Ch 9 Linear Momentum and Collisions 9.1 Linear Momentum and Its Conservation Linear – traveling along a path p = mv single particle dp/dt = d(mv)/dt = mdv/dt = ma F net =dp/dt If F net =0, p=0          

CT1: Two carts of identical inertial mass are put back-to-back on a track. Cart A has a spring loaded piston; cart B is entirely passive. When the piston is released, it pushes against cart B, and A. A is put in motion but B remains at rest. B. both carts are set into motion, with A gaining more speed than B. C. both carts gain equal speed but in opposite directions. D. both carts are set into motion, with B gaining more speed than A. E. B is put in motion but A remains at rest.

CT2: Two carts are put back-to-back on a track. Cart A has twice the mass of cart B. Cart A has a spring loaded piston; cart B is entirely passive. When the piston is released, it pushes against cart B, and A. A is put in motion but B remains at rest. B. both carts are set into motion, with A gaining more speed than B. C. both carts gain equal speed but in opposite directions. D. both carts are set into motion, with B gaining more speed than A. E. B is put in motion but A remains at rest.

Internal Forces as Third Law Pairs

External forces and Momentum Conservation System of Particles (i = 1 to N) F neti = dp i /dt p tot = p i F neti = F ext + F int = dp i /dt = dp tot /dt Internal forces occur in third law pairs so their sum adds to zero. F ext = dp tot /dt If F ext = 0, then p tot = 0 and (p tot is conserved)              

p tot = p i F ext = dp tot /dt If F ext = 0, then p tot = 0 P9.4 (p.260)      

The Average Force During a Collision

Ch9 Linear Momentum and Collisions 9.2 Impulse and Momentum I = Impulse (a vector) I = Area under F vs. t curve I = F av t = (p tot /t)t = p tot       

P9.9 (p.261)

CT3: A ball is dropped onto the floor below. Which of the following statements is true while the ball falls (neglect air resistance)? Consider the system as just the ball. A. Momentum is conserved for the ball, but energy is not. B. Energy is conserved for the ball, but momentum is not. C. Both energy and momentum are conserved for the ball. D. Neither energy nor momentum is conserved for the ball.

A.abc B.acb C.bca D.bac E.cba F.cab Rank greatest to least Concept Question 4

CT6: Two carts are put back-to-back on a track. Cart A has a spring-loaded piston; cart B, which has twice the inertial mass of cart A, is entirely passive. When the piston is released, it pushes against cart B, and the carts move apart. How do the magnitudes of the final momenta and kinetic energies compare? A. p A > p B ; k A > k B B. p A > p B ; k A = k B C. p A > p B ; k A < k B D. p A = p B ; k A > k B E. p A = p B ; k A = k B F. p A = p B ; k A < k B G. p A k B H. p A < p B ; k A = k B I. p A < p B ; k A < k B

Ch 9 Linear Momentum and Collisions 9.3 Collisions in One Dimension A. Inelastic Collisions Momentum is conserved, but energy isn’t. Totally inelastic collision when objects stick together. P9.15 (p.261) P9.16 (p.261) B. Elastic Collisions Momentum and energy are both conserved. P9.19 (p.262)

CT7: A golf ball is fired at a bowling ball initially at rest and sticks to it. Compared to the bowling ball, the golf ball after the collision has A. more momentum but less kinetic energy. B. more momentum and more kinetic energy. C. less momentum and less kinetic energy. D. less momentum but more kinetic energy. E. none of the above

CT8: A golf ball is fired at a bowling ball initially at rest and bounces back elastically. Compared to the bowling ball, the golf ball after the collision has A. more momentum but less kinetic energy. B. more momentum and more kinetic energy. C. less momentum and less kinetic energy. D. less momentum but more kinetic energy. E. none of the above

Ch 9 Linear Momentum and Collisions 9.4 Collisions in Two Dimensions A. Inelastic Collisions Momentum is conserved, but energy isn’t. Totally inelastic collision when objects stick together. P9.28 (p.262) B. Elastic Collisions Momentum and energy are both conserved.

The Center of Mass of Two Objects

Ch 9 Linear Momentum and Collisions 9.4 Collisions in Two Dimensions For several masses x cm = m i x i /M where M = m i y cm = m i y i /M z cm = m i z i /M r cm = m i r i /M The m i /M are the weighting factors. 

Ch 9 Linear Momentum and Collisions 9.4 Collisions in Two Dimensions v cm = dr cm /dt = (m i dr i /dt)/M = m i v i /M Mv cm = m i v i = p tot a cm = dv cm /dt = (m i dv i /dt)/M = m i a i /M Ma cm = F ext + F int = F ext F ext = Ma cm = dp tot /dt If F ext = 0, then p total = 0 and p total is conserved.                    

Consider a system of two particles in the xy plane: m 1 = 2.00 kg is at the location i + 2j m and has a velocity of 3i +0.5j m/s; m 2 = 3.00 kg is at -4i - 3j m and has velocity 3i – 2j m/s. (a) Plot these particles on a grid or graph paper. Draw their position vectors and show their velocities. (b) Find the position of the center of mass of the system and mark it on the grid. (c) Determine the velocity of the center of mass and also show it on the diagram. (d) What is the total linear momentum of the system?

P9.57 (p.266) Find v 0