Physics 111: Lecture 14, Pg 1 Physics 111: Lecture 14 Today’s Agenda l Momentum Conservation l Inelastic collisions in one dimension l Inelastic collisions in two dimensions l Explosions l Comment on energy conservation l Ballistic pendulum

Physics 111: Lecture 14, Pg 2 Center of Mass Motion: Review l We have the following law for CM motion: l This has several interesting implications: l It tells us that the CM of an extended object behaves like a simple point mass under the influence of external forces: FA çWe can use it to relate F and A like we are used to doing. F l It tells us that if F EXT = 0, the total momentum of the system does not change. çThe total momentum of a system is conserved if there are no external forces acting.

Physics 111: Lecture 14, Pg 3 (2) Lecture 14, Act 1 Center of Mass Motion l Two pucks of equal mass are being pulled at different points with equal forces. Which experiences the bigger acceleration? F M M AA1AA1 T T AA2AA2 (a) AA (b) AA (c) AA (a) A 1  A 2 (b) A 1  A 2 (c) A 1 = A 2 (1) Pucks

Physics 111: Lecture 14, Pg 4 AF l We have just shown that MA = F EXT l Acceleration depends only on external force, not on where it is applied! AA l Expect that A 1 and A 2 will be the same since F 1 = F 2 = T = F / 2 AA l The answer is (c) A 1 = A 2. l So the final CM velocities should be the same! Lecture 14, Act 1 Solution

Physics 111: Lecture 14, Pg 5 l The final velocity of the CM of each puck is the same! l Notice, however, that the motion of the particles in each of the pucks is different (one is spinning). V V  This one has more kinetic energy (rotation) Lecture 14, Act 1 Solution

Physics 111: Lecture 14, Pg 6 Momentum Conservation l The concept of momentum conservation is one of the most fundamental principles in physics. l This is a component (vector) equation. çWe can apply it to any direction in which there is no external force applied. l You will see that we often have momentum conservation even when energy is not conserved.

Physics 111: Lecture 14, Pg 7 Elastic vs. Inelastic Collisions l A collision is said to be elastic when kinetic energy as well as momentum is conserved before and after the collision. K before = K after çCarts colliding with a spring in between, billiard balls, etc. vvivvi l A collision is said to be inelastic when kinetic energy is not conserved before and after the collision, but momentum is conserved. K before  K after çCar crashes, collisions where objects stick together, etc.

Physics 111: Lecture 14, Pg 8 Inelastic collision in 1-D: Example 1 l A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V : çWhat is the initial speed of the bullet v? çWhat is the initial energy of the system? çWhat is the final energy of the system? çIs energy conserved? v V beforeafter x

Physics 111: Lecture 14, Pg 9 Example 1... Momentum is conserved in the x direction! l Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction. Momentum is conserved in the x direction! çP x, i = P x, f çmv = (M+m)V v V initialfinal x

Physics 111: Lecture 14, Pg 10 Example 1... l Now consider the kinetic energy of the system before and after: l Before: l After: l So Kinetic energy is NOT conserved! Kinetic energy is NOT conserved! (friction stopped the bullet) However, momentum was conserved, and this was useful.

Physics 111: Lecture 14, Pg 11 Inelastic Collision in 1-D: Example 2 M m M + m V v = 0 v v = ? ice (no friction)

Physics 111: Lecture 14, Pg 12 Example 2... Before the collision: Use conservation of momentum to find v after the collision. After the collision: Conservation of momentum: vector equation Air track

Physics 111: Lecture 14, Pg 13 Example 2... l Now consider the K.E. of the system before and after: l Before: l After: l So Kinetic energy is NOT conserved in an inelastic collision!

Physics 111: Lecture 14, Pg 14 Lecture 14, Act 2 Momentum Conservation l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. çWhich box ends up moving faster? (a) (b) (c) (a) Box 1 (b) Box 2 (c) same 1 2

Physics 111: Lecture 14, Pg 15 Lecture 14, Act 2 Momentum Conservation l Since the total external force in the x-direction is zero, momentum is conserved along the x-axis. l In both cases the initial momentum is the same (mv of ball). l In case 1 the ball has negative momentum after the collision, hence the box must have more positive momentum if the total is to be conserved. l The speed of the box in case 1 is biggest! 1 2 x V1V1 V2V2

Physics 111: Lecture 14, Pg 16 Lecture 14, Act 2 Momentum Conservation 1 2 x V1V1 V2V2 mv init = MV 1 - mv fin V 1 = (mv init + mv fin ) / M mv init = (M+m)V 2 V 2 = mv init / (M+m) V 1 numerator is bigger and its denominator is smaller than that of V 2. V 1 > V 2

Physics 111: Lecture 14, Pg 17 Inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2

Physics 111: Lecture 14, Pg 18 Inelastic collision in 2-D... l There are no net external forces acting. çUse momentum conservation for both components. vv1vv1 vv2vv2 V V = (V x,V y ) m1m1 m2m2 m 1 + m 2

Physics 111: Lecture 14, Pg 19 Inelastic collision in 2-D... l So we know all about the motion after the collision! V V = (V x,V y ) VxVx VyVy 

Physics 111: Lecture 14, Pg 20 Inelastic collision in 2-D... l We can see the same thing using vectors: P pp1pp1 pp2pp2 P pp1pp1 pp2pp2 

Physics 111: Lecture 14, Pg 21 Explosion (inelastic un-collision) Before the explosion: M m1m1 m2m2 v1v1 v2v2 After the explosion:

Physics 111: Lecture 14, Pg 22 Explosion... P l No external forces, so P is conserved. P l Initially: P = 0 Pvv l Finally: P = m 1 v 1 + m 2 v 2 = 0 vv m 1 v 1 = - m 2 v 2 M m1m1 m2m2 vv1vv1 vv2vv2 Rocket Bottle

Physics 111: Lecture 14, Pg 23 Lecture 14, Act 3 Center of Mass l A bomb explodes into 3 identical pieces. Which of the following configurations of velocities is possible? (a) (b) (c) (a) 1 (b) 2 (c) both mm v V v m m m v v v m (1)(2)

Physics 111: Lecture 14, Pg 24 Lecture 14, Act 3 Center of Mass mm v v v m (1) P l No external forces, so P must be conserved. P l Initially: P = 0 P final l In explosion (1) there is nothing to balance the upward momentum of the top piece so P final  0. mvmvmvmv mvmvmvmv mvmvmvmv

Physics 111: Lecture 14, Pg 25 Lecture 14, Act 3 Center of Mass P l No external forces, so P must be conserved. l All the momenta cancel out. P final l P final = 0. (2) m m v v v m mvmvmvmv mvmvmvmv mvmvmvmv

Physics 111: Lecture 14, Pg 26 Comment on Energy Conservation l We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. çEnergy is lost: »Heat (bomb) »Bending of metal (crashing cars) l Kinetic energy is not conserved since work is done during the collision! l Momentum along a certain direction is conserved when there are no external forces acting in this direction. çIn general, momentum conservation is easier to satisfy than energy conservation.

Physics 111: Lecture 14, Pg 27 Ballistic Pendulum H LL LL m M l A projectile of mass m moving horizontally with speed v strikes a stationary mass M suspended by strings of length L. Subsequently, m + M rise to a height of H. Given H, what is the initial speed v of the projectile? M + m v V V=0

Physics 111: Lecture 14, Pg 28 Ballistic Pendulum... l Two stage process: 1. m collides with M, inelastically. Both M and m then move together with a velocity V (before having risen significantly). 2. M and m rise a height H, conserving K+U energy E. (no non-conservative forces acting after collision)

Physics 111: Lecture 14, Pg 29 Ballistic Pendulum... l Stage 1: Momentum is conserved in x-direction: l Stage 2: K+U Energy is conserved Eliminating V gives:

Physics 111: Lecture 14, Pg 30 Ballistic Pendulum Demo H LL LL m M l In the demo we measure forward displacement d, not H: M + m v d L d H L-H

Physics 111: Lecture 14, Pg 31 Ballistic Pendulum Demo... L d H L-H for for d << L Let’s see who can throw fast... Ballistic pendulum

Physics 111: Lecture 14, Pg 32 Recap of today’s lecture l Inelastic collisions in one dimension (Text: 8-6) l Inelastic collisions in two dimensions (Text: 8-6) l Explosions l Comment on energy conservation l Ballistic pendulum (Ex. 8-14) l Look at textbook problems Chapter 8: # 31, 57, 59, 73, 77