform circular motion  remains constant where as for non-uniform motion  varies with respect to time. Sample problems based on angular velocity Problem 85. A scooter is going round a circular road of radius 100 m at a speed of 10 m/s. The angular speed of the scooter will be Motion in Two Dimension 167 (a) 0.01 rad/s Solution : (b)   (b) 0.1 rad/s (c) 1 rad/s (d) 10 rad/s v 10   0.1 rad / sec r 100 Problem 86. The ratio of angular velocity of rotation of minute hand of a clock with the angular velocity (a) 12 (b) 6 Solution : (c)  Minute hand  60 of rotation of the earth about its own axis is (c) 24 (d) None of these  Minute hand 2 rad 2 Rad 2 rad and  Earth     24 : 1 60 min 24 hr 24  60 min  Earth E3 Problem 87. A particle P is moving in a circle of radius ‘a’ with a uniform speed v. C is the centre of the circle and AB is a diameter. When passing through B the angular velocity of P about A and C are in the ratio (a) 1 : 1 [NCERT 1982] (b) 1 : 2 (c) 2 : 1 A  ID Solution : (b) Angular velocity of P about A v 2a  A 1:2 C D YG C U Angular velocity of P about C v  a (d) 4 : 1 A P C a B Problem 88. The length of the seconds hand of a watch is 10 mm. What is the change in the angular speed of the watch after 15 seconds (b) (10 / 2) mms 1 (a) Zero (c) (20 /  ) mms 1 (d) 10 2 mms U Solution : (a) Angular speed of seconds hand of watch is constant and equal to 1 2 rad   rad / sec. So 60 sec 30 ST change in angular speed will be zero. (4) Change in velocity : We want to know the magnitude and direction of the change in velocity of the particle which is performing uniform circular motion as it moves from A to B  during time t as shown in figure. The change in velocity vector is given as v1   v  v2 2   v1 v  v 2  v 1 or    v  v 2  v1  v  v 12  v 22  2v 1 v 2 cos  For uniform circular motion v 1  v 2  v So v   B 2v 2 1  cos    2v sin  2 O   A   v  v 2  v1  v 1 168 Motion in Two Dimension  The direction of  v is shown in figure that can be given as   180 o    90 o   / 2 2  60 Note :  Relation between linear velocity and angular velocity. E3 In vector form v    r Sample problems based on velocity Problem 89. If a particle moves in a circle describing equal angles in equal times, its velocity vector ID [CPMT 1972, 74; JIPMER 1997] (a) Remains constant (b) Changes in magnitude (c) Changes in direction magnitude and direction (d) Changes both in U Solution : (c) In uniform circular motion velocity vector changes in direction but its magnitude always remains constant.  v2 D YG | v1 | | v 2 | | v 3 | | v 4 | = constant  v1  v3  v4 Problem 90. A body is whirled in a horizontal circle of radius 20 cm. It has angular velocity of 10 rad/s. U What is its linear velocity at any point on circular path (a) 10 m/s (b) 2 m/s (c) 20 m/s (d) 2 m/s ST Solution : (b) v  r   0.2  10  2 m/s Problem 91. The linear velocity of a point on the equator is nearly (radius of the earth is 6400 km) (a) 800 km/hr Solution : (b) v  r w  6400 km  (b) 1600 km/hr (c) 3200 km/hr (d) 6400 km/hr 2 rad  1675 km / hr  1600 km / hr 24 hr Problem 92. A particle moves along a circle with a uniform speed v. After it has made an angle of 60o its speed will be [RPMT 1998] (a) v 2 (b) v 2 (c) v (d) v 3 Solution : (d) Uniform speed means speed of the particle remains always constant. Motion in Two Dimension 169 Problem 93. A particle is moving along a circular path of radius 2 m and with uniform speed of 5 ms–1. What will be the change in velocity when the particle completes half of the revolution (c) 10 2 ms 1 (b) 10 ms–1  180 o   Solution : (b) v  2v sin   2  5 sin  2 2  (d) 10 / 2 ms 1     60 (a) Zero 180o  2  5 sin 90  10 m / s o  v1 E3  v2   Problem 94. What is the value of linear velocity, if   3ˆi  4 ˆj  kˆ and r  5ˆi  6ˆj  6kˆ (a) 6ˆi  2ˆj  3kˆ kˆ 1 6 Jˆ 4 6 (c) 4ˆi  13 ˆj  6kˆ (d) 6ˆi  2ˆj  8kˆ ID ˆi    Solution : (b) v    r  3 5 (b) 18ˆi  13 ˆj  2kˆ U  v  (24  6) i  (18  5) Jˆ  (18  20 ) kˆ  18 ˆi  13 Jˆ  2kˆ Problem 95. A particle comes round circle of radius 1 m once. The time taken by it is 10 sec. The average velocity of motion is (b) 2  m/s D YG (a) 0.2  m/s [JIPMER 1999] (c) 2 m/s (d) Zero Solution : (d) In complete revolution total displacement becomes zero. So the average velocity will be zero. Problem 96. Two particles of mass M and m are moving in a circle of radii R and r. If their time-periods are same, what will be the ratio of their linear velocities (a) MR : mr r v1  1 1. v2 r2  2 (c) R : r Time periods are equal i.e.  1   2  U Solution : (c) (b) M : m (d) 1 : 1 v1 r R  1  v2 r2 r ST (5) Time period (T) : In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path. (i) Units : second. (ii) Dimension : [M0L0T] (iii) Time period of second’s hand of watch = 60 second. (iv) Time period of minute’s hand of watch = 60 minute (v) Time period of hour’s hand of watch = 12 hour (6) Frequency (n) : In circular motion, the frequency is defined as the number of revolutions completed by the object on its circular path in a unit time. 170 Motion in Two Dimension (i) Units : s–1 or hertz (Hz). (ii) Dimension : [M0L0T–1] Note :  Relation between time period and frequency : If n is the frequency of revolution of 60 an object in circular motion, then the object completes n revolutions in 1 second. Therefore, the object will complete one revolution in 1/n second. T  1 / n 2  2n T ID  E3  Relation between angular velocity, frequency and time period : Consider a point object describing a uniform circular motion with frequency n and time period T. When the object completes one revolution, the angle traced at its axis of circular motion is 2 radians. It means, when time t = T,   2 radians. Hence, angular  2 velocity    (  T = 1/n)  2n t T U  If two particles are moving on same circle or different coplanar concentric circles in same direction with different uniform angular speeds A and B respectively, the angular velocity of B relative to A will be D YG  rel   B   A So the time taken by one to complete one revolution around O with respect to the other (i.e., time in which B complete one revolution around O with respect to the other (i.e., time in which B completes one more or less revolution around O than A) T 2  rel  TT 2  1 2  2   1 T1  T2 2   as T      U Special case : If  B   A ,  rel  0 and so T = ., particles will maintain their position relative to each other. This is what actually happens in case of geostationary satellite (1 = 2 = constant) ST (7) Angular acceleration () : Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity. (i) If  be the change in angular velocity of the object in time interval t and t + t, while moving on a circular path, then angular acceleration of the object will be d 2  d  2   Lt  t 0 t dt dt (ii) Units : rad. s–2 (iii) Dimension : [M0L0T–2] (iv) Relation between linear acceleration and angular acceleration a    r  Motion in Two Dimension 171 (v) For uniform circular motion since  is constant so   d 0 dt (vi) For non-uniform circular motion   0 Note :  Relation between linear (tangential) acceleration and angular acceleration a   r 60  For uniform circular motion angular acceleration is zero, so tangential acceleration also is equal to zero.  For non-uniform circular motion a  0 (because   0). E3 Sample problems based on angular acceleration Problem 97. A body is revolving with a uniform speed v in a circle of radius r. The angular acceleration of the body is v r (b) Zero ID (a) v2 along the radius and towards the centre (d) r away from the centre v2 along the radius and r U (c) D YG Solution : (b) In uniform circular motion  constant so   d 0 dt Problem 98. The linear acceleration of a car is 10m/s2. If the wheels of the car have a diameter of 1m, the angular acceleration of the wheels will be (a) 10 rad/sec2 Solution : (b) Angular accelerati on  (b) 20 rad/sec2 (c) 1 rad/sec2 (d) 2 rad/sec2 linear accelerati on 10  20 rad / sec 2  radius 0.5 Problem 99. The angular speed of a motor increases from 600 rpm to 1200 rpm in 10 s. What is the U angular acceleration of the motor (a) 600 rad sec 2  2  1 t ST Solution : (d)    (b) 60 rad sec 2 (c) 60 rad sec 2 (d) 2 rad sec 2 2 (n 2  n1 ) 2 (1200  600 ) rad   2 rad / sec 2 t 10  60 sec 2 3.10 Centripetal Acceleration. (1) Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration. (2) It always acts on the object along the radius towards the centre of the circular path. ac (3) Magnitude of centripetal acceleration a  v2 4 2   2 r  4n 2 r  2 r r T v 172 Motion in Two Dimension (4) Direction of centripetal acceleration : It is always the same as that of . When t  decreases,  also decreases. Due to which  becomes more and more perpendicular to . When  t 0,  becomes perpendicular to the velocity vector. As the velocity vector of the particle at an instant acts along the tangent to the circular path, therefore  and hence the 60 centripetal acceleration vector acts along the radius of the circular path at that point and is directed towards the centre of the circular path. E3 Sample problems based on centripetal acceleration Problem 100. If a cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration of the cycle will be Solution : (c) Given r  4 m and T  2 seconds. T2 r  4 2 (2) 2 4  4 2 m / s 2 (d) 8 m/s 2 U 4 2  ac  (c) 4 2 m / s 2 ID (b) 2 2 m / s 2 (a)  2 m / s 2 [Pb. PMT 2001] Problem 101. A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a D YG constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone [Pb. PMT 2000] (a) 493 cm/sec2 Solution : (a) Time period  a c  4 2 T 2.r  (b) 720 cm/sec2 (c) 860 cm/sec2 (d) 990 cm/sec2 20 Total time  2 sec  10 No. of revolution 4 2 (2) 2  (1 / 2) m / s 2  4.93 m / s 2  493 cm / s 2 Problem 102. A particle moves with a constant speed v along a circular path of radius r and completes the U circle in time T. What is the acceleration of the particle (b) ST (a) mg Solution : (b) a c  2v T (c) r 2 T (d) v 2 T v2  2  2 v   2 r  v = v   T r  T  Problem 103. If the speed of revolution of a particle on the circumference of a circle and the speed gained in falling through a distance equal to half the radius are equal, then the centripetal acceleration will be (a) g 2 (b) g 4 (c) g 3 (d) g Motion in Two Dimension 173 Solution : (d) Speed gain by body falling through a distance h is equal to v  2 gh  2g r 2 [As h  r 2 given] gr  v2 g r 60  v Problem 104. Two cars going round curve with speeds one at 90 km/h and other at 15 km/h. Each car experiences same acceleration. The radii of curves are in the ratio of (b) 2 : 1 Solution : (d) Centripetal acceleration = 2 v12 v 22 (given)  r1 r2  36  90       1  15   2 (d) 36 : 1 ID v r  1   1 r2  v 2 (c) 16 : 1 E3 (a) 4 : 1 Problem 105. A wheel of radius 0.20 m is accelerated from rest with an angular acceleration of 1 rad / s 2. (a)  m / s 2 U After a rotation of 90 o the radial acceleration of a particle on its rim will be (b) 0.5  m / s 2 (d) 0.2  m / s 2 (c) 2.0 m / s 2 D YG Solution : (d) From the equation of motion Angular speed acquired by the wheel,  22  12  2  0  2  1   2   22   Now radial acceleration  2 r    0.2  0.2 m / s 2 3.11 Centripetal Force. U According to Newton's first law of motion, whenever a body moves in a straight line with uniform velocity, no force is required to maintain this velocity. But when a body moves along a circular path with uniform speed, its direction changes continuously i.e. velocity keeps on ST changing on account of a change in direction. According to Newton's second law of motion, a change in the direction of motion of the body can take place only if some external force acts on the body. Due to inertia, at every point of the circular path; the body tends to move along the tangent to the circular path at that point (in figure). Since every body has directional inertia, a velocity cannot change by itself and as such we v v have to apply a force. But this force should be such that it changes the direction of velocity and not its magnitude. This is possible only if the force acts perpendicular to the direction of velocity. Because the velocity is along the tangent, this force must be along the radius (because the radius of a circle at any point is perpendicular to the F F F F v v 174 Motion in Two Dimension tangent at that point). Further, as this force is to move the body in a circular path, it must acts towards the centre. This centre-seeking force is called the centripetal force. Hence, centripetal force is that force which is required to move a body in a circular path with uniform speed. The force acts on the body along the radius and towards centre. m 4 2 r mv 2  m  2 r  m 4 2 n 2 r  r T2 60 (1) Formulae for centripetal force : F  (2) Centripetal force in different situation Centripetal Force E3 Situation Tension in the string Vehicle taking a turn on a level road Frictional force exerted by the road on the tyres A vehicle on a speed breaker Weight of the body or a component of weight Revolution of earth around the sun ID A particle tied to a string and whirled in a horizontal circle Gravitational force exerted by the sun Coulomb attraction exerted by the protons in the nucleus U Electron revolving around the nucleus in an atom Magnetic force exerted by the agent that sets up the magnetic field D YG A charged particle describing a circular path in a magnetic field 3.12 Centrifugal Force. It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a ST U circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangential at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body with the same velocity), the body appears to be stationary before it is released. When the body is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. In reality no force is actually seen to act on the body. In absence of any real force the body tends to continue its motion in a straight line due to its inertia. The observer A easily relates this events to be due to inertia but since the inertia of both the observer B and the body is same, the observer B can not relate the above happening to inertia. When the centripetal force ceases to act on the body, the body leaves its circular path and continues to moves in its straight-line motion but to observer B it appears that a real force has actually acted on the body and is responsible for throwing the body radially out-words. This imaginary force is given a name to explain the effects on inertia to the observer who is sharing the circular motion of the body. This inertial force is called centrifugal force. Thus centrifugal force is a fictitious force which has significance only in a rotating frame of reference. Motion in Two Dimension 175 Sample problems based on centripetal and centrifugal force Problem 106. A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string (a) 5 r.p.m. (b) 10 r.p.m. (c) 20 r.p.m. Solution : (a) Tension in the string T  m  2 r  m 4 2 n 2 r  Problem 107. n2  n1 T2  T1 T /4 T [As m and r are constant]  n2  (d) 14 r.p.m. E3 T  n 2 or n  T 60 at an initial speed of 10 r.p.m. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is n1 10   5 rpm 2 2 A cylindrical vessel partially filled with water is rotated about its vertical central axis. (a) Rise equally ID It’s surface will (b) Rise from the sides (c) Rise from the middle (d) U Solution : (b) Due to the centrifugal force. [RPET 2000] Problem 108. A proton of mass 1.6 × 10 –27 kg goes round in a circular orbit of radius 0.10 m under a D YG centripetal force of 4 × 10 –13 N. then the frequency of revolution of the proton is about (a) 0.08 × 108 cycles per sec (b) 4 × 108 cycles per sec (c) 8 × 108 cycles per sec (d) 12 × 108 cycles per sec Solution : (a) F  4  10 13 N ; m  1.6  10 27 kg ; r  0.1 m Centripetal force F  m 4 2 n 2 r  n  F  8  10 6 cycles / sec  0.08  10 8 cycle / sec. 4m  r 2 U Problem 109. Three identical particles are joined together by a thread as shown in figure. All the three ST particles are moving in a horizontal plane. If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is (a) 3 : 5 : 7 O A l B (b) 3 : 4 : 5 C l l (c) 7 : 11 : 6 (d) 3 : 5 : 6 Solution : (d) Let the angular speed of the thread is  For particle ‘C’  T3  m  2 3l For particle ‘B’ T2  T3  m  2 2l  T2  m  2 5l For particle ‘C’ T1  T2  m  2 l  T1  m  2 6l O T1 A T2 B T3 C 176 Motion in Two Dimension  T3 : T2 : T1  3 : 5 : 6 Problem 110. A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle with a uniform angular velocity of 2 rad/s. The tension of the string is (in N) (b) 1 3 (c) 4 (d) 1 4 60 (a) 2 Solution : (c) T  m  2 r  1  (2)2  (1)  4 Newton Problem 111. A cord can bear a maximum force of 100 N without breaking. A body of mass 1 kg tied to (a) 10 m/s E3 one end of a cord of length 1 m is revolved in a horizontal plane. What is the maximum linear speed of the body so that the cord does not break (b) 20 m/s (c) 25 m/s (d) 30 m/s mv2 and for critical condition this r ID Solution : (a) Tension in cord appears due to centrifugal force T  tension will be equal to breaking force (100 N) v max  10 m / s  2 m v max  100  r 2 v max  100  1 1  U Problem 112. A mass is supported on a frictionless horizontal surface. It is attached to a string and D YG rotates about a fixed centre at an angular velocity 0. If the length of the string and angular velocity are doubled, the tension in the string which was initially T0 is now (a) T0 (b) T0/2  T Solution : (d) T  m  l  2   2 T1   1 2    2  l2   l1 (c) 4T0 (d) 8T0  T  2   2l    2       T2  8T0 T0     l   2 Problem 113. A stone is rotated steadily in a horizontal circle with a period T by a string of length l. If the tension in the string is kept constant and l increases by 1%, what is the percentage change in T (b) 0.5% U (a) 1% m 4 l T2  l  T 2 or T  l ST Solution : (b) Tension  2 Percentage change in Time period  (c) 2% (d) 0.25% [Tension and mass are constant] 1 (percentage change in length) 2 [If % change is very small]  1 (1 %)  0.5 % 2 Problem 114. If mass speed and radius of rotation of a body moving in a circular path are all increased by 50%, the necessary force required to maintain the body moving in the circular path will have to be increased by (a) 225% Solution : (b) Centripetal force F  (b) 125% mv2 r (c) 150% (d) 100% Motion in Two Dimension 177 m  v  m   v   2 2 If m, v and r are increased by 50% then let new force F'   r  r   2   9 mv2 4 r  9 F 4 F'  F F 500  100   100 %  %  125 % F F 4 60 Percentage increase in force 2 Problem 115. Two masses m and M are connected by a light string that passes through a smooth hole O E3 at the centre of a table. Mass m lies on the table and M hangs vertically. m is moved round in a horizontal circle with O as the centre. If l is the length of the string from O to m then the frequency with which m should revolvel so that M remains stationary is m O 1 2 Mg ml (b) 1 Mg ml  1 2 (c) ml Mg (d) 1  ml Mg U (a) ID M Solution : (a) ‘m’ Mass performs uniform circular motion on the table. Let n is the frequency of D YG revolution then centrifugal force  m 4 2 n 2 l For equilibrium this force will be equal to weight Mg l m m 4 n l  Mg 2 2 n  1 2 Mg ml M Problem 116. A particle of mass M moves with constant speed along a circular path of radius r under the U action of a force F. Its speed is rF m ST (a) Solution : (a) Centripetal force F  (b) F r mv2 v  r (c) Fm r F mr (d) rF m Problem 117. In an atom for the electron to revolve around the nucleus, the necessary centripetal force is obtained from the following force exerted by the nucleus on the electron (a) Nuclear force (b) Gravitational force (c) Magnetic force (d) Solution : (d) Problem 118. A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be [AFMC 2002] (a) Double (b) Half (c) 4 times (d) 1 times 4 178 Motion in Two Dimension Solution : (c) F  mv2 F  v 2 or r F2  v 2    F1  v 1  2 2  2v      4  F2  4 F1  v  Problem 119. A bottle of soda water is grasped by the neck and swing briskly in a vertical circle. Near which portion of the bottle do the bubbles collect (b) In the middle of the bottle (c) Near the neck (d) Uniformly distributed in the bottle 60 (a) Near the bottom E3 Solution : (c) Due to the lightness of the gas bubble they feel less centrifugal force so they get collect near the neck of the bottle. They collect near the centre of circular motion i.e. near the neck of the bottle. Problem 120. A body is performing circular motion. An observer O1 is sitting at the centre of the circle and another observer O 2 is sitting on the body. The centrifugal force is experienced by the (a) O1 only (b) O 2 only (c) Both by O1 and O 2 (d) None of these ID observer U Solution : (b) Centrifugal force is a pseudo force, which is experienced only by that observer who is attached with the body performing circular motion. 3.13 Work done by Centripetal Force. D YG The work done by centripetal force is always zero as it is perpendicular to velocity and hence instantaneous displacement. S Work done = Increment in kinetic energy of revolving body 90o Work done = 0 Also F W = F. S = F  S cos = FS cos 90o = 0 U Example : (i) When an electron revolve around the nucleus in hydrogen atom in a particular orbit, it neither absorb nor emit any energy means its energy remains constant. ST (ii) When a satellite established once in a orbit around the earth and it starts revolving with particular speed, then no fuel is required for its circular motion. Sample problem based on work done Problem 121. A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal force acting on the particle is 10 N. It’s kinetic energy is (a) 0.1 Joule Solution : (d) (b) 0.2 Joule (c) 2.0 Joule mv2  10 N (given)  m v 2  10  r  10  0.2  2 r (d) 1.0 Joule Motion in Two Dimension 179 Kinetic energy 1 1 m v 2  (2)  1 Joule. 2 2 Problem 122. A body of mass 100 g is rotating in a circular path of radius r with constant velocity. The work done in one complete revolution is (b) (r/100) Joule (c) (100/r) Joule (d) Zero 60 (a) 100r Joule Solution : (d) Because in uniform circular motion work done by the centripetal force is always zero. Problem 123. A particle of mass m is describing a circular path of radius r with uniform speed. If L is the (a) L2 / mr 2 (b) L2 / 2mr 2 L2 L2  2I 2m r 2 (c) 2 L2 / mr 2 (d) mr 2 L (As for a particle I  m r 2 ) ID Solution : (b) Rotational kinetic energy E  E3 angular momentum of the particle about the axis of the circle, the kinetic energy of the particle is given by [CPMT 1995] 3.14 Skidding of Vehicle on a Level Road. When a vehicle turns on a circular path it requires U centripetal force. If friction provides this centripetal force then vehicle can move in circular path safely if D YG Friction force  Required centripetal force  mg   mg m2r mv 2 r v safe  rg This is the maximum speed by which vehicle can turn in a circular path of radius r, where U coefficient of friction between the road and tyre is . ST Sample problem based on skidding of vehicle on a level road Problem 124. Find the maximum velocity for overturn for a car moved on a circular track of radius 100 m. The coefficient of friction between the road and tyre is 0. 2 (a) 0.14 m / s Solution : (d) v max  (b) 140 m / s (c) 1.4 km / s (d) 14 m / s  r g  0.2  100  10  10 2  14 m / s Problem 125. When the road is dry and the coefficient of friction is  , the maximum speed of a car in a circular path is 10 m / s. If the road becomes wet and     2 , what is the maximum speed permitted (a) 5 m / s (b) 10 m / s (c) 10 2 m / s (d) 5 2 m / s 180 Motion in Two Dimension Solution : (d) v    v2  v1 2  1 /2 1 1 10   v2  v1  v 2  5 2m/s  2 2 2 Problem 126. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m with skidding is (assume g = 10 ms– 2 ) [Kerala PMT 2002] (c) 15 ms 1 (b) 20 ms 1 3.15 Skidding of Object on a Rotating Platform. E3 Solution : (d) v max   r g  0. 25  40  10  10 m / s (d) 10 ms 1 60 (a) 40 ms 1 On a rotating platform, to avoid the skidding of an object (mass m) placed at a distance r from axis of rotation, the centripetal force should be provided by force of friction. Centripetal force = Force of friction ID m 2r = mg  max  (g / r) ,  not skid on it. (g / r) , so that object will D YG 3.16 Bending of a Cyclist. U Hence maximum angular velocity of rotation of the platform is A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve. Consider a cyclist of weight mg taking a turn of radius r with velocity v. In order to provide the necessary centripetal force, the cyclist leans through angle  inwards as shown in figure. The cyclist is under the action of the following forces : The weight mg acting vertically downward at the centre of gravity of cycle and the cyclist. U The reaction R of the ground on cyclist. It will act along a line-making angle  with the vertical. ST The vertical component R cos of the normal reaction R will balance the weight of the cyclist, while the horizontal component R sin  will provide the necessary centripetal force to the cyclist. mv 2 R sin   r and R …..(i) R cos  = mg Dividing equation (i) by (ii), we have mv2 r R sin   R cos  mg R cos   …..(ii) R sin mv2/r mg Motion in Two Dimension 181 tan   or v2 rg …..…(iii) v2  Therefore, the cyclist should bend through an angle   tan 1    rg    (i) The radius of the curve is small i.e. the curve is sharper Note :  For E3 (ii) The velocity of the cyclist is large. 60 It follows that the angle through which cyclist should bend will be greater, if the same reasons, an ice skater or an aeroplane has to bend inwards, while taking a turn. ID Sample problem based on bending of cyclist Problem 127. A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres/sec. The U combined mass of the boy and the cycle is 90kg. The angle that the cycle makes with the vertical so that it may not fall is (g  9.8 m / sec 2 ) (b) 63.90 o (c) 26.12 o (d) 30.00 o D YG (a) 60.25 o Solution : (b) r  20 m, v  20 m / s, v2  m  90 kg, g  9.8 m / s 2 (given)  20  20  1   tan 1    tan 1    tan (2)  63.90   20  10  rg Problem 128. If a cyclist moving with a speed of 4.9 m / s on a level road can take a sharp circular turn of radius 4 m , then coefficient of friction between the cycle tyres and road is (a) 0.41 (b) 0.51 (c) 0.71 (d) 0.61 U Solution : (d) v  4.9 m / s, r  4 m and g  9.8 m / s 2 (given) v 2 4.9  4.9  0.61  rg 4  9.8 ST  Problem 129. A cyclist taking turn bends inwards while a car passenger taking same turn is thrown outwards. The reason is [NCERT 1972] (a) Car is heavier than cycle (b) Car has four wheels while cycle has only two (c) Difference in the speed of the two (d) Cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force Solution : (d) 3.17 Banking of a Road. 182 Motion in Two Dimension For getting a centripetal force cyclist bend towards the centre of circular path but it is not possible in case of four wheelers. Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically inclined towards the centre. 60 In the figure (A) shown reaction R is resolved into two components, the component R cos balances weight of vehicle R cos   mg  ……(i)...…(ii) g ...... (iii)  v rg mg h  x Fig. (A) Fig. (B) U  2r l  v2 tan   rg tan    R sin  Dividing (ii) by (i), we have or R cos  R ID Thus mv 2 R sin   r E3 and the horizontal component R sin  provides necessary centripetal force as it is directed towards centre of desired circle...... (iv) [As v = r ] D YG If l = width of the road, h = height of the outer edge from the ground level then from the figure (B) h h.......(v) [since  is very small] tan    x l From equation (iii), (iv) and (v) tan   v 2  2 r v h    g rg l rg U Note :  If friction is also present between the tyres and road then ST  Maximum safe speed on a banked frictional road v    tan  v2  rg 1   tan  rg(  tan  ) 1   tan  Sample problems based on banking of a road Problem 130. For traffic moving at 60 km / hr along a circular track of radius 0.1 km , the correct angle of banking is [MNR 1993] (a) (60 )2 0.1 Solution : (b) v  60 km / hr   (50 / 3)2  (b) tan 1    100  9.8  50 m / s , r  0.1km  100 m, 3  100  9.8  (c) tan 1  2   (50 / 3)  g  9.8 m / s 2 (given) (d) tan 1 60  0.1  9.8 Motion in Two Dimension 183 Angle of banking tan    50 / 3 2   v2  v2 or   tan 1    tan 1   rg rg  100  9.8    Problem 131. A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R. For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is (b) Solution : (a) For Banking of road tan    v2 h v 2l v 2b  h   rg l Rg rg rb Rg (c) vb 2 Rg 60 v 2b Rg (d) v2 h and tan   rg l vb R2g E3 (a) [As l  b and r  R given] Problem 132. The radius of curvature of a road at a certain turn is 50 m. The width of the road is 10 m and (a) 6.5 m / s (b) 8.6 m / s ID its outer edge is 1.5m higher than the inner edge. The safe speed for such an inclination will be (c) 8 m / s (d) 10 m / s v2 h  rg l  v U Solution : (b) h  1.5 m, r  50 m, l  10 m, g  10 m / s 2 (given) 1.5  50  10  8.6 m / s 10 hrg  l D YG Problem 133. Keeping the banking angle same to increase the maximum speed with which a car can travel on a curved road by 10%, the radius of curvature of road has to be changed from 20 m to [EAMCET 1991] (a) 16 m Solution : (c) tan   (b) 18 m v2 rg (d) 30.5 m  r  v 2 (if  is constant) 2   1.1v      = 1.21  r2 = 1.21  r1 = 1.21  20 = 24.2 m  v   2 U r2  v 2  r1  v1 (c) 24.25 m Problem 134. The slope of the smooth banked horizontal road is p. If the radius of the curve be r , the ST maximum velocity with which a car can negotiate the curve is given by (a) prg Solution : (b) tan   (b) prg v v  v p rg rg prg 2 (c) p / rg (d) p / rg 2 3.18 Overturning of Vehicle. When a car moves in a circular path with speed more than maximum speed then it overturns and it’s inner wheel leaves the ground first Weight of the car = mg Speed of the car = v Radius of the circular path = r G R1 h F 2a mg R2 184 Motion in Two Dimension Distance between the centre of wheels of the car = 2a Height of the centre of gravity (G) of the car from the road level = h Reaction on the inner wheel of the car by the ground = R1 Reaction on the outer wheel of the car by the ground = R2 force mv 2 R.......(i) E3 i.e., F  60 When a car move in a circular path, horizontal force F provides the required centripetal For rotational equilibrium, by taking the moment of forces R1, R2 and F about G Fh  R1 a  R 2 a.......(ii) As there is no vertical motion so R1 + R2 = mg 1  v 2h  M g   2  ra  R2  1  v 2h  M g   2  ra .......(iv) U and R1  ID By solving (i), (ii) and (iii).......(iii).......(v) D YG It is clear from equation (iv) that if v increases value of R1 decreases and for R1 = 0 v 2h g ra or v gra h i.e. the maximum speed of a car without overturning on a flat road is given by v  gra h U Sample problems based on overturning of vehicle Problem 135. The distance between two rails is 1.5m. The centre of gravity of the train at a height of 2m ST from the ground. The maximum speed of the train on a circular path of radius 120 m can be (a) 10.5 m / s (b) 42 m / s (c) 21 m / s (d) 84 m / s Solution : (c) Height of centre of gravity from the ground h = 2m, Acceleration due to gravity g = 10 m / s 2 , Distance between two rails 2a = 1.5m, Radius of circular path r = 120 m (given) v max  gra h  v max  10  120  0.75 2  21.2 m / s Problem 136. A car sometimes overturns while taking a turn. When it overturns, it is (a) The inner wheel which leaves the ground first (b) The outer wheel which leaves the ground first (c) Both the wheels leave the ground simultaneously Motion in Two Dimension 185 (d) Either wheel leaves the ground first Solution : (a) Problem 137. A car is moving on a circular path and takes a turn. If R 1 and R 2 be the reactions on the inner and outer wheels respectively, then (b) R1  R 2 Solution : (b) Reaction on inner wheel R1  M 2 (c) R1  R 2 (d) R1  R 2 60 (a) R1  R 2  v 2h M v 2h  g   and Reaction on outer wheel R 2  g   2  r a  r a   E3  R1  R 2. Problem 138. A train A runs from east to west and another train B of the same mass runs from west to east at the same speed along the equator. A presses the track with a force F1 and B presses the track with a force F2 (a) F1  F2 ID (b) F1  F2 (c) F1  F2 U (d) The information is insufficient to find the relation between F1 and F2 Solution : (a) We know that earth revolves about its own axis from west to east. Let its angular speed is  e and the angular speed of the train is  t D YG For train A : Net angular speed = (  e   t ) because the sense of rotation of train is opposite to that of earth So reaction of track R1  F1  m g  m ( e   t )2 R For train B : Net angular speed = ( e   t ) because the sense of rotation of train