# CONSERVATION OF LINEAR MOMENTUM(COLM)-01- Theory

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### CONSERVATION OF LINEAR MOMENTUM(COLM)-01- Theory

• 1. CONSERVATION OF LINEAR MOMENTUM CENTRE OF MASS In translational motion each point on a body undergoes the same displacement as any other point as time goes on. In this way the motion of one particle represents the motion of the whole body. POSITION OF CENTRE OF MASS (a) System of two particles Consider first a system of two particles m1 and m2 at distances x1 and x2 respectively, from some origin O. We define a point C, the centre of mass the system, as a distance xcm from the origin O, where xcm is defined by xcm = 2 1 2 2 1 1 m m x m x m   ...(1) xcm can be regarded as mass -weighted mean of x1 and x2 . (b) System of many particles (i) If m1 , m2 . . . . . , mn , are along a straight line, by definition, xcm =          i i i n n n m x m m m m x m x m x m ..... . ....... 2 1 2 2 1 1 = M x m i i  ...(2) where M is total mass of the system. (ii) If particles do not lie in a straight line but lie in a plane, (suppose x-y plane) the centre of mass C is defined and located by the coordinates xcm and ycm , where xcm =          i i i n n n m x m m m m x m x m x m ..... ...... 2 1 2 2 1 1 = M x m i i  ycm =          i i i n n n m y m m m m y m y m y m ..... ...... 2 1 2 2 1 1 = M y m i i  ...(3) X m1 m2 x2 x1 Y O xn mn (iii) If the particles are distributed in space, xcm = M x m i i  , ycm = M y m i i  , zcm = M z m i i  ...(4) So, position vector of C is given by xcm m1 m2 x1 x2 X Y O C
• 2. k z j y i x r cm cm cm cm ˆ ˆ ˆ    =   i i i m r m = M r m i i  ...(5) A    O1 O O2 Illustration: A circular plate of uniform thickness has a diameter of 56 cm. Acircular portion of diameter 42 cm is removed from one edge of the plate as shown in figure. Find the centre of mass of the remaining portion. Solution: Let O be the center of circular plate and O1 , the center of circular portion removed from the plate. Let O2 be the center of mass of the remaining part. Area of original plate = R2 =  2 2 56       = 282  cm2 Area removed from circular part =  r2 =  2 2 2 ) 21 ( 2 42 cm         × × × m m1 m2 Let  be the mass per cm2 . Then mass of original plate, m = (28)2  mass of the removed part, m1 = (21)2  mass of remaining part, m2 = (28)2  - (21)2  = 343  Now the masses m1 and m2 may be supposed to be concentrated at O1 and O2 respectively. Their combined center of mass is at O. Taking O as origin we have from definition of center of mass, xcm = 2 1 2 2 1 1 m m x m x m   x1 = OO1 = OA - O1 A = 28 - 21 = 7 cm x2 = OO2 = ?, xcm = 0  0 = ) ( 343 7 ) 21 ( 2 1 2 2 m m x       x2 = cm 9 343 7 441 343 7 ) 21 ( 2          This means that center of mass of the remaining plate is at a distance 9 cm from the center of given circular plate opposite to the removed portion. (c) Continuous bodies: xcm =            dm x M dm dm x m x m Lt i i i mi 1 0 Similarly ycm =     dm y M dm ydm 1 and zcm =     dm z M dm zdm 1
• 3.       dm r M dm dm r rcm 1  . . . . . (6) Illustration: Show that the centre of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length. Solution: By symmetry, we see that yCM = zCM = 0 if the rod is placed along the x axis. Furthermore, if we call the mass per unit length λ (the linear mass density), then λ = M/L for a uniform rod. If we divide the rod into elements of length dx, then the mass of each element is dm = λ dx. Since an arbitrary element of each element is at a distance x from the origin, equation gives xCM = M L dx x M dm x M L L 2 1 1 2 0 0       x O x dx dm Because λ = M/L, this reduces to xCM = 2 2 2 L L M M L        One can also argue that by symmetry, xCM = L/2. Brain Teaser: Can the centre of mass of a body be at a point outside the body. DISTINCTION BETWEEN CENTER OF MASS AND CENTER OF GRAVITY The position of the center of mass of a system depends only upon the mass and position of each constituent particles, i.e., CM i i i m r r m      ...(1) The location of G, the center of gravity of the system, depends however upon the moment of the gravitational force acting on each particle in the system (about any point, the sum of the moments for all the constituent particles is equal to the moment for the whole system concentrated at G). Hence, if gi is the acceleration vector due to gravity of a particle P, the position vector rG of the center of gravity of the system is given by ( ) G i i i i i r m g r m g          ...(2) It is only when the system is in a uniform gravitational field, where the acceleration due to gravity (g) is the same for all particles, that equation (2) Becomes CM i i G i m r r r m        In this case, therefore the center of gravity and the center of mass coincide. If, however the gravitational field is not uniform and gi is not constant then, in general equation (2) cannot be simplified and G CM r r    . Brain Teaser: The weight mg of an extended body is generally shownin a diagramto act through the centre ofmass does it mean that the earthdoes not attract other particles. Y
• 4. MOTION OF THE CENTRE OF MASS Assume that the total mass M of the system remains constant with time, then, for our fixed system of particles , ........ 2 2 1 1 n n cm r m r m r m r M     where rcm is the position vector identifying the centre of mass in a particular reference frame. Differentiating this equation with respect to time, we obtain M dt r d m dt r d m dt r d m dt r d n n cm     ...... 2 2 1 1 ...(7) or M cm V  = n n v m v m v m    ..... 2 2 1 1 where v1 is the velocity of the first particle, etc., and drcm /dt (= vcm ) is the velocity of the centre of mass. Differentiating equation (7) with respect to time, we obtain M dt v d m dt v d m dt v d m dt v d n n cm     ........ 2 2 1 1 ...(8) = n n a m a m a m    .......... 2 2 1 1 , where a1 is the acceleration of the first particle, etc., and dvcm /dt (= acm ) is the acceleration of the centre of mass of the system. Now, from Newton's second law, the force F1 acting on the first particle is given by F1 = m1 a1 . Likewise, F2 = m2 a2 , etc. We can then write equation (8) as M cm a  = 1 2 n F F ....... F F F internal external           ...(9) Internal forces are forces exerted by the particles on each other. However, from Newton's third law, these internal forces will occur in equal and opposite pairs, so that they contribute nothing to the sum.  ext cm F a M  This states that the centre of mass of a system of particles moves as through all the mass of the system were concentrated at the centre of mass and all the external forces were applied at that point. Concept: Whatever maybe the rearrangement ofthe bodies inasystem, due to internalforces (such as one part moving awayfromthe other or aninternalexplosion taking place, breaking a bodyinto pieces). (a) If the body was originally at rest, the C.M. will continue to be at rest. (b) If before the change, the body had been moving with a constant velocity, it will continue to move with a constant velocity and In presence of external force if body had been moving with constant acceleration in a par- ticular trajectory, the C.M. will continue to move in the same trajectory, with the same accelera- tion as if it had never experienced any explosion only if there is no change in external force. Illustration : Two blocks of masses m1 and m2 connected by a weightless spring of stiffness k rest on a smooth horizontal plane. Block 2 is shifted a small distance x to the left and released. Find the velocity of the centre of mass of the system after block 1 breaks off the wall. Solution: We know that the potential energy of compression
• 5. = 2 2 1 kx When the block m1 breaks off from the wall the spring has its unstretched length and the kinetic energy of the block m2 is given by 2 2 2 2 2 1 2 1 kx v m  2 2 2 2 m kx v  m1 m2 k v2 = x 2 m k For center of mass xcm = 2 1 2 2 1 1 m m x m x m   The distances x1 and x2 are measured from the wall. dt dx m m m dt dx m m m dt dxcm 2 2 1 2 1 2 1 1     At start 0 1  dt dx  2 2 1 2 2 2 1 2 m k m m x m v m m m dt dxcm     Velocity of center of mass of system = 2 1 2 m m km x  LINEAR MOMENTUM The momentum of a single particle is a vector P defined as the product of its mass and its velocity v . That is P = mv  ...(10) From Newton's second law of motion dt P d v m dt d dt v d m a m F      ) ( Thus, if m is constant, the rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the direction of that force. Suppose that instead of a single particle we have a system of n particles with masses m1 , m2 . . . etc, and their velocities v1 , v2 etc respectively then the total momentum P in a particular reference frame is,
• 6. P = P 1 + P 2 + . . . . . + P n n n 2 2 1 1 v m ...... v m v m        P = cm V M  Also, ext cm cm F a M dt V d M dt P d        ext F  = dt P d CONSERVATION OF LINEAR MOMENTUM If the sum of the external forces acting on a system is zero. Then, ext F  = dt P d = 0 or P = constant. So, when the resultant external force acting on the system is zero, the total vector momentum of the system remains constant. This is called as principle of the conservation of linear momentum. The momen- tum of the individual particles may change, but their sum remains constant if there is no net external force. Brain Teaser: Ameteorite burnsinthe atmosphere beforeit reaches earth’s surface. What happens to its momentum? Brain Teaser: When a ballis thrown up, the magnitude ofits momentumdecreases and then increases. Does thisviolate the conservationofmomentumprinciple. Illustration: A man of mass m climbs a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed v (relative to rope) in upward direction and with what speed (relative to ground) with the balloon move? Solution: Balloon is stationary  No net external force acts on it.  The conservation of linear momentum of the system (balloon + man) is valid  M 0   m b v m v , where m v  = b mb v v  balloon vm M vb m man  M 0 ] [    b mb b v v m v where vmb = velocity of man relative to the balloon (rope)  m M v m v mb b    where vmb = v  vb = m M mv  and directed opposite to that of the motion of the man.
• 7. Illustration: Two identical buggies move one after the other due to inertia (without friction) with the same velocity v0 . A man of mass m rides the rear buggy. At a certain moment the man jumps into the front buggy with a velocity u relative to his buggy. The mass of each buggy is M. Find the velocities with which the buggies will move afterwards. Solution: Initial momentum of rear buggy = (M + m)v0 . The momentum of man when he jumps = m(v1 + u), where v1 is the velocity of buggy as he jumps. By the conservation of linear momentum (M + m)v0 = Mv1 + m(v1 + u)  v1 (M + m) = (M + m)v0 - mu v1 = v0 - u m M m  v1 M v2 M After jumping Initial momentum of front buggy = Mv0 Mv0 + m(v1 + u) = (M + m)v2 Mv0 + m 0 2 mu v u u (M m)v M m            v0 M v0 M Before jumping  Mv0 + m 2 0 ) ( v m M m M Mu v            (M + m)v0 + 2 ) ( v m M m M mMu     v2 = v0 +  2 m M mMu  Brain Teaser: Where is the energy source ? In order to lift a body above the ground work must be done to increase the body's potential energy. This work comes from different sources. For example, an elevator draws its en- ergy from the main power network, an aircraft takes off utilizing the energy developed when fuel is oxidized (burnt) is its engine. But what is the source of the energy used to raise stratospheric and meteorological sounding balloons having no engines ? SYSTEM OF VARIABLE MASS Till now we have studied system with constant mass. Now, here, we will study about the system which gains or loses mass during its motion, e.g. in case of Rocket propulsion, its motion depends upon the constant ejection of fuel from it. Let us choose system of mass M (as shown in figure 1) whose centre of mass is moving with velocity as seen from a particular reference frame. An external force F  ext acts on the system.
• 8. x y cm M v (1) O t x y cm M- M  v+ v  (2) O t+ t  M u cm At a time  t later the configuration has changed to that shown in figure 2. A mass M has been ejected from the system, its centre of mass moving with velocity as seen by our observer. The mass of the body is reduced to M - M and the velocity of the centre of mass of the system is changed to v +  v. . ext F = dP/dt     ext F P t      { for the finite time interval  t} = f i P P t     [ )( ) ] [ ] ext M M v v Mu Mv F t               = [ ( )] v M M u v v t t             as t  0; v t    approaches dv dt  t M   approaches - dt dM and  v  is negligible as compared to v. .  ( ) ext dv dM M F u v dt dt        (as M is decreasing with time) ext rel dv dM M F u dt dt      = ext thrust F F    where rel u  , is the relative velocity of the ejected mass with respect to main body. . v Illustration: A rocket of initial mass m0 (including shell and fuel) is fired vertically at time t = 0. The fuel is consumed at a constant rate q = dm/dt and is expelled at a constant speed u relative to the rocket. Derive an expression for the magnitude of the velocity of the rocket at time t, neglecting the resistance of the air and variation of acceleration due to gravity. Solution: At time t, the mass of the rocket shell and remaining fuel is m = m0 - qt, and the velocity is v. During the time interval  t, a mass of fuel  m = q  t is expelled with a speed u relative to the rocket. Denoting by ve the absolute velocity of expelled fuel, we apply the principle of impulse and
• 9. we write (m0 – qt)v Wt = (m0 – qt - qt)(v + v) mve [Wt = g(m0 – qt)t] [mve = qt(u – v)] + (m0 - qt)v - g(m0 - qt)  t = (m0 - qt - q  t)(v +  v) - q  t(u - v) Dividing throughout by  t and letting  t approach zero, we obtain -g(m0 - qt) = (m0 - qt) qu dt dv  Separating variables and integrating from t = 0, v = 0 to t = t, v = v dv = dt g qt m du           0 dt g qt m au dv t v              0 0 0  v = [u ln(m0 - qt) - t 0 gt]  v = u ln gt qt m m           0 0 IMPULSE AND MOMENTUM We know that the force is related to momentum as dt P d F   P d dt F  We can find the change in momentum of the body during a collision (from P i to P f ) by integrating over the time of collision and assuming that the force during collision has a constant direction,      f f i t t P P i f dt F P d P P 1 ; F(t) O t ti tf in which the subscripts i (= initial) and f(= final) refer to the times before and after the collision. The integral of a force over the time interval during which the forces acts is called the impulse of the force. The impulse of this force, dt F f i t t  , is represented in magnitude by area under the force - time curve. 0 [W t g(m qt) t]     e [ mv q t(u v)]     0 (m qt q t)(v v)     
• 10. Brain Teaser: Can the impulse ofa force be zero, even ifthe force is not zero? Explain whyor whynot Brain Teaser: A man who falls froma height on a cement floor receives more injurythanwhen he falls fromthe sameheight on a heap ofsand. Explain. COLLISIONS In a collision a relatively large force acts on each colliding particle for a relatively short time. CONSERVATION OF MOMENTUM DURING COLLISIONS m1 m2 1 F 2 F Consider now a collision between two particles, such as those of masses m1 and m2 , shown in figure. During the brief collision these particles exert large forces on one another. At any instant F1 is the force exerted on particle 1 by particle 2 and F2 is the force exerted on particle 2 by particle 1. By Newton's third law these forces at any instant are equal in magnitude but oppositely directed. The change in momentum of particle 1 resulting from the collision is 1 p   = f i t 1 1 t F dt F t      in which 1 F is the average value of the force F1 during the time interval of the collision t = tf - ti . The change in momentum of particles 2 resulting from the collision is 2 p   = f i t 2 2 t F dt F t      in which 2 F is the average value of the force F2 during the time interval of the collision t = tf - ti . If no other forces act on the particles, then p1 and p2 gives the total change in momentum for each particle. But we have seen that at each instant 1 F = - 2 F , and therefore 1 2 p p      If we consider the two particles as an isolated system, the total momentum of the system is 1 2 P p p      , And the total change in momentum of the system as a result of the collision is zero, that is, 1 2 P p p 0           1 2 p p     constant Hence, if there are no external forces acting on the system, the total momentum of the system is not changed by the collision. The impulsive forces acting during the collision are internal forces which have no effect on the total momentum of the system.
• 11. TYPES OF COLLISION: Collision between two bodies may be classified in two ways: 1. Elastic collision and inelastic collision. Collision is said to be elastic if both the bodies come to their original shape and size after the collision, i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies, otherwise, it is called an inelastic collision. Thus in addition to the linear momentum, kinetic energy also remains con- served before and after collision. 2. Head on collision or oblique collision. If the directions of the velocity of colliding objects are along the line of action of the impulses, acting at the instant of collision then it is called as head-on or direct collision. Otherwise the impact is said to be oblique or indirect or eccentric. NEWTON’S LAW OF RESTITUTION: Experimental evidence suggests that the ratio of relative speed of separation to relative speed of approach is constant for two given set of objects. e approach of speed lative separation of speed lative  Re Re The ratio e is called the coefficient of restitution and is constant for two particular objects. In general, 0  e  1 e = 0, for completely inelastic collision, as both the objects stick together. e = 1, for an elastic collision. HEAD-ON COLLISION Consider two spheres A and B of mass m1 and m2 , which are moving in the same straight line and to the right with known velocities v1 and v2 as shown in figure. If v2 is larger than v1 , particle B will eventually strike the sphere A. Under the impact, the two spheres will deform and at the end of the period of deforma- tion, they will have the same velocity u as shown in figure. A period of restitution will then place, at the end of which, depending upon the magnitude of the impact forces and upon the materials involved, the two spheres either will have regained their original shape or will stay permanently deformed. The purpose here is to determine the velocities v’1 and v’2 of the spheres at the end of the period of restitution as shown in figure. (i) For elastic collision Considering first the two spheres as a single system, we note that there is no impulsive, external force. Applying law of conservation of linear momentum (COLM). m1 v1 + m2 v2 = m1 1 v + m2 2 v ...(i) m2 v2 m1 v1 (1) B A m2 v2 m1 v1 (2) B A In an elastic collision kinetic energy before and after collision is also conserved. Hence, 2 2 2 2 1 1 2 2 2 2 1 1 ' 2 1 ' 2 1 2 1 2 1 v m v m v m v m    ...(ii)
• 12. Solving eqs. (i) and (ii) for 1 ' v and 2 ' v , we get 1 ' v = 2 2 1 2 1 2 1 2 1 2 v m m m v m m m m                     ...(iii) and 2 ' v = 1 2 1 1 2 2 1 1 2 2 v m m m v m m m m                     ...(iv) SPECIAL CASES : 1. If m1 = m2 , then from eqs. (iii) and (iv), we can see that v1  = v2 and v2  = v1 i.e., when two particles of equal mass collide elastically and the collision is head on, then they exchange their velocities., e.g. 2. If m1 >> m2 and v1 = 0. Then 0 1 2  m m With these two substitutions We get the following two results: v1   0 and v2   -v2 i.e., the particle of mass m1 remains while the particle of mass m2 bounces back with same speed v2 . 3. If m2 >> m1 and v1 = 0 With the substitution and v1 = 0, we get the results v1   2v2 and v2   v2 i.e., the mass m1 moves with velocity 2v2 while the velocity of mass m2 remains unchanged. (ii) For Inelastic Collision The kinetic energy of particles no longer remains conserved. Suppose the velocities of two particles of mass m1 and m2 before collision are v1 and v2 in the directions shown in figure. Let v1  and v2  be the velocities after collision. Applying the law of COLM. m1 v1 + m2 v2 = m1 v1 + m2 v2 ...(i) Applying Newton's Law of Restitution, separation speed = e(approach speed) v1 - v2 = e(v2 - v1 ) ...(ii) Solving eqs. (i) and (ii), we get v1  = 2 2 1 2 2 1 2 1 2 1 v m m em m v m m em m                      ...(iii) and v2  = 1 2 1 1 1 2 2 1 1 2 v m m em m v m m em m                      ...(iv)
• 13. SPECIAL CASES: 1. If collision is elastic, i.e. e = 1, then v1  = 2 2 1 2 1 2 1 2 1 2 v m m m v m m m m                     and v2  = 1 2 1 1 2 2 1 1 2 2 v m m m v m m m m                     which are same as eqs. (iii) and (iv). 2. If collision is perfectly inelastic, i.e., e = 0, then v1 = v2  = 2 1 2 2 1 1 m m v m v m   = v (say) 3. If m1 = m2 and v1 = 0, then 2 2 2 1 2 1 ' 2 1 ' v e v and v e v                 LINE OF IMPACT Line of impact v2 v2 v1 v1 n B A t It is important to know the line of impact during the collision. The lineof impact is the line along which the impul- sive force act on the bodies. To find it draw the tangent at the point of contact of the two bodies. Draw a normal to the tangent at the point. This normal line is known as line of impact. OBLIQUE COLLISION : Let us now consider the case when the velocities of the two colliding spheres are not directed along the line of impact as shown in figure. As already discussed the impact is said to be oblique. Since velocities v1 and v2 of the particles after impact are unknown in direction and magnitude, their determinaion will require the use of four independent equations. We choose as coordinate axes the n-axis along the line of impact, i.e. along the common normal to the surfaces in contact, and the t-axis along their common tangent. Assuming that the sphere are per- fectly smooth and frictionless, we observe that the only impulses exerted on the sphere during the impact are due to internal forces directed along the line of impact i.e. along the n axis. It follows that m2v2 n B A t m1v1 + B A t n n B A t m1v1 m2v2 = (i) The component along the t axis of the momentum of each particle, considered separately, is con- served; hence the t component of the velocity of each particle remains unchanged. We can write. (v1 )t = (v1 )t ; (v2 )t = (v2 )t
• 14. (ii) The component along the n axis of the total momentum of the two particles is conserved. We write. m1 (v1 )n + m2 (v2 )n = m1 (v1 )n + m2 (v2 )n (iii) The component along the n axis of the relative velocity of the two particles after impact is obtained by multiplying the n component of their relative velocity before impact by the coefficient of restitu- tion. (v2 )n - (v1 )n = e[(v1 )n - (v2 )n ] We have thus obtained four independent equations, which can be solved for the components of the velocities of A and B after impact. Note: Definition of coefficient of restitution can be applied along common normal direction in the case of oblique collisions . v v   Illustration: A ball of mass m hits a floor with a speed v making an angle of incidence θ with normal. The coefficient of resti- tution is e. Find the speed of reflected ball and the angle of reflection. Solution: Suppose the angle of reflection is   and the speed after collision is v. It is an oblique impact. Resolving the ve- locity v along the normal and tangent, the components are v cosθ and v sinθ. Similarly, resolving the velocity after reflection along the normal and along the tangent the com- ponents are - vcos   and vsin   . Since there is no tangential action, v sin  = vsin   ...(i) Applying Newton's law for collision, (- vcos   - 0) = -e(v cos - 0)  vcos   = ev cos ...(ii) From equations (i) and (ii), v2 = v2 sin2  + e2 v2 cos2  v =   2 2 2 2 2 cos sin v e v  v = 2 2 2 (v sin e cos )    and tan   = e  tan   = tan-1       e  tan . d B v A Illustration: A disc A of radius r moving on perfectly smooth surface at a speed v undergoes an elastic collision with an identical stationary disc B. Find the veloc- ity of the disk B after collision if theimpact param- eter is d as shown in figure.
• 15. Solution: One of the discs is at rest before impact. After the impact its velocity will be in the direction of the center line at the moment of contact because this is the direction in which the force acted on it. Thus, sin  2 = r d 2  1 +  2 = 2  Since the masses of both disks are equal, the triangle of momenta turns into triangle of velocities. we have V1 = v cos  1 = v sin  2 1 v1 v2 2 v1 900 v2 v = r vd 2 v2 = v cos  2 = v 2 2 4 1 r d 