More Conservation of Momentum Contents: Whiteboards:

A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, what speed are the second block and bullet going? (.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg+3.5kg)v 69.156kgm/s = 22.75kgm/s + (3.653kg)v 47.406kgm/s = (3.653kg)v (47.406kgm/s)/(3.653kg) = 12.7 m/s 12.7 m/s

A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, What is the bullet’s velocity between the blocks? (.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s + (.153kg)v 46.406kgm/s = (.153kg)v (46.406kgm/s)/(.153kg) = 303.31 = 303 m/s 303 m/s

A 153 gram bullet going 452 m/s goes through two stationary 3 A 153 gram bullet going 452 m/s goes through two stationary 3.50 kg blocks of wood. After this, the first block is traveling at 6.50 m/s and the second is going 8.30 m/s in the same direction. What is the bullet’s velocity after this? (.153kg)(452m/s)= (3.5kg)(6.5m/s)+(3.5kg)(8.3m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s+29.05kgm/s+(.153kg)v 17.356kgm/s = (.153kg)v (17.356kgm/s)/(.153kg) = 113.44 = 113 m/s 113 m/s

A 153 gram bullet going 452 m/s is heading toward two stationary 3 A 153 gram bullet going 452 m/s is heading toward two stationary 3.50 kg blocks of wood. What is the velocity if the bullet sticks in the first block, and the two blocks then slide and stick together? (.153kg)(452m/s)= (3.5kg+3.5kg+.153kg)v 69.156kgm/s = (7.153kg)v (69.156kgm/s)/(.153kg) = 9.668 = 9.67 m/s 9.67 m/s

A B 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 1. What is the velocity of car A when he is in midair? 0 = (60kg)(+5.2 m/s) + (560 kg)v v = -.557 m/s -.56 m/s

A B 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 2. What is the velocity of car B when he leaves it? (60 kg)(5.2 m/s) + 0 = (60 kg)(3.4 m/s) + (780 kg)v v = +.138 m/s +.14 m/s

A B 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 3. What would have been the velocity of car B had he remained there, and not jumped off? (60 kg)(5.2 m/s) + 0 = (60 kg+ 780 kg)v v = +.371 m/s +.37 m/s

A B 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 4. What would the velocity of car B have been had he jumped off the back of it to give himself a velocity of zero? (60 kg)(5.2 m/s) + 0 = (60 kg)(0 m/s)+ (780 kg)v v = +.40 m/s +.40 m/s