m A =6.0 kg v A =5.0 m/s m B =4.0 kg v B =10.0 m/s v - ?

1. Example: An eagle of mass, mA = 6.0 kg moving with speed vA = 5.0 m/s is on collision course with a second eagle of mass mB = 4.0 kg moving at vB = 10.0 m/s in a direction perpendicular to the first. After they collide, they hold onto one another. At what speed they moving after the collision? mA=6.0 kg vA=5.0 m/s mB=4.0 kg vB=10.0 m/s v - ?

2. Example: Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the speed of the cart • Increases • Does not change • Decreases Mass is increasing P = mv must be conserved (Fext = 0) Speed must decrease

3. Ballistic Pendulum A simple device for measuring the speed of projectile 1. Before collision: 2. After collision: 3. At the highest point: p1=mv1 p2=(m+M)v2 p3=0 p1=p2 mv1 = (m+M)v2

4. Example: m1 m2 h1 h2

5. Example: You want to knock down a large bowling pin by throwing a ball at it. You can choose between two balls of equal mass and size. One is made of rubber and bounces back when it hits the pin. The other is made of putty and sticks to the pin. Which ball do you choose? p’ p+p’ putty p p rubber A. The rubber ball. B. The putty ball. C. It makes no difference.

6. Elastic Collisions (1D) The kinetic energy of the system is conserved: after the collision it is the same as that before Before Before After After

7. Elastic Collisions (1D) (1) (2) (2a)

8. Example: A steel ball with mass m1 = 1 kg and initial speed v0 collides head-on with another ball of mass m2 = 2 kg that is initially at rest. What are the final speeds of the balls? 1)Conservation of momentum: 2) Conservation of energy: We have 2 equations and 2 unknown: v1 and v2. It is more convenient to use eq. (2a) instead of eq.(2) 2a) Relative velocity: Velocity of 1 relative to 2 before the collision Velocity of 1 relative to 2 after the collision Adding eq. 1and 2a:

9. Example: Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M>>m) and drop these from some height h. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with small rubber ball? Relative speed is 2v v v v 3v m Here too! V? M v v Remember thatrelative velocityhas to be equal before and after collision! Before the collision, the basketball bounces up withvand the rubber ball is coming down withv, so their relative velocity is–2v. After the collision, it therefore has to be+2v!! This means that, after collision,the velocity of the smaller ball after is 3v.

10. Newton’s craddle (a row of adjacent steel-ball ) If two balls on the left are pulled to a certain height h and released, what happens? • One ball rises on the right, but higher than h pi = pf 2mvi = mvf vf = 2vi , but then kinetic energy would not be conserved: KEi = 2½mvi2 = mvi2 KEf =½mvf2 = 2mvi2 B. Two balls rise on the right at height h pi = pf 2mvi = 2mvf vf = vi. KEi =2½mvi2= mvi2 KEf =2 ½mvf2 = mvi2Ok! C. A ball will rise on each side to height h Same height means |vi|= |vf| (from conservation of energy), but this violates conservation of momentum: pi =2mvi ; pf = mvi – mvi = 0

11. Elastic Collisions in 2D and 3D  In 2D and 3D the initial velocities of the two particles dose not determine the final velocities. You also need to know the direction of one of the final particles. Kinetic energy:

12. 2D Elastic: Nuclear scattering A particle of unknown mass M is initially at rest. A particle of known mass m is “shot” against it with initial momentum pi. After the collision, the momentum of the particle of known mass is measured again, and it is pf. If the collision is elastic, that’s all we need to determine M and the final momentum of the target, P. pf M (at rest) m pi 3 unknowns: M, Px, Py P 3 equations: conservation of momentum in the x direction conservation of momentum in the y direction conservation of kinetic energy