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Conservation Of Momentum

Home > Study Notes > Conservation Of Momentum
Published in: Physics
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notes on conservation of momentum and collision

Anshul J / Pune

5 years of teaching experience

Qualification: b.tech

Teaches: Olympiad Exam Preparation, Physics, NTSE

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  1. Conservation of Linear Momentum of the System of Particles dV2 dvn = Mäcm Since m n dt dt dt (rn1V1 + m V dt d ext dt where PI, P2, .. pn are the linear momenta of the particles. If = 0, then PI +P2 constant. Hence, in the absence of any external force, the total linear momentum of a system of particles remains constant. Exercise 4: (i) You are marooned on a big frictionless surface with your school bag in you hand. How will you get out of it? (ii) When a ball is thrown up, the magnitude of its momentum first decreases, then increases. Does this violate conservation of momentum principle? Exercise 5 Two men, each of mass m, stand on the edge of a stationary car and jump off with a horizontal velocity u relative to the car, first simultaneously and then one after the other. If friction be negligible, in which case will they impart greater speed to the car? Illustration 7: A man of mass m is standing on a stationary wooden board of mass M kept on smooth ice. The man starts running on the board and acquires a speed u relative to the board. Find the speed of the man relative to a stationary observer. The board is long enough. The system is initially at rest. Hence, its momentum is zero. Since no net external force is acting on the system along x-axis, therefore momentum of the system will remain zero. Considering in x-direction, the momentum of the system (man + board) should be zero. Solution: Illustration 8: Given velocity of the man with respect to board, vmb ui Let velocity of the board, üb = Vbi Hence, velocity of the man, vm x mu M Vbi + -k V — Velocity of the man, vm mu u A gun is mounted on a stationary rail road car. The mass of the car, the gun, the shells and the operator is 50m, where m is the mass of one shell. Two shells are fired one after the other along same horizontal line in same direction. If the muzzle velocity (velocity with respect to gun) of the shells is 200 m/s, then find the speed of the car after second shot.
  2. Solution: Collision Let Ü be the muzzle velocity of the shell and be the velocity of the car after first shot. As Pi - = 49mv + + v) 50mv + mu = 0 u 50 (1) [initial momentum of the system] (2) [final momentum of the system] (3) Negative sign shows that ü and v are oppositely directed. For the second shot: Momentum of the system before second shot is p'i = 49 mü (4) If v be the velocity of the car after second shot, then p; = 48mv'+ m(u + v From COM, we get 49V'+u = 49v (5) 49 49V' = 49V -Cl = -—u - 50 1 1 50 49 u as u 50 Collision is said to occur if two bodies interact strongly for very short interval of time. Types of Collision 1. 2. Direct collision (Head-on collision) When the velocities at the point of contact coincide with the common normal (line of impact). Oblique collision When the velocities at the point of contact do not coincide with the line of impact. Examples of Direct Collision Examples of Oblique Collision O Rod O Ball Line of impact all —Line. of impact o o Ball Rod Line of impact Line of impact Newton's Experimental Law Coefficient of restitution,
  3. relative velocity of points of contact along the common normal after impact relative velocity of points of contact along the common normal before impact On the basis of coefficient of restitution, collision is divided into three types. If e = 1, collision is called elastic if in this collision mechanical energy of the whole system is conserved. If 0 < e < 1, collision is called inelastic. If e = 0, collision is called perfectly inelastic. Exercise 6 (i) Is it possible to have a collision in which all the kinetic energy is lost ? If so, cite an example. (ii) Two objects collide and one is initially at rest. (a) Is it possible for both to be at rest after collision? (b) Is it possible for one of them to be at rest after collision? Explain. Velocities of colliding bodies after collision In the following discussion, we will assume all the movements and velocities along a straight line. Let there be two bodies with masses ml and moving with velocities I-Il and 112. They collide at an instant and acquire velocities VI and after collision. For sake of feasibility we assume that ul > 112. Let the coefficient of restitution of the colliding bodies be 'e'. Then, applying Newton's experimental law and the law of conservation of momentum, we can find the value of velocities VI and '.12. ml 1.12 Before collision ml After collision Conserving momentum of the colliding bodies before and after the collision, milli + = ml\/l + m2'v/2 Applying Newton's experimental law, we have -ul V2 = VI —e (1.12 — 111) Putting (ii) in (i), we obtain milli + = + {VI —e (1.12 — ul)} VI u (ml -em2) 1 ml + m2 ml + rn2 From (ii), V2 = VI —e (1.12 — Lil) (m2 —mie) + 1.12 .. .. (ii) ... (iii) (iv) ml + rn2 ml + m2 When the collision is elastic; e = 1 ml —m 2 Finally, VI ml +m and 2 2m1 ml + m2 2m2 ml +m 2 m2 —ml ml + rn2 (i) If ml = m2, then VI = 1.12 and = ul When the two bodies of equal mass collide head on elastically, their velocities are mutually exchanged. (ii) If ml = nu and = O, then (iii) If target particle is massive (m2 ml) and = O, then I-Il and = 0
  4. The light particle recoils with same speed while the heavy target remains practically at rest. If then VI -I-Il + 2112 and (iv) If particle is massive (ml C] m2), then VI = ul and = 2111 - If the target is initially at rest (112 = 0), then VI = ul and = 2111 The motion of heavy particle is unaffected while the light target moves apart at a speed twice that of the particle. (v) When collision is perfectly inelastic, e = O ml VI = V2 = ml + m2 ml +1712 Lilml +1.12m2 ml + m2
  5. Exercise 7 A ball of mass m moving with velocity v strikes head on elastically with a number of balls of the same mass in a line. Only one ball from other side moves with the same velocity. Explain why not two balls move simultaneously each with velocity v/2. Impulse If a very large force acts on a body for a very short interval of time, then product of the force and the time is known as impulse. Impulse is a vector quantity. —dt = Ai Hence, impulse is equal to the change in linear momentum > . Exercise 8: A heavy book is suspended with a length of thread. Another piece of same thread is tied to the lower end of the book as shown in the figure. (a) The lower thread is pulled gradually, harder and harder in downward direction, so as to apply force on the upper thread. Which of the threads will break? Explain. (b) In case the lower thread is pulled with a sharp jerk, will the same thread break as in case (a). Explain. ///ustration 9: A pan being connected by a string passing over Solution: a pulley counter balances a block of mass M. A mass m falls on the pan at rest from a height h above it and sticks to it. What is the speed of the block and the pan soon after the body hits the pan. Find the impulse due to tension in the string (a) connecting the pan and mass M. (b) connecting the pulley and the ceiling. During the impact, the impulses of the external forces are shown on the respective FBD. mgdt Mass m 2 Mgdt Pan Pulley 2 Mgdt Counter weight M Mg dt Mg dt Pulley mass system For each FBD, the impulse momentum theorem says that net Impulse = Change in momentum During the impact, the impulses of the weights (mg dt) and (Mg dt) can be neglected compared to the impulses of the forces due to impact. mg dt, Mg dtD Il, 12, 13
  6. assuming that the net impulse Il = -13=0 Consider the FBD of 'Pan + mass m' system. Momentum before impact = m N[äöfi Momentum after impact = ( m + M)VI For the counter weight, momentum before impact Momentum after impact = MV2t 12 = -O As the string is inextensible, VI = m 2gh From equations (i) and (ii), VI = = Impulsive tension in the string connecting the pan and mass M Impulsive tension in the string connecting the pulley and the ceiling = 13 = 212 = Illustration 10: A disc of mass m is moving with constant speed For any system (here the pulley—mass system), the momentum cannot be conserved by ... (ii) O 1 2 d O' 2 Solution: vo on a smooth horizontal table. Another disc of mass M is placed on the table at rest as shown in the figure. If the collision is elastic, find the velocity of the discs after collision. Both discs lie on the same horizontal plane of the table. d sin ...Line of impact 2 P For mass m: Momentum before impact Momentum after impact For mass M: Momentum before impact = 0 Momentum after impact mvo 9 mvl mv2
  7. mv4 mv3 Net impulse on the system = 0 Momentum of the system is conserved. mvo cos 0 + 0 = mvl + MV3 mvo sin 0 +0 = mv2 + MV4 ... (ii) Individually for mass m and M, the impulse along the tangent = 0 . For mass m, mvo sino = mv2 ... (iii) For mass M, 0 = MV4 ... (iv) As the collision is elastic 1 0 -vo cos0 V3 — VI = VO COS 0 Solving equations (i) through (v) = vo sin 0, 2mvo cos0 It can be verified that, vo cos0 1 = —m(V12 + V 2) + 2 1 2 The kinetic energy is conserved in an elastic impact. Special case: If m = M VI = 0, = vo sin 0 = vo cos 0, w = 0 Note: The impulse along the common tangent during an impact may not be always negligible. For example, when normal reaction is impulsive and the effect of friction is to be considered. Elastic collision in two dimensions A particle of mass ml moving along x-axis with speed vo makes elastic collision with another stationary particle of mass m2. After collision, the two particles move in the directions shown in the figure with speeds ml VI and '.12. Initial momentum of the system is mlV0i After collision, momentum of the system is = (ml\/l cos0 + m2v2 cosß)i +(m1V1 sino Conserving momentum, we get ml'./o = ml\/l COS 0 + m2'.J2 COS and sin 0 - m2'./2 sin p = 0 From conservation of K.E. x m2V2 sin B) j 1 2 —mv o 2 1 1 —mlV12 2 2 Knowing the value of 0 or 13, we can get the values of VI and by solving the above equations.
  8. Exercise 9 A ball of mass ml experiences a perfectly elastic collision with a stationary ball of mass m2. What fraction of the K.E. does the striking ball lose, if (a) it recoils at right angle to the original direction of motion, and (b) the collision is head-on one. Exercise 10: Two particles A and B, with different but unknown masses, collide. A is initially at rest while B has speed v. After collision, B has a speed v/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after collision. Can you determine the speed of A from the information given ? Explain.

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