Physics Notes for NEET Chapter 4 PDF

177 60 Newton's Laws of motion E3 Chapter 4 Newton's Laws of Motion (1) An object can be considered as a point object if during motion in a given time, it covers distance much greater than its own size. (2) Object with zero dimension considered as a point mass. (7) If two objects of different masses...

177 60 Newton's Laws of motion E3 Chapter 4 Newton's Laws of Motion (1) An object can be considered as a point object if during motion in a given time, it covers distance much greater than its own size. (2) Object with zero dimension considered as a point mass. (7) If two objects of different masses have same momentum, the lighter body possesses greater velocity. ID Point Mass p  m1v1  m 2v 2 = constant (3) Point mass is a mathematical concept to simplify the problems. U i.e. v  Inertia D YG v1 m 2  v 2 m1 1 m [As p is constant] (8) For a given body p  v (1) Inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform motion along a straight line by their own is called inertia. (2) Inertia is not a physical quantity, it is only a property of the body which depends on mass of the body.  (9) For different bodies moving with same velocities p  m p m = constant p v = constant (3) Inertia has no units and no dimensions (4) Two bodies of equal mass, one in motion and another is at rest, possess same inertia because it is a factor of mass only and does not depend upon the velocity. Linear Momentum Fig : 4.2 v Fig : 4.3 m Newton’s First Law (2) It is measured in terms of the force required to stop the body in unit time. A body continue to be in its state of rest or of uniform motion along a straight line, unless it is acted upon by some external force to change the state. ST U (1) Linear momentum of a body is the quantity of motion contained in the body. (3) It is also measured as the product of the mass of the body and its velocity i.e., Momentum = mass × velocity. If a body of mass m is moving with velocity v then its linear momentum p is given by p  m v (4) It is a vector quantity and it’s direction is the same as the direction of velocity of the body. (5) Units : kg-m/sec [S.I.], g-cm/sec [C.G.S.] (6) Dimension : [MLT 1 ] v p = constant Fig : 4.1 m (1) If no net force acts on a body, then the velocity of the body cannot change i.e. the body cannot accelerate. (2) Newton’s first law defines inertia and is rightly called the law of inertia. Inertia are of three types : Inertia of rest, Inertia of motion and Inertia of direction. (3) Inertia of rest : It is the inability of a body to change by itself, its state of rest. This means a body at rest remains at rest and cannot start moving by its own. Example : (i) A person who is standing freely in bus, thrown backward, when bus starts suddenly. When a bus suddenly starts, the force responsible for bringing bus in motion is also transmitted to lower part of body, so this part of the body 178 Newton's Laws of Motion :  (i) If the motion of the bus is slow, the inertia of Hole by the bullet Fig : 4.4 (iv) In the arrangement shown in the figure : D YG (a) If the string B is pulled with a sudden jerk then it will experience tension while due to inertia of rest of mass M this force will not be transmitted to the string A and so the string A B will break. (b) If the string B is pulled steadily the force applied to it will be transmitted from string B to A through the mass M and as tension in A will be greater than in B by Mg (weight of mass M), the string A will break. (iii) When a car goes round a curve suddenly, the person sitting inside is thrown outwards. Newton’s Second Law (1) The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the applied force.  (2) If a body of mass m, moves with velocity v then its linear   momentum can be given by p  m v and if force F is applied on a body, then   d p dp F FK dt dt   dp or F  (K = 1 in C.G.S. and S.I. units) dt   d   dv or F  (m v )  m  ma dt dt  dv (As a   acceleration produced in the body) dt    F  ma U Cracks by the ball (ii) The rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional inertia. ID motion will be transmitted to the body of the person uniformly and so the entire body of the person will come in motion with the bus and the person will not experience any jerk. (ii) When a horse starts suddenly, the rider tends to fall backward on account of inertia of rest of upper part of the body as explained above. (iii) A bullet fired on a window pane makes a clean hole through it, while a ball breaks the whole window. The bullet has a speed much greater than the ball. So its time of contact with glass is small. So in case of bullet the motion is transmitted only to a small portion of the glass in that small time. Hence a clear hole is created in the glass window, while in case of ball, the time and the area of contact is large. During this time the motion is transmitted to the entire window, thus creating the cracks in the entire window. Example : (i) When a stone tied to one end of a string is whirled and the string breaks suddenly, the stone flies off along the tangent to the circle. This is because the pull in the string was forcing the stone to move in a circle. As soon as the string breaks, the pull vanishes. The stone in a bid to move along the straight line flies off tangentially. 60 Note (5) Inertia of direction : It is the inability of a body to change by itself it's direction of motion. E3 comes in motion along with the bus. While the upper half of body (say above the waist) receives no force to overcome inertia of rest and so it stays in its original position. Thus there is a relative displacement between the two parts of the body and it appears as if the upper part of the body has been thrown backward. M B ST U (v) If we place a coin on smooth piece of card Fig : 4.5 board covering a glass and strike the card board piece suddenly with a finger. The cardboard slips away and the coin falls into the glass due to inertia of rest. (vi) The dust particles in a carpet falls off when it is beaten with a stick. This is because the beating sets the carpet in motion whereas the dust particles tend to remain at rest and hence separate. (4) Inertia of motion : It is the inability of a body to change by itself its state of uniform motion i.e., a body in uniform motion can neither accelerate nor retard by its own. Example : (i) When a bus or train stops suddenly, a passenger sitting inside tends to fall forward. This is because the lower part of his body comes to rest with the bus or train but the upper part tends to continue its motion due to inertia of motion. Force = mass  acceleration Force (1) Force is an external effect in the form of a push or pull which (i) Produces or tries to produce motion in a body at rest. (ii) Stops or tries to stop a moving body. (iii) Changes or tries to change the direction of motion of the body. Table 4.1 : Various condition of force application F u=0 v=0 Body remains at rest. Here force is trying to change the state of rest. F u=0 v>0 Body starts moving. Here force changes the state of rest. (ii) A person jumping out of a moving train may fall forward. (iii) An athlete runs a certain distance before taking a long jump. This is because velocity acquired by running is added to velocity of the athlete at the time of jump. Hence he can jump over a longer distance. F u0 v>u In a small interval of time, force increases the magnitude of speed and direction of motion remains same. Newton's Laws of motion F vg) from the floor of the lift and stick to the ceiling of the lift. ST U D YG U ID E3 60 R = – ve Newton's Laws of motion Acceleration of Block on Horizontal Smooth Surface R R = mg (1) When inclined plane is at rest a m F and F = ma  a = F/m mg Fig : 4.22 (2) When a pull is acting at an angle () to the horizontal (upward) F sin R + F sin  = mg R  R = mg – F sin  m F cos  m R a  F cos mg   mg R F R = mg + F sin a  and F cos = ma F cos  m m F cos Fig : 4.25 Normal reaction R = mg cos + mb sin and ma = mg sin  – mb cos   a = g sin – b cos Note :  The condition for the body to be at rest relative to the ID F sin  b = g tan Fig : 4.24 Free body diagram Equation U Condition mg cos +mb sin  inclined plane : a = g sin – b cos = 0 mg Motion of Blocks In Contact mb b E3 Fig : 4.23 (3) When a push is acting at an angle () to the horizontal (downward) a Normal reaction R = mg cos Force along a inclined plane F = mg sin ; ma = mg sin a = g sin (2) When a inclined plane given a horizontal acceleration ‘b’ Since the body lies in an accelerating frame, an inertial force (mb) acts on it in the opposite direction. F and F cos = ma  a Acceleration of Block on Smooth Inclined Plane 60 (1) When a pull is horizontal 185 Force and acceleration a F A F m1 m1 D YG B f F  f  m1a a F m1  m 2 f  m 2a f m2F m1  m 2 f  m1a a F m1  m 2 F  f  m 2a f m1 F m1  m 2 a m2 f m2 a m1 B A m1 m2 a F U f B F m1 m2 m2 F a F C m1 m3 a ST A f f1 m2 f1 f2 a f2 f1 a A B m1 m2 m3 f1 F m2 f2 a f2 a F m1  m 2  m 3 f1  f2  m 2a f1  (m 2  m 3 )F m1  m 2  m 3 f2  m 3 a f2  m3 F m1  m 2  m 3 f1  m1a a F m1  m 2  m 3 f2  f1  m 2a f1  m1 F m1  m 2  m 3 m3 a m1 C F  f1  m1a m3 F 186 Newton's Laws of Motion F  f2  m 3 a f2  (m1  m 2 )F m1  m 2  m 3 Motion of Blocks Connected by Mass Less String Condition Free body diagram Equation Tension and acceleration a A F m2 a T F  T  m 2a F m2 a F T m1 B a A T m1 m2 T m2 a B T1 m1 T2 m2 T1 m2 F m3 m1 T1 a A m1 B T1 m2 T2 T1 T2 m2 m3 U F C a T2 T m2F m1  m 2 a F m1  m 2  m 3 T2  T1  m 2a T1  m1 F m1  m 2  m 3 F  T2  m 3 a T2  (m1  m 2 )F m1  m 2  m 3 F  T1  m1a a F m1  m 2  m 3 T1  T2  m 2a T1  (m 2  m 3 )F m1  m 2  m 3 T2  m 3 a T2  m3 F m1  m 2  m 3 Equation Tension and acceleration F m3 a F T  m 2a T2 a T2 F m1  m 2 T1  m1a T1 D YG A a C m1 F m1  m 2 a U m1 T F  T  m1a ID F F m1  m 2 E3 T m1 a 60 B T  m1a T m1 m3 ST Motion of Connected Block Over A Pulley Condition Free body diagram T1 m1 a m1g T1 T2 a m2 P m2g T1 T1 a m1 A m2 B a T2 m1a  T1  m1 g T1  2m 1 m 2 g m1  m 2 Newton's Law of motion 187 T2  m 2 a  m 2 g  T1 T1 m1 m1a  T1  m1 g a m1g T1 T1 T1 m1 A m2g + T2 m2 B a T2 a C D YG m3 m 2 a  m 2 g  T2  T1 U a m2 ID T3 p T2 60  m  m1  a 2 g  m1  m 2  E3 T2  2T1 m 3 a  m 3 g  T2 T1  2m1 [m 2  m 3 ] g m1  m 2  m 3 T2  2m 1 m 3 g m1  m 2  m 3 T3  4 m1 [m 2  m 3 ] g m1  m 2  m 3 a m3 4 m 1m 2 g m1  m 2 m3 g ST U T3 Condition When pulley have a finite mass M and radius R then tension in two segments of string are different T1 a Equation Tension and acceleration T1 Free body diagram T1 a m1 m1a  m1 g  T1 m1g T2 M a m2 T2 m2g R T2 a T1 m2  R B m1 [(m 2  m 3 )  m1 ]g m1  m 2  m 3 T3  2T1 a A T2 T1 m 2a  T2  m 2 g a m1  m 2 m1  m 2  M 2 M  m1  2m 2   2  T1  g M m1  m 2  2 188 Newton's Laws of Motion Torque  (T1  T2 )R  I (T1  T2 )R  I (T1  T2 )R  1 a MR 2 2 R T1  T2  M  m 2  2m1   2  T2  g M m1  m 2  2 Ma 2 m1a T m1 T  m1a P m1 T T T a m2 B m 2a  m 2 g  T a m2g T a m1 m1a  T  m1 g sin ID m1g sin  P  T T a m1 m2  a T B m 2a  m 2 g  T U A T m 1m 2 g m1  m 2  m  m1 sin  a 2 g  m1  m 2  T a m2 m2 g m1  m 2 E3 m2 a 60 a A a R m1m 2 (1  sin ) g m1  m 2 D YG m2g T a a T T m1 A a m2   m1 m1g sin  a U ST  m1 m1g sin   P m1 a m2 B T m1m 2 (sin  sin  ) g m1  m 2 T Equation Tension and acceleration T a A m 2a  m 2 g sin   T m2g sin Free body diagram T  (m 2 sin   m1 sin ) g m1  m 2 B m2 a a  T Condition T  m1 g sin  m1a m2 m1 g sin  T  m1a a m1 g sin m1  m 2 Newton's Law of motion 189 a1 A a P T m1 T m2 dt 2 B 2 1 d (x 1 ) 2 dt 2 a2  a2 2T m2 a1 2 (a/2) m2 m2g a1  acceleration of block A a 2  acceleration of block B T1 T2 C M a T1 T 2m1m 2 g 4 m1  m 2 m1a  m1 g  T1 a (m1  m 2 ) g [m1  m 2  M ] m 2a  T2  m 2 g T1  m1 (2m 2  M ) g [m1  m 2  M ] T1  T2  Ma T2  m 2 (2m 2  M ) g [m1  m 2  M ] a m2 A m2g ST U B m2g 4 m1  m 2 T2 a m1 m2 a2  m1g T1 T2 a m1 D YG a a  m 2 g  2T 2 ID  d 2 (x 2 ) 2m 2 g 4 m1  m 2 a1  a  U As T  m1a T 2 m 1m 2 g 4 m1  m 2 E3 m1 T 60 T  m 2a Ma M T2 T1 Table 4.3 : Motion of massive string Free body diagram Condition Equation a M F  (M  m)a T1 Tension and acceleration a F M m T1  Ma T1  force applied by the string on the a block m M F m/2 M T2 T1  M F (M  m ) 190 Newton's Laws of Motion m  T2   M   a 2  T2  T2  Tension at mid point of the rope L m m = Mass of string 60 a x m [(L – x)/L] from the end where the force is applied A x A (M/L)x F1  T  F1 a B M = Mass of uniform string B T M F2 F1 D YG T A A a T'  F1  F2 M x  x T  F1  1    F2   L  L F1  F2  Ma U a L = Length of string Mxa L ID F1 E3 a L Lx  T  F  L  Lx  T m a  L  T T = Tension in string at a distance x F2 a  F/m F  ma F F T (2 M  m ) F 2(M  m ) T '  F  Mg M (L  x )g  T L L–x L B T B T T x T B C F U M Mass of segment BC    x  L  x C F Spring Balance and Physical Balance ST (1) Spring balance : When its upper end is fixed with rigid support and body of mass m hung from its lower end. Spring is stretched and the weight of the body can be measured by the reading of spring balance R  W  mg M xg L T F M xg L (2) Physical balance : In physical balance actually we compare the mass of body in both the pans. Here we does not calculate the absolute weight of the body. a b A B O The mechanism of weighing machine is same as that of spring balance. Effect of frame of reference : In inertial frame of reference the reading of spring balance shows the actual weight of the body but in noninertial frame of reference reading of spring balance increases or decreases in accordance with the direction of acceleration T F R X m Fig : 4.26 Fig : 4.27 Y Here X and Y are the mass of the empty pan. (i) Perfect physical balance : Weight of the pan should be equal i.e. X = Y and the needle must in middle of the beam i.e. a = b. Newton's Law of motion 191 Effect of frame of reference : If the physical balance is perfect then there will be no effect of frame of reference (either inertial or non-inertial) on the measurement. It is always errorless. a b A B O According to Newton, time and space are absolute. The velocity of observer has no effect on it. But, according to special theory of relativity Newton’s laws are true, as long as we are dealing with velocities which are small compare to velocity of light. Hence the time and space measured by two observers in relative motion are not same. Some conclusions drawn by the special theory of relativity about mass, time and distance which are as follows : (1) Let the length of a rod at rest with respect to an observer is L0. If the rod moves with velocity v w.r.t. observer and its length is L, then L  L0 1  v 2 / c 2 Y X X  Y and a  b For rotational equilibrium about point ‘O’ Xa  Yb …(i) In this physical balance if a body of weight W is placed in pan X then to balance it we have to put a weight W1 in pan Y. (X  W )a  (Y  W1 )b …(ii) Now if the pans are changed then to balance the body we have to put a weight Now, as v increases L decreases, hence the length will appear shrinking. (2) Let a clock reads T for an observer at rest. If the clock moves with velocity v and clock reads T with respect to observer, then T0 T  v2 1 2 c Hence, the clock in motion will appear slow. 0 (3) Let the mass of a body is m 0 at rest with respect to an observer. Now, the body moves with velocity v with respect to observer and m0 its mass is m, then m  v2 1 2 c m is called the rest mass. Hence, the mass increases with the increases of velocity. ID For rotational equilibrium about point ‘O’ 60 (a) If the beam of physical balance is horizontal (when the pans are empty) but the arms are not equal where, c is the velocity of light. E3 4.28 (ii) False balance : When Fig the: masses of the pan are not equal then balance shows the error in measurement. False balance may be of two types W2 in pan X. For rotational equilibrium about point ‘O’ U 0 (X  W2 )a  (Y  W )b …(iii) D YG From (i), (ii) and (iii) True weight W  W1 W2 Note :  If v  c, i.e., velocity of the body is very small w.r.t. velocity of light, then m  m 0. i.e., in the practice there will be no change in the mass.  If v is comparable to c, then m > m i.e., mass will increase. (b) If the beam of physical balance is not horizontal (when the pans are empty) and the arms are equal i.e. X  Y and a  b In this physical balance if a body of weight W is placed in X Pan then to balance it. We have to put a weight W in Y Pan 1 U For equilibrium X  W  Y  W1 a A …(i) b  If v  c, then m  or m  2 m0  . Hence, the 0 v v2 mass becomes infinite, which is not possible, thus the speed cannot be equal to the velocity of light.  The velocity of particles can be accelerated up to a certain limit. Even in cyclotron the speed of charged particles cannot be increased beyond a certain limit. 1 B ST O X Y Fig : 4.29 Now if pans are changed then to balance the body we have to put a weight 0 m0 W2 in X Pan. For equilibrium X  W2  Y  W From (i) and (ii) True weight W  W1  W2 2 Modification of Newton’s Laws of motion …(ii)     Inertia is proportional to mass of the body. Force cause acceleration. In the absence of the force, a body moves along a straight line path. A system or a body is said to be in equilibrium, when the net force acting on it is zero.     If a number of forces F1 , F2 , F3 ,......... act on the body, then it is     in equilibrium when F1  F2  F3 .........  0 192 Newton's Laws of Motion  A body in equilibrium cannot change the direction of motion.  Four types of forces exist in nature. They are – gravitation (Fg ) , electromagnetic (Fem ) , weak force (Fw ) and nuclear force (Fn ). (Fg ) : (Fw ) : (Fem ) : (Fn ) : : 1 : 10 : 10 : 10 25 36 2l g sin t  1 sin 2h g (iv) If the angle of inclination is changed keeping the height constant then 38  If a body moves along a curved path, then it is certainly acted upon by a force. t1 sin 2  t2 sin 1  For an isolated system (on which no external force acts), the total  Forces in nature always occur in pairs. momentum remains conserved (Law of conservation of momentum).  Newton's first law of the motion defines the force.  The change in momentum of a body depends on the  Absolute units of force remains the same throughout the universe magnitude and direction of the applied force and the period of time over which it is applied i.e. it depends on its impulse.  Newton's second law of motion gives the measure of force i.e. F = ma.  Force is a vector quantity. SI. m  v M ID 5  Gravitational units of force are gf (or gwt) in CGS system and kgf   1 gf = 980 dyne and 1 kgf = 9.8 N  The beam balance compares masses.  The rocket pushes itself forwards by pushing the jet of exhaust gases backwards. D YG  Acceleration of a horse-cart system is velocity of bullet. U (or kgwt) in SI. where H = Horizontal component of reaction; F = force of friction; M = mass of horse; m = mass of cart.  The weight of the body measured by the spring balance in a lift is equal to the apparent weight.  Apparent weight of a freely falling body = ZERO, (state of weightlessness).  Recoil velocity of the gun is V   where m = mass of bullet, M = mass of gun and v = muzzle  1 N = 10 dyne. HF M m momentum. The positive momentum gained by the bullet is equal to negative recoil momentum of the gun and so the total momentum before and after the firing of the gun is zero.   Absolute units of force are dyne in CGS system and newton (N) in a  Guns recoil when fired, because of the law of conservation of E3 while gravitational units of force varies from place to place as they depend upon the value of ‘g’. 60  A single isolated force cannot exist. U  If the person climbs up along the rope with acceleration a, then  Upthrust on the rocket = u  dm. dt where u = velocity of escaping gases relative to rocket and dm  rate dt of consumption of fuel.  Initial thrust on rocket = m(g + a), where a is the acceleration of the rocket.  Upward acceleration of rocket =  u dm . m dt   Impulse, I  F  t  change in momentum tension in the rope will be m(g+a)  Unit of impulse is N-s.  If the person climbs down along the rope with acceleration, then  Action and reaction forces never act on the same body. They act on ST tension in the rope will be m(g – a)  When the person climbs up or down with uniform speed, tension different bodies. If they act on the same body, the resultant force on the body will be zero i.e., the body will be in equilibrium. in the string will be mg.  Action and reaction forces are equal in magnitude but opposite in  A body starting from rest moves along a smooth inclined plane of direction. length l, height h and having angle of inclination . (i) Its acceleration down the plane is g sin . (ii) Its velocity at the bottom of the inclined plane will be 2 gh  2 gl sin. (iii) Time taken to reach the bottom will be  Action and reaction forces act along the line joining the centres of two bodies.  Newton's third law is applicable whether the bodies are at rest or in motion.  The non-inertial character of the earth is evident from the fact that a falling object does not fall straight down but slightly deflects to the