Chapter 34 Dynamics PDF

60 116 Dynamics 4.1 Introduction. ID E3 Dynamics is that branch of mechanics which deals with the study of laws governing motions of material system under the action of given forces. (1) Displacement : The displacement of a moving point is its change of position. To know the displacement of a moving...

60 116 Dynamics 4.1 Introduction. ID E3 Dynamics is that branch of mechanics which deals with the study of laws governing motions of material system under the action of given forces. (1) Displacement : The displacement of a moving point is its change of position. To know the displacement of a moving point, we must know both the length and the direction of the line joining the two positions of the moving point. Hence the displacement of a point involves both magnitude and direction i.e., it is a vector quantity. (2) Speed : The speed of a moving point is the rate at which it describes its path. The speed of a moving particle or a body does not give us any idea of its direction of motion; so it is a scalar quantity. In M.K.S. or S.I. system, the unit of speed is metre per second (m/s). U Average speed : The average speed of moving particle in a time-interval is defined as the distance travelled by the particle divided by the time interval. Note : D YG If a particle travels a distance s in time-interval t, then Average speed = s t  The average speed of a particle gives the overall “rapidity” with which the particle moves in the given interval of time. Instantaneous speed : The instantaneous speed or simply speed of a moving particle is defined as the rate of change of distance along its path, straight or curved. If s is the distance travelled by a particle along its path, straight or curved, in time t, then ds ds or speed at time t  dt dt U Instantaneous speed = Note :  The average speed is defined for a time interval and the instantaneous speed is ST defined at a particular instant. 4.2 Velocity and Acceleration. (1) Velocity : The velocity of a moving point is the rate of its displacement. It is a vector quantity. Let a particle starting from the fixed point O and moving along the straight line OX describes the distance OP = x in time t. If the particle x O x X describes a further distance PQ  x in time t , then is x Q P t called the average velocity of the particle in time t. x dx   v is called the instantaneous velocity (or simply velocity) of the particle t 0 t dt And lim at time t. Dynamics 117 Note :  Average velocity in time t is the mean of the initial and final velocity. (2) Acceleration : The rate of change of velocity of a moving particle is called its acceleration. It is a vector quantity. The acceleration of a moving point at time t at distance x is given by, v 2  v1 dv d x dv f  2  v. t dt dt dx In M.K.S. or S.I. system, the unit of acceleration is m/sec2. It should be noted that a  0 if v increases with time and a  0 if v decreases with time. A negative acceleration is called retardation. 2 E3 60 f Clearly, retardation means decrease in the velocity. Example: 1 The average speed of a bicycle over a journey of 20 km; if it travels the first 10 km. at 15 km/hr and the second 10 km. at 10 km/hr, is Solution: (a) (b) 10 km/hr (c) 15 km/hr (d) None of these ID (a) 12 km/hr Time taken by the bicycle to travel the first 10 km = 10 2 hour = hour 15 3 Time taken by the bicycle to travel the second 10 km = 10 hour = 1 hour 10 2 5 hour. 1  3 3 D YG Total time taken = U Total distance travelled = 10 + 10 = 20 km  Average speed of bicycle = Example: 2 If a particle moves in a straight line according to the formula, x  t 3  6 t 2  15 t , then the time interval during which the velocity is negative and acceleration is positive, is (a) [0, 5] Solution: (c) 20 60 Distance   12 km/hr. = 5/3 5 Time (b) (2, ) (c) (2, 5) (d) None of these We have x  t 3  6 t 2  15 t  v dx d2x  3 t 2  12 t  15 and a  2  6 t  12 dt dt U Now v  0 and a  0  3(t 2  4 t  5)  0 and 6(t  2)  0  1  t  5 and t  2  t  (2, 5). ST Example: 3 A particle moves in a fixed straight path so that S  1  t. If v is the velocity at any time t, then its acceleration is (a)  2v 3 We have s  1  t  Example: 4 ds 1 d 2s 1 1  1     2  4  1  t dt dt 4 (1  t)3 / 2 2 1t     3 3  d 2 s  1  ds  d 2s   2   2  2v 3. 2 4  dt  dt dt Two particles are moving with uniform velocities u and v respectively along X and Y axes, each directed towards the origin. If the particles are at distances a and b from the origin, the time at which they will be nearest to each other will be equal to (a) Solution: (c) (d) 2v 3 au u2  v2 (b) bu u2  v2 (c) au  bv u2  v2 (d) au bv Let the two particles be at A and B respectively at t  0 and at time t their positions be P and Q respectively. Then, AP  ut and BQ  vt Y B vt Q vt Solution: (a) (c)  v 2 (b)  v 3 118 Dynamics  OP  a  ut and OQ  b  vt Now, PQ 2  OP 2  OQ 2   PQ 2  (a  ut )2  (b  vt)2 d d2 (PQ 2 )  2u(a  ut )  2v(b  vt) and (PQ 2 )  2(u 2  v 2 ) dt dt 2 For maximum or minimum value of PQ 2 , we must have Clearly, 60 au  bv d (PQ 2 )  0  2u(a  ut )  2v(b  vt)  0  t  2 dt u  v2 d2 au  bv. (PQ 2 )  0 for all t. Hence, PQ is least at t  2 u  v2 dt 2 E3 4.3 Resultant and Components of Velocities. (1) Resultant of velocities : The resultant of two velocities possessed by a particle is given by the law of parallelogram of velocities. Parallelogram law of velocities : If a moving particle has two simultaneous velocities U parallelogram passing through that point. ID represented in magnitude and direction by the two sides of a parallelogram drawn from an angular point, then their resultant is represented in magnitude and direction by the diagonal of the D YG Let a particle at O has simultaneously two velocities u and v represented in magnitude and direction by OA and OB respectively of the parallelogram OACB. The diagonal OC represents the B C v R   u O A resultant velocity. Let R be the resultant velocity. If AOB   and AOC   , then the magnitude of their resultant velocity is R  u 2  v 2  2uv cos  and, the direction of this resultant velocity makes an angle  with the v sin  u  v cos  U direction of u such that tan   Case I : When   0 , then R max  u  v Case II : When    , then R min | u  v | ST Case III : When    / 2 , i.e., u and v are at right angle to each other, then v  R  u 2  v 2 and   tan 1  . u  Case IV : When u  v , then R  2u cos  2 and    2. Note :  The angle made by the direction of the resultant velocity with the direction of v is  u sin   given by tan 1  .  v  u cos   Dynamics 119  If the direction of resultant velocity R makes an angle  with the direction of u , then sin   v sin  u  v cos  and cos   R R 60 (2) Components of a velocity in two given directions : If components of a velocity V in two given directions making angles  and  with it, then by the law of parallelogram of velocities the required components of V are represented by AB and AC = BD along AX Yand AY respectively. AB  AD. sin  sin  and BD  AD. sin(   ) sin(   ) D  V   –(+) A ID  C E3 AB BD AD AB BD AD   i.e.,   sin  sin  sin(   ) sin ADB sin BAD sin ABD Hence, the components of velocity V in the directions making angle  and  are X V sin  and sin(   ) U V sin  respectively. sin(   ) B Important Tips If components of velocity V are mutually perpendicular, then   90 o  . Thus components of V are V cos  along D YG  AX and V sin  along AY. Example: 5 Drops of water falling from the roof of the tunnel seems from the window of the train to be falling 1 from an angle making tan 1 from the horizontal. It is known that their velocities are 24 2 decimeter/sec. Assuming the resistance of air negligible, what will be the velocity of the train (a) 42 decimeter/sec. (c) 45 decimeter/sec. (d) 44 decimeter/sec. Let the velocity of train is v decimeter/sec. U Solution: (b) (b) 48 decimeter/sec. Let the real velocity of drops in magnitude and direction may be represented by OB and the velocity of train in opposite direction is represented by OD, then OB  24 , ST OD  BC  v Completing parallelogram OBCD, OC will represent the relative velocity of drops. D If the drops are seen in a direction making an angle  with the 1 1 horizontal, i.e., COD   , then   tan 1   or tan   or cot   2 2 2 In OBC , from sine formula, v O  Train Directio n of 90– drop 24 C B 24 v OB BC  or or  sin  cos  sin sin(90   ) v  24 cot   24  2  48 decimeter/second. Example: 6 If u and v be the components of the resultant velocity w of a particle such that u  v  w , then the angle between the velocities is (a) 60 o (b) 150 o (c) 120 o (d) 30 o 120 Dynamics Solution: (a) Let  be the angle between the components of the resultant velocity w. Then w 2  u 2  v 2  2uv cos   u 2  2u 2 (1  cos  )  4 cos 2 Example: 7  2  1  cos 2  1    60 o. 2 A man swims at a speed of 5 km/hr. He wants to cross a canal 120 metres wide, in a direction perpendicular to the direction of flow. If the canal flows at 4 km/hr, the direction and the time taken 3 (a) tan 1   , 2.4 min. 4 3 (b)   tan 1   , 144 sec. 4 60 by the man to cross the canal are Solution: (b)  1 (c) tan 1   , 100 sec. (d) None of these 2 Suppose the man swims in a direction making an angle  with the direction of current. Therefore, tan  2 5 sin  4  4  5 cos   0  cos   4  5 cos  5  Let V be the resultant velocity, then V  4  5  2  4  5 cos  2 V 2  16  25  40  4  9  V  3km / hr. 5 D B 5 km/hr C V 120 m  O 4 km/hr A U 2 ID 4 3  4     cos 1       tan 1       cos 1 5 5 4   2 E3 Since the man wants to cross the canal in a direction perpendicular to the direction of flow. D YG Time taken by the man to cross the canal = 120 1 60 hr  hr  min.  2.4 min.  144 sec. 3  1000 25 25 4.4 Relative Velocity. Let A and B are two bodies in motion. The velocity of A relative to B is the velocity with which A appears to move as viewed by B i.e., velocity of A relative to B = Resultant of velocity of A and reversed velocity of B. Apparent velocity of A as seen from B means velocity of A relative to B. U The relative velocity of A with respect to B is also called the velocity of A relative to B and is denoted by V AB and V AB  V A  V B. ST If two bodies A and B are moving with velocities of magnitudes VA and VB respectively inclined at an angle  with VA VAB VB VB each other, then V AB  V A2  VB2  2 V A VB cos  If the direction of the relative velocity of A with respect to B makes an angle  with the direction of velocity of B, then tan   VA sin  VB  VA cos  Case I : When the particles A and B move parallel to each other in the same direction with VA = u velocities u and v respectively. Here   0 , then VAB  (u  v) in the direction ofA A VBA  (v  u) in the direction of B B VB = v Dynamics 121 Case II : When the particles A and B move parallel to each other in opposite directions with velocities u and v respectively. Here    , then VAB  (u  v) in the direction of A.A VA = u VBA  (v  u) in the direction of B. :  True velocity of A = Resultant of the relative velocity of A with respect to B and the true velocity of B [Since V AB  V A  V B ] i.e., V A  V AB  V B Example: 8 A train of length 200 m travelling at 30 m/sec. overtakes another of length 300 m travelling at 20 (a) 30 sec. E3 m/sec. The time taken by the first train to pass the second is Solution: (b) VB = v 60 Note B (b) 50 sec. (c) 10 sec. (d) 40 sec. Distance covered by the first train to pass the second = (200 + 300)m = 500 metres. ID Since both the trains are travelling in the same direction.  Velocity of first train relative to second = (30 – 20) = 10 m/sec.  To a man running at a speed of 20 km/hr, the rain drops appear to be falling at an angle of 30 o from the vertical. If the rain drops are actually falling vertically downwards, their velocity in km/hr. is (a) 10 3 (b) 10 (c) 20 3 Resolving horizontally and vertically, V cos 30 o  u and V sin 30 o = 20 i.e., 40 Example: 10 (d) 40 Velocity of rain relative to man = Actual velocity of rain – Velocity of man D YG Solution: (c) 500  50 second. 10 U Example: 9 Time taken by the first train to pass the second = 20(Rev.) V  20 ,  V  40 2 O 20(Act.) 30° 3  u  u  20 3 km / hr. 2 Actua l V(Rel.) A man is walking at the rate of 3 km/hr and the rain appears to him falling vertically with a velocity of 3 3 km / hr. If the actual direction of the rain makes an angle  with vertical, then  = (b) 30 o (a) 15 o (c) 45 o Let V M denotes the velocity of man. Then, VM | V M |  3km / hr. It appears to the man that rain is falling U Solution: (d) (d) 60 o vertically at the rate of 3 3 km / hr. This means that the relative ST velocity of rain with respect to man is 3 3 km / hr , in a direction O Vm= 3km/hr.  A perpendicular to the velocity of the man. Let V RM denote the relative velocity of rain with respect to man and V R be the velocity of rain. Let OA  V R , OB  V RM. Complete the parallelogram OACB. VRm=33Km/ hr B Then, the diagonal OC represents true velocity VR of the rain. 2  2 VM.VRM cos 90 o  VR2  VM2  VRM [Using w 2  v 2  u 2  2uv cos  ]  VR  3 2  (3 3 )2  2  3  3 3  0  6 km / hr. Let  be the angle between the direction of velocity of man and velocity of rain. Then, tan   VRM sin 90 o VM  VRM cos 90 o  tan   3 3  3 3 VR C 122 Dynamics Example: 11 Thus, the actual velocity of the rain is 6 km/hr in a direction making an angle of 60 o with the direction of motion of the man. The particles start simultaneously from the same point and move along two straight lines, one with uniform velocity u and the other from rest with uniform acceleration f. Let  be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time [AIEEE 2003] Solution: (d) (b) f cos  u (c) u sin  (d) After t, velocity = f  t A f 2 t 2  u 2  2 fut cos  For max. and min., u cos  f O u  f B ID t d 2 (VBA )  2 f 2 t  2 fu cos   0 dt E3 VBA  f t  (u) VBA  u cos  f 60 u sin  f (a) 4.5 Rectilinear motion with uniform Acceleration. D YG U A point or a particle moves in a straight line, starting with initial velocity u, and moving with constant acceleration f in its direction of motion. If v be its final velocity at the end of time t and s be its distance at the instant, from its starting point, then the equations describing the motion of the particle are, s t=0 t Velocity = u (i) v  u  ft (ii) s  ut  1 2 ft 2 X P (iii) v 2  u 2  2 fs 1 f (2n  1) 2 Important Tips If a particle moves in a straight line with initial velocity u m/sec. and constant acceleration f m/sec2, then distance 1 1 1 u v  travelled in t seconds is given by, s  ut  ft 2  [2ut  ft 2 ]  [u  (u  ft )] t =  . t, where v  u  ft 2 2 2  2  ST  U (iv) If s n is the distance travelled by the particle in nth second, then sn  u  = (Average velocity) × t {Average velocity = u v } 2 Example: 12 A body is moving in a straight line with uniform acceleration. It covers distances of 10 m and 12 m in third and fourth seconds respectively. Then the initial velocity in m/sec. is (a) 2 (b) 3 (c) 4 (d) 5 Solution: (d) Let initial velocity is u m/sec. and acceleration is f m / sec 2. So, u  1 5 f (2  3  1)  10 or u  f  10 2 2.....(i) and u  1 7 f (2  4  1)  12 or u  f  12 2 2 2 f  2 or f  2m / sec 2 2 5 Substituting value of f in equation (i) , u   2  10 or u  5  10 or u  5m / sec. 2 Subtracting (i) from (ii), we get 0 ......(ii) Dynamics 123 Solution: (d) Two trains A and B, 100 kms apart, are travelling to each other with starting speed of 50 km/hr for both. The train A is accelerating at 18 km/hr2 and B is decelerating at 18 m/h2. The distance where the engines cross each other from the initial position of A is (a) 50 kms (b) 68 kms (c) 32 kms (d) 59 kms Let engine of train A travel x km and cross engine of train B in t hours. u  50 km / hr , 1 Acceleration f  18 km / hr 2 , so s  50 t   18 t 2 , s  50 t  9 t 2.....(i) 2 So distance travelled by engine of train B will be (100  s) in t hour. 60 Example: 13 u  50 km / hr., Acceleration = –18 km/hr 1  18 t 2  100  s  50 t  9 t 2 2 …..(ii) E3  100  s  50 t  Adding (i) and (ii), we get 100  100 t, t  1 hour. From (i), s  50  1  9  1 2  50  9  59 kms. Example: 14 A particle is moving with a uniform acceleration. If during its motion, it moves x, y and z distance in pth, qth and rth seconds respectively, then (a) (q  r)x  (r  p)y  ( p  q)z  1 (b) (q  r)x  (r  p)y  ( p  q)z  1 (d) (q  r)x  (r  p)y  ( p  q)z  0 ID (c) (q  r)x  (r  p)y  ( p  q)z  0 Let u be the initial velocity and f be the acceleration of the particle. 1 1 1 Then s pth  u  f (2 p  1)  x , sqth  u  f (2 q  1)  y , srth  u  f (2r  1)  z 2 2 2 Now, multiplying (i) by (q  r), (ii) by (r  p ) , (iii) by ( p  q ) x (q  r)  y(r  p)  z ( p  q)  0. 1 1 (a) 2 s    f r Solution: (d) adding, we get A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by [AIEEE 2003] D YG Example: 15 and U Solution: (c) (b) 2s 1 1  f r (c) 2 s( f  r) 1 1 2 s    f r (d) Portion OA, OB corresponds to motion with acceleration ‘f’ and retardation ‘r’ respectively. Area of OAB  s and OB  t. Let OL  t1 , LB  t2 and AL  v U 1 2s 1 s  OB. AL = t.v and v  2 2 t v v 2s v v 2s Also, f  , t1   and r  , t2   t2 r tr t1 f tf ST t  t1  t2  2s 2s ;  tf tr  1 1  2s 1 1  t  2 s  . t     f r t  f r   A v v O t1 L t2 B t 4.6 Motion Under Gravity. When a body is let fall in vacuum towards the earth, it will move vertically downward with an acceleration which is always the same at the same place on the earth but which varies slightly from place to place. This acceleration is called acceleration due to gravity. Its value in M.K.S. system is 9.8 m / sec 2 , in C.G.S. system 981 cm / sec 2 and in F.P.S. system 32 ft / sec 2. It is always denoted by g. (1) Downward motion: If a body is projected vertically downward from a point at a height h above the earth’s surface with velocity u, and after t second its velocity becomes v, the equation of its motion are v  u  g t , 124 Dynamics 1 1 2 gt , v 2  u 2  2 gh , s t  u  g (2t  1). In particular, if the body starts from rest or is 2 2 1 simply let fall or dropped, then, v  gt , h  gt 2 , v 2  2 gh. ( u  0) 2 (2) Upward motion : When a body be projected vertically upward from a point on the earth’s surface with an initial velocity u and if the direction of the upward motion is regarded as +ve, the direction of acceleration is –ve and it is, therefore, denoted by – g. The body thus moves with a retardation and its velocity gradually becomes lesser and lesser till it is zero. Thus, for upward motion, the equations of motion are, 1 1 v  u  gt , h  ut  gt 2 , v 2  u 2  2 gh , s t  u  g (2t  1). 2 2 (3) Important deductions (i) Greatest height attained : Let H be the greatest height. From the result, v 2  u 2  2 gh. E3 60 h  ut  u2 u2.. Hence greatest height (H )  2g 2g (ii) Time to reach the greatest height : Let T be the time taken by the particle to reach the greatest height. From the result, v  u  gt u u 0  u  gT i.e., T . Therefore time to reach the greatest height (T )  We have, g g (iii) Time for a given height : Let t be the time taken by the body to reach at a given height  H  U ID We have, 0  u 2  2 gH u  u 2  2 gh u  u 2  2 gh 1 2 and t 2  gt  gt 2  2ut  2h  0. Hence, t1  g g 2 D YG h. Then, h  ut  ST U Clearly t 1 and t 2 are real; if u 2  2 gh. If u 2  2 gh , A u then t1  t 2  , which is the time taken to reach the highest g P point. t1 h u So, let u 2  2 gh. O The lesser value of t 1 gives the time when the body is going up i.e., it is time from O to P and the larger value t 2 gives the total time taken by the body to reach the highest point and then coming back to the given point, i.e., it is the time from O to A and then A to P. (iv) Time of flight : It is the total time taken by the particle to reach the greatest height and then return to the starting point again. When the particle returns to the starting point, 1 h  0. Therefore, from the result, h  ut  gt 2. 2 2u 1 We have, 0  ut  gt 2 or t = 0 or g 2 t = 0 corresponds to the instant with the body starts, and t = 2 u/g corresponds to the time when the particle after attaining the greatest height reaches the starting point.  Time of flight 2u . g Dynamics 125 Solution: (c) A stone is dropped from a certain height which can reach the ground in 5 sec. If the stone is stopped after 3 seconds of its fall and then time taken by the stone to reach the ground for the remaining distance is [MNR 1985; UPSEAT 2000] (a) 2 seconds (b) 3 seconds (c) 4 seconds (d) None of these 1 25 Let distance travelled in 5 sec. is h1 metre.  h1  0  g  5  5 or h1  g 2 2 Let distance travelled in 3 seconds is h2 metre. So, h2  0  So remaining distance h  h1  h2 or h  1 9 g  3  3 or h2  g 2 2 16 25 9 g  g or h  g  8g. 2 2 2 60 Example: 16 Let time taken by the stone to reach the ground for the remaining distance 8g is t second. Solution: (a) 1 2 gt or t 2  16 ,  t  4 sec. 2 A man in a balloon, rising vertically with an acceleration of 4.9 m/sec2 releases a ball 2 seconds after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (a) 14.7 m (b) 19.6 m (c) 9.8 m (d) 24.5 m Velocity after 2 seconds = (4.9)2 = 9.8 m/sec. Distance covered in 2 seconds = ID Example: 17 8g  0  E3  1 2 1 ft  (4.9 )(2)2  9.8 m 2 2 Again after the release of ball, velocity of the ball = 9.8 m/sec. U v  0 ; Using v 2  u 2  2 gh , we get 0  (9.8)2  2(9.8)h  h  9.8  4.9 metre. 2 Hence greatest height attained = 9.8 + 4.9 = 14.7 m. A particle is dropped under gravity from rest from a height h(g  9.8 m / sec 2 ) and then it travels a distance D YG Example: 18 9h in the last second. The height h is 25 (a) 100 metre Solution: (b) Sn  u  (b) 122.5 metre 1 1 f (2n  1)  g(2n  1) 2 2 (c) 145 metre (d) 167.5 metre (u  0, f  g) 9h 1 1  g(2n  1) and h  gn 2 25 2 2 9 1 2 1 9 2. gn  g(2n  1)  n  (2n  1) 25 2 2 25 U  50  2500  900 50  40 5  ; n  5 or 18 18 9 ST  9 n 2  50 n  25  0  n   9 n 2  50 n  25 Since 5 1  1 ,  rejecting this value. We have n  5 ;  h  (9.8 )  25  122.5 metre. 9 2 4.7 Laws of Motion. (1) Newton’s laws of motion (i) First law : Every body continues in its state of rest or of uniform motion in a straight line except when it is compelled by external impressed forces to change that state. (ii) Second law : The rate of change of momentum is proportional to the impressed force, and takes place in the direction of the straight line in which the force acts. d dv (mv )  F  m F dt dt  m dv  KF  F  mf dt (for K  1) 126 Dynamics Force = mass  acceleration. The S.I. unit of force is Newton and C.G.S. unit of force is dyne. 1 Newton = 1 kg-m/sec2, 1 Dyne = 1 gm-cm/sec2 60 (iii) Third law : To every action there is an equal and opposite reaction, or the actions and reaction are always equal and opposite. Note :  1 Newton = 105 Dynes.  The action and reaction do not act together on the same body or the same part of E3 the body. (2) Weight : The weight of a body is the force with which it is attracted by the earth towards its centre. For a body of mass m, the weight W is given by W  mg (By Newton's second law of motion) and W2 be the weights of two bodies of masses m 1 and m 2 respectively at a ID Note :  Let W1 place on the earth then, W1  m 1 g and W2  m 2 g  W1 m 1  W2 m 2 A train weighing W tons is moving with an acceleration f ft/sec2. When a carriage of weight w tons is suddenly detached from it, then the change in the acceleration of the train is w Wf W wf ft / sec 2 ft / sec 2 ft / sec 2 ft / sec 2 (a) (b) (c) (d) W w W w W w W w Mass of train = W × 2240 lbs.  Pull of the engine = W × 2240 × f Poundals When a carriage of mass W tons is detached, mass of the train = (W – w) tons = (W – w). 2240 lbs Pull is the same as before i.e., W × 2240 × f Poundals. W  2240  f  New acceleration = f1 = (W  w) 2240 U Example: 19 D YG U  1 gm.wt. = g dynes = 981 dynes 1 kg. wt. = g Newtons = 9.81 N. (3) Momentum of a body : It is the quantity of motion in a body and is equal to the product of its mass (m) and velocity (v) with which it moves. Thus, momentum of the body is mv. The units of momentum are gm-cm/sec or kg-m/sec. Momentum = Mass × Velocity = m.v. (4) Impulse of a body : The impulse of a force in a given time is equal to the product of the force and the time during which it acts. The impulse of a force F acting for a time t is therefore F.t. ST Solution: (c) f1  Wf ft / sec 2 W w wf Wf W W w  f = . f = W w W w  W w  A hockey stick pushes a ball at rest for 0.01 second with an average force of 50 N. If the ball weighs 0.2 kg, then the velocity of the ball just after being pushed is (a) 3.5 m/sec. (b) 2.5 m/sec. (c) 1.5 m/sec. (d) 4.5 m/sec. If v m/sec. is the velocity of the ball just after being pushed, then 50 × 0.01 = 0.2 × v[  Impulse = change of mom v = 2.5 m/sec. Change in acceleration = f1  f  Example: 20 Solution: (b) Dynamics 127 A mass m is acted upon by a constant force P lb.wt., under which in t seconds it moves a distance of x feet and acquires a velocity v ft/sec. Then x is equal to gP 2mt 2 (a) gt 2 2 Pm (d) mv 2 2 gP 60 v 2  0 mv 2 v2  u2  =. Pg 2 Pg 2f 2 m An engine and train weight 420 tons and the engine exerts a force of 7 tons. If the resistance to motion be 14 lbs. wt. per ton, then the time, the train will take to acquire a velocity of 30 m/hr. from rest is [BIT Ranchi 1994] (a) 2.2 m Solution: (a) (c) Force = P lb.wt.= Pg poundals and mass = m lbs F Pg  f  ft / sec 2 m m Now initial velocity = 0, final velocity = v ft/sec. Distance = x and time = t Hence x = Example: 22 mg 2v 2 P E3 Solution: (d) (b) (b) 2.6 m (c) 2.8 m Effective force = Pull of the engine – Resistance (d) 3 m ID Example: 21 = 7 tons – 420 × 14 lbs = 7 × 2240 lbs – 5880 lbs = 9800 lbs = 9800 × 32 Poundals P 9800  32 1   ft / sec 2 m 420  2240 3 U Also acceleration = Again initial velocity = 0 and final velocity = 30 m.p.h. = 44 ft/sec. 1 t 3 D YG Use v  u  ft, we get 44  0   t  132 second = 2 1 12 minute; t  2 minute = 2.2 minute. 5 60 4.8 Motion of a Body released from a Balloon or a Lift. (1) When a lift is ascending with uniform acceleration of f m/sec2 and after t second a body is dropped from it, then at the time when the body is dropped: U (i) Initial velocity of the body is same as that of the lift and is in the same direction. So, the velocity of the body is ft m/sec. (ii) Initial velocity of the body relative to the lift = Velocity of the body – Velocity of the lift = ft – ft = 0 ST (iii) Acceleration of the body = g m/sec2 in downward direction. (iv) Acceleration of the lift = f m/sec2 in upward direction. (v) Acceleration of the body relative to the lift = Acceleration of the body – Acceleration of the lift = g – (– f) = f + g in downward direction. (2) When a lift is ascending with uniform acceleration of f m / sec 2 and after t second a body is thrown vertically upward with velocity v m / sec , then at that time, we have the following : (i) Initial velocity of the body = v + velocity of lift = v + ft, in upward direction (ii) Initial velocity of the body relative to the lift = Velocity of the body – Velocity of lift = (v + ft) – ft = v m/sec. 128 Dynamics (iii) Acceleration of the body relative to the lift in vertically downward direction is (f + g) m/sec2. (3) When a lift is descending with uniform acceleration f m / sec 2 and after time t a body is dropped from it. Then at that time, we have the following 60 (i) Velocity of the body = Velocity of the lift = ft m/sec in downward direction (ii) Acceleration of the body relative to the lift in downward direction = Acceleration of the body – Acceleration of lift = g  f m / sec 2 E3 4.9 Apparent weight of a Body resting on a moving Horizontal plane or a Lift. Let a body of mass m be placed in a lift moving with an acceleration f and R is the normal reaction, then  mf  R  Mg  R  m ( f  g) ID (1) When the lift is rising vertically upwards : Effective force in upward direction = Sum of the external forces in the same direction. U Clearly, the pressure R exerted by the body on the plane is greater than the actual weight mg of the body. This pressure is also known as apparent weight. D YG If a man of mass m is standing on a lift which is moving vertically upwards with an  f R acceleration, then R  m (g  f )  R  mg  1   g  f ST U  f  R = (weight of the man)  1   g  f mg  R = Weight of the man + (weight of the man) g f Thus, the apparent weight of the man is times more than the actual weight. g (2) When the lift is descending vertically downwards : Effective force in downward direction = Sum of the forces in the same direction. R  mf  mg  R  R  m (g  f )......(ii) f mg Clearly, the pressure exerted by the body on the plane is less than its actual weight when the plane moves vertically downwards. If a man of mass m in is standing on a lift, which is moving vertically downward with an  f f acceleration f, then the pressure R  m (g  f ) = mg  1    R = weight of the man – (weight g g   of the man) R Sand m g Dynamics 129 Thus, the apparent weight of the man is Note :  f times less than his actual weight. g Effective force stopping a falling body = F  mg Important Tips If the plane moves vertically upward with retardation equal to g i.e., f   g , then from R  m( f  g) , we get R  0. E3  60  If R be the resistance of sand on a body of mass m falling in sand, then effective force = R  mg. Thus there is no pressure of the body on the plane when the plane rises vertically with retardation equal to g.  If the plane moves down freely under gravity i.e., with acceleration equal to g, then from R  m(g  f ) , we get R  0. Thus there is no pressure of the body on the plane, when it moves vertically downwards with an acceleration equal to g. ID Solution: (c) A paracute weighing 112 lbs. wt. falling with a uniform acceleration from rest, describes 16 ft. in the first 4 sec. Then the resultant pressure of air on the parachute is (a) 85 lbs. wt. (b) 95 lbs. wt. (c) 105 lbs. wt. (d) 115 lbs. wt. Let f be the uniform acceleration. Here u  0 , s  16 , t  4 sec. 1 1 Since s  ut  ft 2 ,  16  0  f (4 )2.  f  2 ft / sec 2 2 2 Equation of motion is mf  mg  R (Downward) ST U D YG  R  m (g  f ) = 112 (32 – 2) U Example: 23 = 112 × 30 Poundals  112  30 lbs. wt. = 105 lbs. wt. 32 128 Dynamics Example: 24 A man falls vertically under gravity with a box of mass ‘m’ on his head. Then the reaction force between his head and the box is (a) mg Solution: (c) (b) 2 mg (c) 0 (d) 1.5 mg Let reaction be R. Since the man falls vertically under gravity  f  g From mg  R  mf , we have mg  R  mg.  R  0. A man of mass 80 kg. is travelling in a lift. The reaction between the floor of the lift and the man when the lift is ascending upwards at 4 m/sec2 is (a) 1464.8 N (b) 1784.8 N (c) 1959.8 N 60 Example: 25 (d) 1104.8 N Solution: (d) When the lift is ascending, we have R  mg  ma  m (g  a) = 80 (9.81  4 )  80 (13.81) = 1104.8 N. Example: 26 A particle of mass m falls from rest at a height of 16 m and is brought to rest by penetrating (a) 2 kg (b) 3 kg (c) 4 kg (d) 8 kg Let f m / sec 2 be the retardation produced by the thrust exerted by the ground and let v m / sec. be the ID Solution: (c) E3 1 m into 6 the ground. If the average vertical thrust exerted by the ground be 388 kg. wt, then the mass of the particle is [ v 2  u 2  2 gh] velocity with which the mass m penetrates into the ground. Then, v 2  0 2  2 g  16  v  2  9.8  16  4 19.6  17.70 m / sec. U u=0 1 Since the mass m penetrates through m in the ground. 6 16 m D YG Therefore, its final velocity is zero. So, 0 2  v 2  2 fs  0  (4 19.6 )2  2 f   f 16  19.6  6 940.8 m  940.8 m / sec 2  (388  m )  2 9.8  388  m  96 m  m  1/6 m 1 6 v=0 f [ g  9.8 m / sec 2 ] 388  4. Hence, the mass of the particle is 4 kg. 97 4.10 Motion of two particles connected by a String. U (1) Two particles hanging vertically : If two particles of masses m 1 and m 2 (m 1  m 2 ) are ST suspended freely by a light inextensible string which passes over a smooth fixed light pulley, (m  m 2 )g then the particle of mass m 1 ( m 2 ) will move downwards, with Acceleration f  1 m1  m 2 Tension in the string T  2m 1 m 2 g m1  m 2 4m1m 2 g Pressure on the pulley  2T  m1  m 2 f f T T m 2g m1g (2) One particle on smooth horizontal table : A particle of mass m 1 attached at one end of light inextensible string is hanging vertically, the string passes over a pulley at the end of a smooth Dynamics 129 horizontal table, and a particle of mass m 2 placed on the table is attached at the other end of the string. Then for the system, m1 g m1  m 2 Tension in the string T  f T m1m 2 g m1  m 2 f T 2m 1 m 2 g Pressure on the pulley  T 2 . m1  m 2 T 60 Acceleration f  T m 2g E3 m1g (3) One particle on an inclined plane : A particle of mass m 1 attached at one end of a light inextensible string is hanging vertically, the string passes over a pulley at the end of a smooth plane inclined at an angle  to the horizontal, and a particle of mass m 2 placed on this inclined ID plane is attached at the other end of the string. If the particle of mass m 1 moves vertically downwards, then for the system, U (m  m 2 sin  ) Acceleration f  1 g m1  m 2 D YG m m (1  sin  ) Tension in the string T  1 2 g m1  m 2 Pressure on the pulley  {2(1  sin  )}T  Example: 27 f m2g sin  f m 2g m1g 2m 1 m 2 g(1  sin  )3 / 2 m1  m 2 The sum of two weights of an Atwood's machine is 16 lbs. The heavier weight descends 64 ft in 4 seconds. The value of weights in lbs are (a) 10, 6 Solution: (a) (b) 9, 7 (c) 8, 8 (d) 12, 4 Let the two weights be m1 and m 2 (lbs.) U (m1  m 2 ),  m1  m 2  16......(i) Now, the distance moved by the heavier weight in 4 seconds from rest is 64 ft. ST If acceleration = f, then 64  0  But f  1 f (4 )2  f  8 ft / sec 2 2 m1  m 2 m  m2 g  1  32  8  m1  m 2  4......(ii) m1  m 2 16 From (i) and (ii), m 1  10 lbs, m 2  6 lbs. Example: 28 Two weights W and W  are connected by a light string passing over a light pulley. If the pulley moves with an upward acceleration equal to that of gravity, the tension of the string is (a) Solution: (d) WW  W  W (b) 2W W  W  W (c) 3W W  W  W (d) 4 WW  W  W Suppose W descends and W  ascends. If f is the acceleration of the weights relative to the pulley, then since the pulley ascends with an acceleration f, the actual acceleration of W and W  are ( f  g) (downwards) and ( f  g) (upwards) respectively. Let T be tension in the string. 130 Dynamics  Wg  T  W ( f  g) and T  W g  W ( f  g) or g  Subtracting, We get 4 WW  T T kg. wt.   2 g  2 g or T  W W W  W To one end of a light string passing over a smooth fixed pulley is attached a particle of mass M, and the other end carries a light pulley over which passes a light string to whose ends are attached particles of masses m 1 and m 2. The mass M will remain at rest or will move with a uniform velocity, if Solution: (c) 1 1 1   M m1 m 2 (b) 2 1 1   M m1 m 2 If the mass M has no acceleration, then Mg  T1  2m 1 m 2  4 m 1 m 2 But T1  2T2  2  g   g  m1  m 2  m1  m 2 4 m 1m 2 m  m2 4 1 1  1   g or M m1m 2 m1 m 2 m1  m 2 4.11 Impact of Elastic Bodies. 4 1 1   M m1 m 2 ID  Mg  (c) (d) 8 1 1   M m1 m 2 E3 (a) [MNR 1993] 60 Example: 29 T T g  f g  f  g and W W T2 m 2g m1g T1 T2 mg D YG U (1) Elasticity : It is that property of bodies by virtue of which they can be compressed and after compression they recover or tend to recover their original shape. Bodies possessing this property are called elastic bodies. (2) Law of conservation of momentum : It states “the total momentum of two bodies remains constant after their collision of any other mutual action.” Mathematically m 1.u 1  m 2.u 2  m 1.v 1  m 2.v 2 where m 1  mass of the first body, u 1  initial velocity of the first body, v 1  final velocity of the first body. m 2 , u 2 , v 2  Corresponding values for the second body. ST U (3) Coefficient of restitution : This constant ratio is denoted by e and is called the coefficient of restitution or coefficient of elasticity. The values of e varies between the limits 0 and 1. The value of e depends upon the substances of the bodies and is independent of the masses of the bodies. If the bodies are perfectly elastic, then e  1 and for inelastic bodies e  0. (4) Impact : When two bodies strike against each other, they are said to have an impact. It is of two kinds : Direct and Oblique. (5) Newton's experimental law of impact : It states that when two elastic bodies collide, their relative velocity along the common normal after impact bears a constant ratio of their relative velocity before impact and is in opposite direction. If u 1 and u 2 be the velocities of the two bodies before impact along the common normal at their point of contact and v 1 and v 2 be their velocities after impact in the same direction, v 1  v 2  e (u 1  u 2 ) Case I : If the two bodies move in direction shown in diagram given below, then (a) v 1  v 2  e (u 1  u 2 ) and (b) m 1 v 1  m 2 v 2  m 1 u 1  m 2 u 2 Dynamics 131 u1 m1 v1 u2 A m2 v2 B 60 Case II : If the direction of motion of two bodies before and after the impact are as shown below, then by the laws of direct impact, we have (a) v 1  v 2  e[u 1  (u 2 )] or v 1  v 2  e (u 1  u 2 ) and (b) m 1 v 1  m 2 v 2  m 1 u 1  m 2 (u 2 ) or m 1 v 1  m 2 v 2  m 1 u 1  m 2 u 2 m1 v1 u2 m2 v2 E3 u1 Case III : If the two bodies move in directions as shown below, then by the laws of direct impact, we have (a) v 1  (v 2 )  e[u 1  (u 2 )] or v 1  v 2  e (u 2  u 1 ) u1 m1 A ID and (b) m 1 v 1  m 2 (v 2 )  m 1 (u 1 )  m 2 (u 2 ) or m 1 v 1  m 2 v 2  (m 1 u 1  m 2 u 2 ) v1 u2 m2 v2 B U (6) Direct impact of two smooth spheres : Two smooth spheres of masses m 1 and m 2 moving along their line of centres with velocities u 1 and u 2 (measured in the same sense) impinge D YG directly. To find their velocities immediately after impact, e being the coefficient of restitution between them. Let v 1 and v 2 be the velocities of the two spheres immediately after impact, measured along their line of centres in the same direction in which u 1 and u 2 U are measured. As the spheres are smooth, the impulsive action and reaction between them will be along the common normal at the point of contact. From the principle of conservation of momentum, m 1v1  m 2 v 2  m 1u1  m 2 u 2 …..(i) Also from Newton’s experimental law of impact of two bodies, v 2  v 1  e[u 1  u 2 ] …..(ii) ST Multiplying (ii) by m 2 and subtracting from (i), we get u1 u2 m1 m 2 v2 v1  (m 1  m 2 )v 1  (m 1  em 2 )u 1  m 2 u 2 (1  e )  v1  (m 1  em 2 )u1  m 2 u 2 (1  e ) m1  m 2 …..(iii) (7) Oblique impact of two spheres : A smooth spheres of mass m 1 , impinges with a velocity u 1 obliquely on a smooth sphere of mass m 2 moving with a velocity u 2. If the direction of motion before impact make angles  and  respectively with the line joining the centres of the spheres, and if the coefficient of restitution be e, to find the velocity and directions of motion after impact, 132 Dynamics Let the velocities of the sphere after impact be v 1 and v 2 in directions inclined at angles  and  respectively to the line of centres. Since the v 2 sin   u 2 sin  …..(ii) u2 u1   m1 O  m2  O 60 spheres are smooth, there is no force perpendicular to the line joining the centres of the two balls and therefore, velocities in that direction are unaltered. v 1 sin   u 1 sin  …..(i) v2 v1 E3 And by Newton’s law, along the line of centres, v 2 cos   v 1 cos   e (u 2 cos   u 1 cos  ) …..(iii) Again, the only force acting on the spheres during the impact is along the line of centres. Hence the total momentum in that direction is unaltered.  m 1 v 1 cos   m 2 v 2 cos   m 1 u 1 cos   m 2 u 2 cos  …..(iv) U ID The equations (i), (ii), (iii) and (iv) determine the unknown quantities. (8) Impact of a smooth sphere on a fixed smooth plane : Let a smooth sphere of mass m moving with velocity u in a direction making an angle  with the vertical strike a fixed smooth horizontal plane and let v be the velocity of the sphere at an angle  to the vertical after impact. Since, both the sphere and the plane are smooth, so there is no change in velocity parallel to the horizontal plane. …..(i) D YG  v sin   u sin  N And by Newton’s law, along the normal CN, velocity of separation v u = e.(velocity of approach)   v cos   0  eu cos  o v cos   eu cos  m ….(ii) Dividing (i) by (ii), we get cot   e cot  c Particular case : If   0 then from (i), v sin   0  sin   0 U   0 ;  v  0 and from (ii) v  eu ST Thus if a smooth sphere strikes a smooth horizontal plane normally, then it will rebound along the normal with velocity, e times the velocity of impact i.e. velocity of rebound = e.(velocity before impact). Rebounds of a particle on a smooth plane : If a smooth ball falls from a height h upon a fixed smooth horizontal plane, and if e is the coefficient of restitution, then whole time before  2h  1  e the rebounding ends   .  g  1e And the total distance described before finishing rebounding  Example: 30 1e2 h. 1e2 A ball is dropped from a height h on a horizontal plane and the coefficient of restitution for the impact is e, the velocity with which the ball rebounds from the floor is (a) eh (b) egh (c) e gh (d) e 2 gh Dynamics 133 Solution: (d) Let u be the velocity with which the ball strikes the ground. Then, u 2  2 gh  u  2 gh Suppose the ball rebounds with velocity v, then V  e (u  0)  v  ue  v  e 2 gh , Example: 31.....(i) [from (i)]. A sphere impinges directly on a similar sphere at rest. If the coefficient of restitution is 1 , the 2 velocities after impact are in the ratio Solution: (c) (b) 2 : 3 (c) 1 : 3 From principle of conservation of momentum , mv1  mv1  mu1  mu1......(i) (d) 3 : 4 60 (a) 1 : 2 From Newton's rule for relative velocities before and after impact. v1  v1  e(u1  u1 ) 1 and let u1  0 2......(iii) 1 u1 2.......(iv) Adding (iii) and (iv) , 2v1  1 1 u1 or v1  u1 4 2 From (iii) and (v), v1  u1  1 3 u1  u1. Hence v1 : v1  1 : 3. 4 4 U......(v) A ball impinges directly upon another ball at rest and is itself reduced to rest by the impact. If half of the K.E. is destroyed in the collision, the coefficient of restitution, is 1 1 1 3 (a) (b) (c) (d) 4 3 4 2 Let masses of the two balls be m1 and m 2. D YG Solution: (d) ID Then from (i) mv1  m v1  mu1 or v1  v1  u1 From (ii), v1  v1  Example: 32 E3 Here m  m , e ......(ii) By the given question, we have m2 m1 u1 u2  0 v1  0 v2  m1u1  m2.0  m1.0  m2v2 or m1u1  m2v2.....(i) and 0  v2  e(0  u1 ) i.e., v2  eu1.....(ii) U Again for the given condition, K.E. (before impact) = 2 × K.E. after impact  1 1   m1u12  0  2  0  m 2v 22  2 2   ST But from (i), m1u1  m2v2 , .....(iii) or 1 1 m1u12  2. m 2v 22 2 2 1 1 u1  v 2 i.e., u1  2v2. Putting in (ii), we get v 2  e.2v 2 ,  e . 2 2 4.12 Projectile Motion. When a body is thrown in the air, not vertically upwards but making an acute angle  with the horizontal, then it describes a curved path and this path is a Parabola. The body so projected is called a projectile. The curved path described by the body is called its trajectory. The path of a projectile is a parabola whose equation is y  x tan   gx 2 Y v sin  2u 2 cos 2  u  v v cos  O L B X 134 Dynamics  u 2 sin  cos  u 2 sin 2   . Its vertex is A ,  g 2g   (ii) Range of flight i.e., Horizontal Range R  (iii) Greatest height  u 2 sin 2 g u2  and this happens, when   g 4 ID Maximum horizontal Range  E3 60  u 2 sin 2  u 2 cos 2   Focus is S  ,  2g 2g   (1) Some important deductions about projectile motion: Let a particle is projected with velocity u in a direction making angle  with OX. Let the particle be at a point P( x , y ) after time t and v be its velocity making an angle  with OX. 2u sin  (i) Time of flight  g u 2 sin 2  2g u sin  g U Time taken to reach the greatest height  D YG (iv) Let P( x , y ) be the position of the particle at time t. Then x  (u cos  ) t , y  (u sin  ) t  (v) Horizontal velocity at time t  Vertical velocity at time t  (vi) Resultant velocity 1 2 gt 2 dx  u cos  dt dy  u sin   gt dt at time t  u 2  2ugt sin   g 2 t 2. And its direction is  u sin   gt    u cos     tan 1  U (vii) Velocity of the projectile at the height h ST  u 2 sin 2   2 gh   is   tan 1  u cos      (2) Range and time of flight on an inclined plane Y u  O R  A (R cos ,R sin ) X  u 2  2 gh and its direction Dynamics 135 Let OX and OY be the coordinate axes and OA be the inclined plane at an angle  to the horizon OX. Let a particle be projected from O with initial velocity u inclined at an angle  gx 2. 2u 2 cos 2  (i) Range and time of flight on an inclined plane with angle of inclination  are given by 2u 2 cos  sin(   ) and maximum range up the plane  g cos 2  (ii) Time of flight T  2u sin(   ) g cos  (iii) Range down the plane  2u 2 cos  sin(   ) g cos 2  (v) Time of flight  u2  , where 2    g(1  sin  ) 2 ID (iv) Maximum range down the plane  u2  , where 2    g(1  sin  ) 2 E3 R 60 with the horizontal OX. The equation of the trajectory is y  x tan   2u sin(   ) , down the plane g cos  Two stones are projected from the top of a cliff h metres high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle  to the horizontal then tan  equals (a) Solution: (d) D YG Example: 33 U (vi) Condition that the particle may strike the plane at right angles is cot   2 tan(    ). 2u gh (b) 2 g 2h 1  (u cos  )  t ; t  g cos  R u Now h  (u sin  )t  u sin  cos  (c) 2h 2h g.....(i) u g h  u 2h 1  2h  2h  g tan   h sec 2   , h  u g 2  g cos 2   g 2 2h 2. tan   h tan 2   h , tan 2   u tan   0 ;  tan   u hg hg g (a) 9.8 m Solution: (a) Horizontal range = u h O R X (c) 2.45 m (d) 19.6 m u 2 sin 2 u 2 sin 90 o u2  39.2   39. 2  39. 2  g g g u 2 sin 2  39.2 1 = sin 2 45 o = (19.6)  9.8 m. 2g 2 2 If the range of any projectile is the distance equal to the height from which a particle attains the velocity equal to the velocity of projection, then the angle of projection will be (a) 60 o Solution: (b)  [AMU 1999] (b) 4.9 m Greatest height attained = Example: 35 Y 2 gh A gun projects a ball at an angle of 45 o with the horizontal. If the horizontal range is 39.2 m, then the ball will rise to ST Example: 34 (d) u ‘t’ from (i) U h 1 2 gt , 2 u h Range = (b) 75 o (c) 36 o u 2 sin 2 , where  is the angle of projection. g (d) 30 o 136 Dynamics Also, u 2  2 gh  u2 h 2g By the given condition, u 2 u 2 sin 2 1    sin 2 ,  2  30 o or 150 o 2g g 2    15 o or 75 o. Hence, required angle of projection = 75 o. (a) Solution: (c) 2v 2 3g (b) 4v2 3g (c) 4v2 5g Range = 2  (maximum height)  60 A particle is thrown with the velocity v such that its range on horizontal plane is twice the maximum height obtained. Its range will be (d) v2 7g E3 Example: 36 v 2 sin 2 v 2 sin 2  2  sin 2  sin2  g 2g  2 sin cos   sin 2   0  sin  (2 cos   sin  )  0  sin   0 or tan   2 Example: 37 v 2 2.2 v 2 sin 2 v 2 2 tan  4v 2.  . . 2 g 14 g g 1  tan  5g ID But   0 , sin  0 , tan   2. Now R  Let u be the velocity of projection and v1 be the velocity of striking the plane when projected so that the range up the plane is maximum, and v 2 be the velocity of striking the plane when projected so U that the range down the plane is maximum, then u is the (a) A.M. of v1 , v 2 Solution: (b) (b) G.M. of v1 , v 2 (c) H.M. of v1 , v 2 (d) None of these Let u be the velocity of projection and  be the inclination of the plane with the horizon. Let R be the u2 g(1  sin  ) D YG maximum range up the plane. Then, R  The coordinates of the point where the projectile strikes the plane are ( R cos  , R sin  ). It is given that v1 is the velocity of striking the plane when projected so that the range up the plane is maximum.   v12  u 2  2 gR sin  v12  u 2  2 g sin .  1  sin  u2  v12  u 2  g(1  sin  )  1  sin     U  1  sin    Replacing  by   in (i) , we get v 22  u 2   1  sin  .....(i).....(ii) ST From (i) and (ii), we get v12.v 22  u 4  v1v 2  u 2 i.e., u is the G.M. of v1, v2. 4.13 Work, Power and Energy. (1) Work : Work is said to be done by a force when its point of application undergoes a displacement. In other words, when a body is displaced due to the action of a force, then the force is said to do work. Work is a scalar quantity. d Work done = force  displacement of body in the direction of force    W  F.d W  F d cos  , where  is the angle between F and d. A F (2) Power : The rate of doing work is called power. It is the amount of work that an agent is capable of doing in a unit time. 1 watt = 107 ergs per sec = 1 joule per sec., 1 H.P. = 746 watt. (3) Energy : Energy of a body is its capacity for doing work and is of two kinds : Dynamics 137 Example: 39 U Solution: (b) E3 Solution: (b) Two bodies of masses m and 4m are moving with equal momentum. The ratio of their K.E. is [MNR 1990; UPS (a) 1 : 4 (b) 4 : 1 (c) 1 : 1 (d) 1 : 2 Momentum of first body = mu, where u is the velocity of the first body.  Momentum of second body = 4mv, where v is the velocity of the second body. By the given condition, mu  4 mv  u  4 v 1 mu 2 (4 v)2 2  Ratio of K.E. =   4. Hence, the ratio of K.E. is 4 : 1. 1 4v 2.4 mv 2 2 A bowler throws a bumper with a speed of 24 m / sec. The moment the ball touches the ground, it losses its energy by 1.5 kgm. If the weight of the ball is 225 gm, then speed of the ball at which it hits the bat is [UPSEAT 2000] (a) 2.22 m/sec. (b) 22.2 m/sec. (c) 4.44 m/sec. (d) 44.4 m/sec. Weight of the ball = w  225 gm = 0.225 kg Velocity of the ball = v = 25 m/sec. Loss of energy = 1.5 kg.m ID Example: 38 60 (i) Kinetic energy : Kinetic energy is the capacity for doing work which a moving body possesses by virtue of its motion and is measured by the work which the body can do against 1 any force applied to stop it, before the velocity is destroyed. K. E.  mv 2 2 (ii) Potential energy : The potential energy of a body is the capacity for doing work which it possesses by virtue of its position of configuration. P. E.  mgh. D YG  K.E. of the ball before it touches the ground = 1 wv 2 1 0. 225 .  (25 )2  7. 17 kgm. 2 g 2 9. 81 K.E. of the ball after bumping = K.E. before touching the ground – loss of K.E. = 7.17 – 1.5 = 5.67 kgm. Let v1 be the velocity of the ball at which it hits the bat. Now K.E. of the ball at the time of hitting the bat  1w 2 v1  2 g 5.67  1 0.225 2  v1 2 9.81  v12  494  v1  22.2 m / sec. Example: 40 A particular starts from rest and move with constant acceleration. Then the ratio of the increase in U the K.E. in mth and (m  1)th second is (a) m : m Required ratio = ST Solution: (c) (b) m  1 : m  1 (c) 2m  1 : 2m  1 th (d) None of these Increase in K.E. in m second Work done against the force in m th second = th Increase in K.E. in (m  1) second Work done against the force in (m  1)th second 1   P  0  f (2m  1) 2   = 2m  1  2m  1. = 1 2m  2  1 2m  1   P  0  f (2(m  1)  1) 2  