Centre of Mass & Collisions Impulse and Momentum

CENTRE OF MASS & COLLISIONS INTRODUCTION Until now, we have studied kinematics of particles, Newton’s laws of motion and methods of work and energy. Newton’s laws of motion describe relation between forces acting on a body at an instant and acceleration of the body at that instant. Therefore, it only helps us do analyze what is happening at an instant. The work kinetic r energy theorem is obtained by integrating equation of motion ( F  m ar ) over a path. Therefore, methods of work and energy help us to in exploring change in speed over a position interval. Now, we direct our attention on r r another principle – principle of impulse and momentum. It is obtained when equation of motion ( F  m a ) is integrated with respect to time. Therefore, this principle facilitates us with method to explore what is happening over a time interval. Impulse of a Force Net force applied on a rigid body changes momentum i.e. amount of motion of that body. A net force for a longer duration cause more change in momentum than the same force acting for shorter duration. Therefore duration in which a force acts on a body together with magnitude and direction of the force decide effect of the force on the change in momentum of the body. Linear impulse or simply impulse of a force is defined as integral of the force with respect to time. r If a force F acts on a body, its impulse in a time interval from ti to tf is given by the following equation. r tf r I m p   Fdt ti r If the force is constant, its impulse equals to product of the force vector F and time interval t. r r I m p  F  t  F For one-dimensional force, impulse equals to area between force-time graph and the time axis. In the given figure is shown how a force F along x-axis varies with time t. Impulse of this force in time interval ti to tf equals to area of the shaded portion. tf ti t r r r r r If several forces F1 , F2 , F3 ..... Fn act on a body in a time interval, the total impulse Imp of all these forces equals to impulse of the net force. r tf r tf r tf r I m p   F1 dt   F2 dt  ..............   Fn dt  ti ti Impulse is measured in newton-second. Dimensions of impulse are MLT1 ti r  F tf ti 1 r r  F2  .............  Fn dt  Ex. r Calculate impulse of force F  3 t 2 ˆi   2 t  1  ˆj  2 kˆ N over the time interval from t = 1 s to t = 3 s. Sol. r tf r r 3 3 3 3 I m p   Fdt  Im p   3 t 2 iˆ  2 t  1  ˆj  2 kˆ dt  t 3 iˆ  t 2  t  ˆj  2 tkˆ 1  iˆ t 3 1  ˆj  t 2  t  1  kˆ 2 t 13 ti 1        2 6 iˆ  6 ˆj  4 kˆ N-s Ex. A one-dimensional force F varies with time according to the given graph. Calculate impulse of the force in following time intervals. (a) From t = 0 s to t = 10 s. (b) From t = 10 s to t = 15 s. (c) From t = 0 s to t = 15 s. Sol. F (N) 10 5 10 15 t (s) C 10 15 t (s) For one-dimensional force, impulse equals to area between force-time graph and the time axis. (a) I010 = Area of trapezium OABC = 75 N-s (b) I1015 = – Area of triangle CDE = – 25 N-s (c) I015 = Area of trapezium OABC – Area of triangle CDE = 50 N-s Impulse Momentum Principle :- F (N) A 10 O B 5 E D r Consider body of mass m in translational motion. When it is moving with velocity v , net external force acting on r it is F . Equation of motion as suggested by Newton’s second law can be written in the form r r Fdt  d (m v ) r r If the force acts during time interval from ti to tf and velocity of the body changes from v i to v f , integrating the above equation with time over the interval from ti to tf, we have tf r r r  Fdt  m v f  m v i ti r r Here left hand side of the above equation is impulse Imp of the net force F in time interval from ti to tf, and r r quantities m v i and m v f on the right hand side are linear momenta of the particle at instants ti and tf. If we denote r r them by symbols p i and p f , the above equation can be written as r r r Im p  p f  p i The idea expressed by the above equation is known as impulse momentum principle. It states that change in the momentum of a body in a time interval equals to the impulse of the net force acting on the body during the concerned time interval. For the ease of application to physical situations the above equation is rearranged as r r r p i  Im p  p f This equation states that impulse of a force during a time interval when added to momentum of a body at the beginning of an interval of time we get momentum of the body at the end of the interval concerned. Since impulse and momentum both are vector quantities, the impulse momentum theorem can be expressed by there scalar equations making use of Cartesian components. p 1 x   Im p , x  p 2 x p 1 y   Im p ,y  p 2 y p 1 z   Im p , z  p 2 z CENTRE OF MASS & COLLISIONS The impulse momentum principle is deduced here for a single body moving relative to an inertial frame, therefore impulses of only physical forces are considered. If we are using a non-inertial reference frame, impulse of corresponding pseudo force must also be considered in addition to impulse of the physical forces. How to apply Impulse Momentum Principle :The impulse momentum principle is deduced here for a single body, therefore it is recommended at present to use it for a single body. To use this principle the following steps should be followed. r r (i) Identify the initial and final positions as position 1 and 2 and show momenta p 1 and p 2 of the body at these instants. (ii) Show impulse of each force acting on the body at an instant between positions 1 and 2. r r r (iii) Use the impulse obtained in step (ii) and momenta obtained in step (i) into equation p i  I m p  p f . r Consider a particle moving with momentum p 1 in beginning. It is acted upon by two forces, whose impulses in a r r r time interval are Im p 1 and Im p 2 . As a result, at the end of the time interval, momentum of the particle becomes p 2 . This physical situation is shown in the following diagram. Such a diagram is known as impulse momentum diagram. I mp 1 p2 p1 Ex. I mp 1 p1 I mp 2 I mp 2 r A particle of mass 2 kg is moving with velocity v o   2 iˆ  3 ˆj  m/s in free space. Find its velocity 3 s after a r constant force F   3 iˆ  4 ˆj  N starts acting on it. r r r r r r p f  p i  Im p  m v f  mv o  F t Sol. Substituting given values, we have r 2 v f  2  2 iˆ  3 ˆj    3 ˆi  4 ˆj   3  13 iˆ  6 ˆj r v f   6.5 ˆi  3 ˆj  m/s Ex. r A particle of mass 2 kg is moving in free space with velocity v o  2 ˆi  3 ˆj  kˆ m/s is acted upon by force   r F  2 iˆ  ˆj  2 kˆ N. Find velocity vector of the particle 3 s after the force starts acting.  Sol.  r r r r r r p f  p i  Im p  m v f  mv o  F t Substituting given values, we have r 2 v f  2 2 ˆi  3 ˆj  kˆ  2 iˆ  ˆj  2 kˆ  3  1 0 iˆ  3 ˆj  4 kˆ  r vf     5 ˆi  1.5 ˆj  2 kˆ  m/s  Ex. Sol. A box of mass m = 2 kg resting on a frictionless horizontal ground is acted upon by a horizontal force F, which varies as shown. Find speed of the particle when the force ceases to act. F 20 N 10 N tf r r r r r r p f  p i  Im p  m v f  m v i  t Fdt 4 s t 2 s i 2 v  2  0  12  2 0  4  v = 20 m/s Ex. Sol. Two boxes A and B of masses m and M interconnected by an ideal rope and ideal pulleys are held at rest as shown. When it is released, box B accelerates downwards. Find velocities of box A and B as function of time t after system has been released. A B We first explore relation between accelerations aA and aB of the boxes A and B, which can be written either by using constrained relation or method of virtual work or by inspection. vA = 2vB v Mv A  0  Tt  m gt zero ...(ii) Tt A Applying impulse momentum principle to box B p 2 y  p 1 y   Im p , y  A A B T vB ...(i) Applying impulse momentum principle to box A p 2 y  p 1 y   Im p , y  T T m v B  0  Mgt  2Tt ...(iii) zero From equations (i), (ii) and (iii), we have x mvA mgt 2Tt B y Mgt y x MaB  M  2m   M  2m  vA  2 gt and v B   gt  M  4 m   M  4 m  Impulsive Motion Sometimes a very large force acts for a very short time interval on a particle and produces finite change in momentum. Such a force is known as impulsive force and the resulting motion as impulsive motion. When a batsman hits a ball by bat, the contact between the ball and the bat lasts for a very small duration t, but the average value of the force F exerted by the bat on the ball is very large, and the resulting impulse Ft is large enough to change momentum of the ball. During an impulsive motion, some other forces of magnitudes very small in comparison to that of an impulsive force may also act. Due to negligible time interval of the impulsive motion, impulse of these forces becomes negligible. These forces are known as non-impulsive forces. Effect of non-impulsive forces during an impulsive motion is so small that they are neglected in analyzing impulsive motion of infinitely small duration. Non-impulsive forces are of finite magnitude and include weight of a body, spring force or any other force of finite magnitude. When duration of the impulsive motion is specified, care has to be taken in neglecting any of the nonimpulsive force. In analyzing motion of the ball for very small contact duration (usually in milliseconds), impulse of the weight of the ball has to be neglected. Unknown reaction forces may be impulsive or non-impulsive; their impulse must therefore be included. CENTRE OF MASS & COLLISIONS Ex. A 100 gm ball moving horizontally with 20 m/s is struck by a bat, as a result it starts moving with a speed of 35 m/s at an angle of 37° above the horizontal in the same vertical plane as shown in the figure. (a) Find the average force exerted by the bat if duration of impact is 0.30 s. (b) Find the average force exerted by the bat if duration of impact is 0.03 s. (c) Find the average force exerted by the bat if duration of impact is 0.003 s. 35 m/s 20 m/s (d) What do you conclude for impulse of weight of the ball as duration of contact decreases? Sol. The impulse momentum diagram of the ball is shown in the figure below. Here F, mg, and Dt represent the average value of the force exerted by the bat, weight of the ball and the time interval. y mgt = 1.0 t Ft p ix= 2 Fxt Fy  t x Pf = 3.5 37 ° p fx = 2.8 p fy = 2.1 Applying principle of impulse and momentum in x- direction, we have p fx  p ix   Im p , x  Fx  4.8 N t 2  Fx t  2.8 ...(i) Applying principle of impulse and momentum in y- direction, we have p fy  p iy   I m p , y  0 .0  Fy t  1 .0 t  2 .1  2 .1  Fy    1.0  N  t  ...(ii) (a) Substituting t = 0.30 s, in equations (i) and (ii), we find r F  16 iˆ  8 ˆj N (b) Substituting t = 0.03 s, in equations (i) and (ii), we find r F  160 iˆ  71 ˆj N (c) Substituting t = 0.003 s, in equations 1 and 2, we find r F  1600 ˆi  701ˆj N (d) It is clear from the above results that as the duration of contact between the ball and the bat decreases, effect of the weight of the ball also decreases as compared with that of the force of the bat and for sufficiently short time interval, it can be neglected. Momentum and Kinetic Energy A moving particle possesses momentum as well as kinetic energy. If a particle of mass m is moving with velocity v, magnitude of its momentum p and its kinetic energy K bear the following relation. K  p2 1  pv 2m 2 Ex. An object is moving so that its kinetic energy is 150 J and the magnitude of its momentum is 30.0 kg-m/s. With what velocity is it traveling? Sol. K  2  1 50 p2 1  10.0 m/s  pv  v  3 0.0 2m 2 Internal and external Forces and System of interacting Particles Bodies applying forces on each other are known as interacting bodies. If we consider them as a system, the forces, which they apply on each other, are known as internal forces and all other forces applied on them by bodies not included in the system are known as external forces. Consider two blocks A and B placed on a frictionless horizontal floor. Their weights W1 and W2 are counterbalanced by normal reactions N1 and N2 on each of them from the floor. Push F by the hand is applied on A. The forces of normal reaction Non-A and Non-B constitute Newton’s third law action-reaction pair, therefore are equal in magnitude and opposite in direction. Among these forces weights W1 and W2 applied by the earth, normal reactions N1 and N2 applied by the ground and the push F applied by the hand are external forces and normal reactions Non-A and Non-B are internal forces. A B W2 W1 F Non -A A N1 Non B- B N2 If the blocks are connected by a spring and the block A is either pushed or pulled, the forces W1, W2, N1 and N2 still remain external forces for the two block system and the forces, which the spring applies on each other are the internal forces. Here force of gravitational interaction between them being negligible has been neglected. We can conceive a general model of two interacting particles. In the figure is shown a system of two particles of masses m1 and m2. r Particle m1 attracts m2 with a force F12 and m2 attracts (or pulls) m1 r r r with a force F21 . These forces F12 and F21 are the internal forces of this two-particle system and are equal in magnitude and opposite in directions. Instead of attraction may repeal each other. Such a system of two particles repealing each other is also shown. In similar way we may conceive a model of a system of n interacting particles having masses m1,m2,...mi....mj.....and mn respectively. The r r forces of interaction Fij and Fji between mi and mj are shown in the figure. Similar to these other particles may also interact with each other. These forces of mutual interaction between the particles are internal forces of the system. Any of the two interacting particles always apply equal and opposite forces on each other. Here fore r r simplicity only the forces Fij and Fji are shown. CENTRE OF MASS & COLLISIONS Principle of Conservation of linear momentum The principle of conservation of linear momentum or simply conservation of momentum for two or more interacting bodies is one of guiding principles of the classical as well as the modern physics. To understand this principle, we first discuss a system of two interacting bodies, and then extend the ideas developed to a system consisting of many interacting bodies. Consider a system of two particles of mass m 1 and m 2. Particle m1 r r attracts m2 with a force F12 and m2 attracts m1 with a force F21 . These forces have equal magnitudes and opposite directions as shown in the figure. If the bodies are let free i.e. without any external force acting on any of them, each of them move and gain momentum equal to the impulse of the force of interaction. Since equal and opposite interaction forces act on both of them for the same time interval, the momenta gained by them are equal in magnitude and opposite in direction resulting no change in total momentum of the system. However, if an external force acts on any one of them or different forces with a nonzero resultant act on both of them, the total momentum of the bodies will certainly change. If the system undergoes an impulsive motion, total momentum will change only under the action of external impulsive force or forces. Internal impulsive forces also exist in pairs of equal and opposite forces and cannot change the total momentum of the system. Non-impulsive forces if act cannot change momentum of the system by appreciable amount. For example, gravity is a nonimpulsive force, therefore in the process of collision between two bodies near the earth the total momentum remains conserved. The total momentum of a system of two interacting bodies remains unchanged under the action of the forces of interaction between them. It can change only if a net impulse of external force is applied. In similar way we may conceive a model of a system of n interacting particles having masses m1, m2, ....mi....mj.....and mn respectively. The r r forces of interaction Fij and Fji between mi and mj are shown in the figure. Since internal forces exist in pairs of equal and opposite forces, in any time interval of concern each of them have a finite impulse but their total impulse is zero. Thus if the system is let free, in any time interval momentum of every individual particle changes but the total momentum of the system remains constant. It can change only if external forces are applied to some or all the particles. Under the action of external forces, the change in total momentum of the system will be equal to the net impulse of all the external forces. Thus, total momentum of a system of particles cannot change under the action of internal forces and if net impulse of the external forces in a time interval is zero, the total momentum of the system in that time interval will remain conserved. r p initial r   p final The above statement is known as the principle of conservation of momentum. It is applicable only when the net impulse of all the external forces acting on a system of particles becomes zero in a finite time interval. It happens in the following conditions. (i) (ii) (iii) When no external force acts on any of the particles or bodies. When resultant of all the external forces acting on all the particles or bodies is zero. In impulsive motion, where time interval is negligibly small, the direction in which no impulsive forces act, total component of momentum in that direction remains conserved. Since force, impulse and momentum are vectors, component of momentum of a system in a particular direction is conserved, if net impulse of all external forces in that direction vanishes. Ex. Two blocks of masses m and M are held against a compressed spring M on a frictionless horizontal floor with the help of a light thread. When the thread is cut, the smaller block leaves the spring with a velocity u relative to the larger block. Find the recoil velocity of the larger block. Sol. m When the thread is cut, the spring pushes both the block, and impart them momentum. The forces applied by the spring on both the block are internal forces of the two-block system. External forces acting on the system are weights and normal reactions on the blocks from the floor. These external forces have zero net resultant of the system. In addition to this fact no external force acts on the system in horizontal direction, therefore, horizontal component of the total momentum of the system remains conserved. v u v Velocities of both the objects relative to the ground (inertial frame) are M m shown in the adjoining figure. Since before the thread is cut system was at rest, its total momentum was zero. Principle of conservation of momentum for the horizontal direction yields n p i 1 horizontal 0 v  Ex. Sol. Mv  m (u  v )  0 mu M m A shell fired vertically up, when reaches its highest point, explodes into three fragments A, B and C of masses mA = 4 kg, mB = 2 kg and mC = 3kg. Immediately after the explosion, A is observed moving with velocity vA = 3 m/s towards north and B with a velocity vB = 4.5 m/s towards east as shown in the figure. Find the velocity vC of the piece C. North 3 m/s A 4.5 m/s B East C Explosion takes negligible duration; therefore, impulse of gravity, which is a finite external force, can be neglected. The pieces fly off acquiring above-mentioned velocities due to internal forces developed due to expanding gases produced during the explosion. The forces applied by the expanding gases are internal forces; hence, momentum of the system of the three pieces remains conserved during the explosion and total momentum before and after the explosion are equal. r Assuming the east as positive x-direction and the north as positive y-direction, the momentum vectors p A and r p B of pieces A and B become r r p A  m A v A ˆj  12 kg-m/s and p B  m B v B ˆi  9 kg-m/s Before the explosion, momentum of the shell was zero, therefore from the principle of conservation of momentum, the total momentum of the fragments also remains zero. r r r r r p A  p B  pC  0  p C  (9 iˆ  12 ˆj ) From the above equation, velocity of the piece C is r p r v C  C  (3 ˆi  4 ˆj ) = 5 m/s, 53° south of west. mC CENTRE OF MASS & COLLISIONS Ex. In free space, three identical particles moving with velocities v o iˆ , 3 v o ˆj and 5 v o kˆ collide successively with each other to form a single particle. Find velocity vector of the particle formed. Sol. Let m be the mass of a single particle before any of the collisions. The mass of particle formed after collisions must be 3m. In free space, no external forces act on any of the particles, their total momentum remains conserved. Applying principle of conservation of momentum, we have r p initial r   p final  r m v o ˆi  3 m v o ˆj  5 m v o kˆ  3 m v r v  13 v o ˆi  3 ˆj  5 kˆ m/s  Ex.  A bullet of mass 50 g moving with velocity 600 m/s hits a block of mass 600 m/s 1.0 kg placed on a rough horizontal ground and comes out of the block with a velocity of 400 m/s. The coefficient of friction between the block and the ground is 0.25. Neglect loss of mass of the block as the bullet pierces through it. Block (a) In spite of the fact that friction acts as an external force, can you apply principle of conservation of momentum during interaction of the bullet with the block? (b) Find velocity of the block immediately after the bullet pierces through it. (c) Find the distance the block will travel before it stops. Sol. (a) There is no net external force in the vertical direction and in the horizontal direction, only external force friction is non-impulsive, therefore momentum of the bullet-block system during their interaction remains conserved. (b) Let us denote velocities of the bullet before it hits the block and immediately after it pierces through the block by vbo and vb, velocity of the block immediately after the bullet pierces through it is vB and masses of the bullet and the block by m and M respectively. These are shown in the adjacent figure. vB vbo m M vb m M Immediately before the bullet hits the block Immediately after the bullet pieces the block Applying principle of conservation of momentum for horizontal component, we have m v bo  m v b  Mv B  vB  m  v bo  v b  M Substituting the given values, we have vB = 10 m/s (c) To calculate distance traveled by the block before it stops, work kinetic energy theorem has to be applied. K 1  12 Mv B2 M vB K 2= 0 Mg Fk = Mg M x N = Mg During sliding of the box on the ground only the force of kinetic friction does work. W 1 2  K 2  K 1   Mgx  0  12 Mv B2 Substituting given values, we have x = 20 m  x v B2 2 g Ex. Ballistic Pendulum : A ballistic pendulum is used to measure speed of bullets. It consists of a wooden block suspended from fixed support. A wooden block of mass M is suspended with the help of two threads to prevent rotation while swinging. A bullet of mass m moving horizontally with velocity vo hits the block and becomes embedded in the block. Receiving momentum from the bullet, the bullet-block system swings to a height h. Find expression for speed of the bullet in terms of given quantities. Sol. vo When the bullet hits the block, in a negligible time interval, it becomes embedded in the block and the bullet-block system starts moving with horizontally. During this process, net force acting on the bullet-block system in vertical direction is zero and no force acts in the horizontal direction. Therefore, momentum of the bullet-block system remains conserved. Immediately after the bullet becomes embedded in the block Before the bullet hits the block vo p Let us denote momentum of the bullet-block system immediately after the bullet becomes embedded in the block by p and apply principle of conservation of momentum to the system for horizontal component of momentum. p = mvo Using equation K = p2 / 2m, we can find kinetic energy K1 of the bullet-block system immediately after the bullet becomes embedded in the block. K1   m v o 2 2 M  m  During swing, only gravity does work on the bullet-block system. Applying work-kinetic energy theorem during swing of the bullet-block system, we have Immediately after the bullet becomes embedded in the block W 1 2  K 2  K 1  K2 = 0 K1 p   M  m  gh  0  Rearranging terms, we have vo  M  m  m 2 gh h  m v o 2 2 M  m  CENTRE OF MASS & COLLISIONS Impact between two bodies Impact or collision is interaction of very small duration between two bodies in which the bodies apply relatively large forces on each other. Interaction forces during an impact are created due to either direct contact or strong repulsive force fields or some connecting links. These forces are so large as compared to other external forces acting on either of the bodies that the effects of later can be neglected. The duration of the interaction is short enough as compared to the time scale of interest as to permit us only to consider the states of motion just before and after the event and not during the impact. Duration of an impact ranges from 1023 s for impacts between elementary particles to millions of years for impacts between galaxies. The impacts we observe in our everyday life like that between two balls last from 103 s to few seconds. Central and Eccentric Impact The common normal at the point of contact between the bodies is known as line of impact. If mass centers of the both the colliding bodies are located on the line of impact, the impact is called central impact and if mass centers of both or any one of the colliding bodies are not on the line of impact, the impact is called eccentric impact. A B Common Normal or Line of Impact Common Tangent B A Common Normal or Line of Impact Common Tangent Eccentric Impact Central Impact Central impact does not produce any rotation in either of the bodies whereas eccentric impact causes the body whose mass center is not on the line of impact to rotate. Therefore, at present we will discuss only central impact and postpone analysis of eccentric impact to cover after studying rotation motion. Head–on (Direct) and Oblique Central Impact If velocities vectors of the colliding bodies are directed along the line of impact, the impact is called a direct or head-on impact; and if velocity vectors of both or of any one of the bodies are not along the line of impact, the impact is called an oblique impact. uA A t B A uB n n Central Impact B uA uB Oblique Impact In this chapter, we discuss only central impact, therefore the term central we usually not use and to these impacts, we call simply head-on and oblique impacts. Furthermore, use of the line of impact and the common tangent is so frequent in analysis of these impacts that we call them simply t-axis and n-axis. TYPES OF COLLISIONS On the basis of kinetic energy On the basis of direction One dimensional collision or head on collision The collision, in which the particles move along the same straight line before and after the collision, is defined as one dimensional collision Two dimensional collision or oblique collision Elastic collision In-elastic collision A collision is said to be In-elastic, if the total kinetic energy does not remains constant A collision is said to be elastic, if the total kinetic energy before and after collision remains the same The collision, in which the particles move along the same planeat different angles before and after the collision, is defined as oblique collision Perfectly in elastic collision The collision, in which particle gets sticked together after the collision, is called perfectly inelastic collision. In this type of inelastic collision, loss of energy is maximum. Head–on (Direct) Central Impact To understand what happens in a head-on impact let us consider two balls A and B of masses mA and mB moving with velocities uA and uB in the same direction as shown. Velocity uA is larger than uB so the ball A hits the ball B. During impact, both the bodies push each other and first they get deformed till the deformation reaches a maximum value and then they tries to regain their original shape due to elastic behaviors of the materials forming the balls. u A B A uB u A B u Deformation Period Instant when impact starts vA A B vB Restitution Period Instant of maximum deformation Instant when impact ends The time interval when deformation takes place is called the deformation period and the time interval in which the ball try to regain their original shapes is called the restitution period. Due to push applied by the balls on each other during period of deformation speed of the ball A decreases and that of the ball B increases and at the end of the deformation period, when the deformation is maximum both the ball move with the same velocity say it is u. Thereafter, the balls will either move together with this velocity or follow the period of restitution. During the period of restitution due to push applied by the balls on each other, speed of the ball A decrease further and that of ball B increase further till they separate from each other. Let us denote velocities of the balls A and B after the impact by vA and vB respectively. Equation of Impulse and Momentum during impact Impulse momentum principle describes motion of ball A during deformation period.  Ddt mAuA mA u mA uA  D dt  m u A  ...(i) Impulse momentum principle describes motion of ball B during deformation period. mB u B  Ddt mB u m B u B   Ddt  mB u ...(ii) CENTRE OF MASS & COLLISIONS Impulse momentum principle describes motion of ball A during restitution period.  Rdt mA u m A vA m A u   R dt  mA vA ...(iii) Impulse momentum principle describes motion of ball B during restitution period. mB u  Rdt mB v B mB u   Rdt  mB vB ...(iv) Conservation of Momentum during impact From equations, (i) and (ii) we have m A uA  m BuB   m A  m B  u ...(v) From equations, (iii) and (iv) we have m A  m B u  m A v A  m Bv B ...(vi) From equations, (v) and (vi) we obtain the following equation. m A v A  m B v B  m A u A  m Bu B ...(vii) The above equation elucidates the principle of conservation of momentum. Coefficient of Restitution Usually the force D applied by the bodies A and B on each other during period differs from the force R applied by the bodies on each other during period of restitution. Therefore, it is not necessary that magnitude of impulse  Ddt of deformation equals to the magnitude of impulse  Rdt restitution. The ratio of magnitudes of impulse of restitution to that of deformation is called the coefficient of restitution and is denoted by e. e  Rdt  Ddt ...(viii) vB  vA uA  uB ...(ix) Now from equations (i), (ii), (iii) and (iv), we have e Coefficient of restitution depend on various factors as elastic properties of materials forming the bodies, velocities of the contact points before impact, state of rotation of the bodies and temperature of the bodies. In general, its value ranges from zero to one but in collision where kinetic energy is generated its value may exceed one. Depending on values of coefficient of restitution, two particular cases are of special interest. Perfectly Plastic or Inelastic Impact For these impacts e = 0, and bodies undergoing impact stick to each other after the impact. Perfectly Elastic Impact For these impacts e = 1. Strategy to solve problems of head-on impact Write momentum conservation equation m A v A  m B v B  m A u A  m Bu B ...(A) Write rearranging terms of equation of coefficient of restitution v B  v A  e u A  u B  Use the above equations A and B. ...(B) Ex. A ball of mass 2 kg moving with speed 5 m/s collides directly with another of mass 3 kg moving in the same direction with speed 4 m/s. The coefficient of restitution is 2/3. Find the velocities after collision. Sol. Denoting the first ball by A and the second ball by B velocities immediately before and after the impact are shown in the figure. uB = 4 uA = 3 Applying principle of conservation of momentum, we have A B vA vB A B Immediately after impact ends Immediately before impact starts m B v B  m A v A  m A u A  m Bu B  3v B  2v A  2  5  3  4 3v B  2v A  2 2 ...(i) Applying equation of coefficient of restitution, we have v B  v A  e u A  u B   v B  v A  23  5  4  3v B  3v A  2 ...(ii) From equation (i) and (ii), we have vA = 4 m/s and vB = 4.67 m/s Ex. Sol. Ans. A block of mass 5 kg moves from left to right with a velocity of 2 m/s and collides with another block of mass 3 kg moving along the same line in the opposite direction with velocity 4 m/s. (a) If the collision is perfectly elastic, determine velocities of both the blocks after their collision. (b) If coefficient of restitution is 0.6, determine velocities of both the blocks after their collision. Denoting the first block by A and the second block by B velocities immediately before and after the impact are shown in the figure. Applying principle of conservation of momentum, we have m B v B  m A v A  m A u A  m B u B  3 v B  5 v A  5  2  3   4  3 v B  5 v A  2 u =2 A A B uB = 4 Immediately before impact starts vA A B v B Immediately after impact ends ...(i) Applying equation of coefficient of restitution, we have v B  v A  e u A  u B   v B  v A  e 2   4  v B  v A  6e (a) ...(ii) For perfectly elastic impact e = 1. Using this value in equation (ii), we have vB – v A = 6 ...(i) Now from equation (i) and (iia), we obtain vA = – 2.5 m/s and vB = 3.5 m/s (b) For value e = 0.6, equation 2 is modified as vB – vA = 3.6 ...(ii) Now from equation (i) and (iib), we obtain vA = – 1.6 m/s and vB = 2.0 m/s Block A reverse back with speed 1.6 m/s and B also move in opposite direction to its original direction with speed 2.0 m/s. CENTRE OF MASS & COLLISIONS Ex. Two identical balls A and B moving with velocities uA and uB in the same direction collide. Coefficient of restitution is e. (a) Deduce expression for velocities of the balls after the collision. (b) If collision is perfectly elastic, what do you observe? Sol. Equation expressing momentum conservation is v A  v B  u A  uB ...(A) Equation of coefficient of restitution is v B  v A  eu A  eu B (a) (b) ...(B) From the above two equations, velocities vA and vB are 1  e  1  e  vA   u  u  2  A  2  B ...(i) 1  e  1  e  vB   u  u  2  A  2  B ...(ii) For perfectly elastic impact e = 1, velocities vA and vB are vA = u B ...(iii) vB = u A ...(iv) Identical bodies exchange their velocities after perfectly elastic impact. Conservation of kinetic energy in perfectly elastic impact For perfectly elastic impact equation for conservation of momentum and coefficient of restitution are m A v A  m B v B  m A u A  m Bu B ...(A) v B  v A  u A  uB ...(B) Rearranging the terms of the above equations, we have m A v A  u A   m B u B  v B  uA  v A  v B  uB Multiplying LHS of both the equations and RHS of both the equations, we have m A v A2  u 2A   m B u B2  v B2  Multiplying by ½ 1 2 and rearranging term of the above equation, we have m A u 2A  12 m B u B2  12 m A v 2A  12 m B v B2 In perfectly elastic impact total kinetic energy of the colliding body before and after the impact are equal. In inelastic impacts, there is always loss of kinetic energy. Oblique Central Impact In oblique central impact, velocity vectors of both or of any one of the bodies are not along the line of impact and mass center of bodies are on the line of impact. Due to impact speeds and direction of motion of both the balls change. In the given figure is shown two balls A and B of masses mA and mB moving with velocities uA and uB collide obliquely. After the collision let they move with velocities vA and vB as shown in the nest figure. uB vA uA A vB B A B Immediately after Impact Immediately before Impact To analyze the impact, we show components of velocities before and after the impact along the common tangent and the line of impact. These components are shown in the following figure. t A uAn uAt t B u Bn uB uA B A vBn vAn n uBt vAt Immediately before Impact n vB vA vBt Immediately before Impact Component along the t-axis If surfaces of the bodies undergoing impact are smooth, they cannot apply any force on each other along the t-axis and component of momentum along the t-axis of each bodies, considered separately, is conserved. Hence, tcomponent of velocities of each of the bodies remains unchanged. vAt = uAt and vBt = uBt ...(A) Component along the n-axis For components of velocities along the n-axis, the impact can be treated same as head-on central impact. The component along the n-axis of the total momentum of the two bodies is conserved ...(B) m B v Bn  m A v An  m B u Bn  m A u A n Concept of coefficient of restitution e is applicable only for the n-component velocities. v Bn  v A n  e u An  u Bn  ...(C) The above four independent equation can be used to analyze oblique central impact of two freely moving bodies. Ex. Sol. A disk sliding with velocity u on a smooth horizontal plane strikes A u another identical disk kept at rest as shown in the figure. If the impact between the disks is perfectly elastic impact, find velocities of the B disks after the impact. (a) We first show velocity components along the t and the n-axis immediately before and after the impact. angle that the line of impact makes with velocity u is 30°. u t An 30º u vAt A vAn vA B t A B n Immediately before Impact vBn n Immediately after Impact CENTRE OF MASS & COLLISIONS Component along t-axis Components of momentum along the t-axis of each disk, considered separately, is conserved. Hence, t-component of velocities of each of the bodies remains unchanged. v At  u At  u and v Bt  u Bt  0 2 ...(i) Component along n-axis The component along the n-axis of the total momentum of the two bodies is conserved m B v Bn  m A v An  m B u Bn  m A u An  m v Bn  m v An  m  0  m v Bn  v An  u 3 2 u 3 2 ...(ii) Concept of coefficient of restitution e is applicable only for the n-component velocities. v Bn  v A n  e u An  u Bn   v Bn  v A n  From equations (ii) and (iii), we have u 3 2 ...(iii) v An  0 and v Bn  u 3 2 ...(iv) From equations (i) and (iv) we can write velocities of both the disks. Ex. A ball collides with a frictionless wall with velocity u as shown in the figure. Coefficient of restitution for the impact is e. (a) Find expression for the velocity of the ball immediately after the impact. (b) If impact is perfectly elastic what do you observe? Sol. (a) Let us consider the ball as the body A and the wall as the body B. Since the wall has infinitely large inertia (mass) as compared to the ball, the state of motion of the wall, remains unaltered during the impact i.e. the wall remain stationary. Now we show velocities of the ball and its t and n-components immediately before and after the impact. For the purpose we have assumed velocity of the ball after the impact v. t vt un n ut  u Immediately before Impact n vn v ' t Immediately after Impact Component along t-axis Components of momentum along the t-axis of the ball is conserved. Hence, t-component of velocities of each of the bodies remains unchanged. v t  u t  u sin  ...(i) Component along n-axis Concept of coefficient of restitution e is applicable only for the n-component velocities. v Bn  v A n  e u An  u Bn   v n  eu n v n  eu cos  ...(ii) From equations (i) and (ii), the t and n-components of velocity of the ball after the impact are v t  u sin  and v n  eu sin  (b) If the impact is perfectly elastic, we have v t  u sin  , v n  u sin  and  ' =  The ball will rebound with the same speed making the same angle with the vertical at which it has collided. In other words, a perfectly elastic collision of a ball with a wall follows the same laws as light follows in reflection at a plane mirror. Oblique Central Impact when one or both the colliding bodies are constrained in motion In oblique collision, we have discussed how to analyze impact of bodies that were free to move before as well as after the impact. Now we will see what happens if one or both the bodies undergoing oblique impact are constrained in motion. Component along the t-axis If surfaces of the bodies undergoing impact are smooth, the t-component of the momentum of the body that is free to move before and after the impact remain conserved. If both the bodies are constrained, the t-component of neither one remains conserved. Momentum Conservation We may find a direction in which no external force acts on both the bodies. The component of total momentum of both the bodies along this direction remains conserved. Coefficient of restitution Concept of coefficient of restitution e is applicable only for the n-component velocities. v Bn  v A n  e u An  u Bn  Ex. A 250 g ball moving horizontally with velocity 10.0 m/s strikes inclined surface of a 720 g smooth wedge as shown in the figure. The wedge is placed at rest on a frictionless horizontal ground. If the coefficient of restitution is 0.8, calculate the velocity of the wedge after the impact. 10 m/s 37 ° CENTRE OF MASS & COLLISIONS Sol. Let us consider the ball as the body A and the wedge as the body B. After the impact, the ball bounces with velocity vA and the wedge advances in horizontal direction with velocity v B. These velocities and their t and n-components are immediately before and after the impact are shown in the following figures. n n vA uAn uAt t vAn t vAt vB 37 ° 53° 37 ° 37 ° vBn Immediately before Impact n Immediately after Impact Component along t-axis The ball is free to move before and after the impact, therefore its t-component of momentum conserved. Hence, tcomponent of velocities of the ball remains unchanged. v A t  u A t  10 cos 3 7   8 m/s ...(i) Momentum Conservation In the horizontal direction, there is no external force on both the bodies. Therefore horizontal component of total momentum of both the bodies remain conserved. m A u A  m A v At cos 37   v An cos 53   m B v B 4 3v   0.25  10  0.25  8   An   0.72 v B  5 5  ...(ii) Coefficient of restitution Concept of coefficient of restitution e is applicable only for the n-component velocities. v Bn  v A n  e u An  u Bn   3v B  v An  0 .8  6  0  5 From equations (i), (ii) and (iii), we obtain vB = 2.0 m/s ...(iii)