1. 1. A train car of mass 4.00 x 10 3 kg is moving at +6.0 m/s. It collides with a stationary car of mass 6.00 x 10 3 kg. The cars couple together. Find the speed of the cars after the collision. 4000 kg6000 kg v 1b = + 6.0 m/s v a = ? Before After p before = p after The momentum before the collision is contained only in Car 1, since Car 2 is stationary So p before = m 1 v 1 = ( 4000 kg )( + 6.0 m/s )= + 24 000 kg m/s

2. 4000 kg6000 kg v 1b = + 6.0 m/s v a = ? Before After p before = + 24 000 kg m/sM = 4000 kg + 6000 kg = 10 000 kg p before = p after, so p after = + 24 000 kg m/s The momentum after the collision is contained in the combined Cars 1 and 2 (they stick together) p after = M v a = ( 10 000 kg ) v a So ( 10 000 kg ) v a = + 24 000 kg m/s 10 000 kg v a = + 2.4 m/s

3. m 1 v 1b = M v a MM v a = m 1 v 1b M = v a = + 2.4 m/s= ( 4000 kg )( + 6.0 m/s ) 10 000 kg 4000 kg6000 kg v 1b = + 6.0 m/s v a = ? Before After M = 4000 kg + 6000 kg = 10 000 kg Another way to do problem: p before = p after m 1 v 1 + m 2 v 2 = M v a m 1 v 1 + 0 = M v a v 2 = 0

4. 2. A train car of mass 2.50 x 10 3 kg moving at +4.0 m/s collides with a stationary car of mass 5.00 x 10 3 kg. The cars couple together. Find the speed of the cars after the collision. 2500 kg5000 kg v 1b = + 4.0 m/s v a = ? Before After p before = p after The momentum before the collision is contained only in Car 1, since Car 2 is stationary So p before = m 1 v 1 = ( 2500 kg )( + 4.0 m/s )= + 10 000 kg m/s

5. 2500 kg5000 kg v 1b = + 4.0 m/s v a = ? Before After p before = + 10 000 kg m/sM = 2500 kg + 5000 kg = 7500 kg p before = p after, so p after = + 10 000 kg m/s The momentum after the collision is contained in the combined Cars 1 and 2 (they stick together) p after = M v a = ( 7500 kg ) v a So ( 7500 kg ) v a = + 10 000 kg m/s 7500 kg v a = + 1.3 m/s

6. 2500 kg5000 kg v 1b = + 4.0 m/s v a = ? Before After M = 2500 kg + 5000 kg = 7500 kg m 1 v 1b = M v a MM v a = m 1 v 1b M = v a = + 1.3 m/s= ( 2500 kg )( + 4.0 m/s ) 7500 kg p before = p after m 1 v 1 + m 2 v 2 = M v a m 1 v 1 + 0 = M v a v 2 = 0 OR:

7. 3. A train car of mass 3.00 x 10 3 kg moving at 3.6 m/s collides with a stationary car of mass M. The cars couple together and move at a velocity of +1.2 m/s. Find M. 3000 kgM v 1b = + 3.6 m/s v a = + 1.2 m/s Before After M + 3000 kg p before = p after m 1 v 1 + M v 2 = ( M + 3000 kg ) v a m 1 v 1 + 0 = ( M + 3000 kg ) v a v 2 = 0 m 1 v 1 = ( M + 3000 kg ) v a vava vava

8. 3000 kgM v 1b = + 3.6 m/s v a = + 1.2 m/s Before After M + 3000 kg m 1 v 1 = ( M + 3000 kg ) v a vava vava M + 3000 kg = m 1 v 1 vava = ( 3000 kg )( + 3.6 m/s ) + 1.2 m/s M + 3000 kg = 9000 kg - 3000 kg M = 6000 kg

9. 4. A team of police officers is investigating the scene of a fatal car crash. A car of mass 1.50 x 10 3 kg drifted off the highway and struck a car of mass 2.00 x 10 3 kg parked on the side of the road. Apparently the driver fell asleep. The cars stuck together and traveled a distance of 4.5 m before stopping. Assuming a coefficient of friction of 1.4 between the tires and road, estimate the speed of impact. 1500 kg v 1b = ? 4.5 m BeforeAfter 2000 kg 3500 kg μ = 1.4 v a = ? We can solve this by using p before = p after, if we knew the velocity v a of the cars after impact To find v a, recall problems we did earlier, in the Forces Unit that involve friction Friction accelerates the cars (negatively) to a stop; so the final velocity = 0; let’s try to find the initial velocity (the velocity just after they collide)

10. 1500 kg v 1b = ? 4.5 m BeforeAfter 2000 kg 3500 kg μ = 1.4 v a = ? v f = 0 d = 4.5 m Need one more known v i = ? ; use friction to find the acceleration F f = μ F N = μ Wt. = μ m g = ( 1.4 )( 3500 kg )( 9.8 m/s 2 ) = 48 020 N So F = m aa = F m F f m a = 48 020 N 3500 kg = a = 13.72 m/s 2 a = - 13.72 m/s 2 negative, since cars slow down

11. 1500 kg v 1b = ? 4.5 m BeforeAfter 2000 kg 3500 kg μ = 1.4 v a = ? v f = 0 d = 4.5 m v i = ?a = - 13.72 m/s 2 Then we can use v f 2 = v i 2 + 2 a d 0 = v i 2 + 2 a d - v i 2 - v i 2 = 2 a d= 2 ( - 13.72 m/s 2 )( 4.5 m ) = - 123.48 m 2 /s 2 vi2vi2 = 123.48 m 2 /s 2 v i = 123.48 m 2 /s 2 = v i = 11.1 m/s This is the initial velocity of the cars as they begin to slow down (immediately after they collide) ; so v a = 11.1 m/s

12. 1500 kg v 1b = ? 4.5 m BeforeAfter 2000 kg 3500 kg μ = 1.4 v a = 11.1 m/s Incidentally, we could also get this result (11.1 m/s) by using the Work-Energy Theorem Friction does work on the cars after impact, taking away all of their Kinetic Energy So set the work done by friction, ( F f d ) equal to the KE ( ½ m v a 2 ) and solve for v a This may actually be an easier way to find v a than by using the motion equations ( Try it and find out )

13. 1500 kg v 1b = ? 4.5 m BeforeAfter 2000 kg 3500 kg μ = 1.4 v a = 11.1 m/s m 1 v 1b = M v a v 1b = 26 m/s ( 1500 kg ) v 1b = ( 3500 kg )( 11.1 m/s ) 1500 kg p before = p after m 1 v 1 + m 2 v 2 = M v a m 1 v 1 + 0 = M v a v 2 = 0 Then, to find v 1b :

14. 5. A ball of mass 3.0 kg is moving with a velocity of +1.5 m/s towards a stationary ball of mass 2.0 kg. After colliding elastically, the 2-kg ball moves with a velocity of +1.8 m/s. Find the velocity of the 3-kg ball after the collision. v 1b = +1.5 m/s v 2a = + 1.8 m/sv 1a = ? Before After m 1 = 3.0 kgm 2 = 2.0 kg p before = p after The momentum before the collision is contained only in Ball 1, since Ball 2 is stationary So p before = m 1 v 1b = ( 3.0 kg )( + 1.5 m/s ) = + 4.5 kg m/s p before = p after, so p after = + 4.5 kg m/s Both balls move after the collision, so p after is contained in both

15. v 1b = +1.5 m/s v 2a = + 1.8 m/sv 1a = ? Before After m 1 = 3.0 kgm 2 = 2.0 kg p before = + 4.5 kg m/s = p after p after = m 1 v 1a + m 2 v 2a + 4.5 kg m/s = ( 3.0 kg ) v 1a + ( 2.0 kg )( + 1.8 m/s ) + 4.5 kg m/s = ( 3.0 kg ) v 1a + ( 3.6 kg m/s ) - 3.6 kg m/s + 0.9 kg m/s = ( 3.0 kg ) v 1a 3.0 kg v 1a = + 0.30 m/s

16. 6. A bowling ball of mass 7.0 kg moving at +3.5 collides elastically with a bowling pin of mass 2.6 kg. After the elastic collision, the ball moves with a velocity of +1.6 m/s. Find the velocity of the pin after the collision. Before After m 1 = 7.0 kg m 2 = 2.6 kg v 2a = ? v 1b = + 3.5 m/s v 1a = + 1.6 m/s p before = p after The momentum before the collision is contained only in the bowling ball, since the pin is stationary So p before = m 1 v 1b = ( 7.0 kg )( + 3.5 m/s ) = + 24.5 kg m/s p before = p after, so p after = + 24.5 kg m/s Both objects move after the collision, so p after is contained in both

17. Before After m 1 = 7.0 kg m 2 = 2.6 kg v 2a = ? v 1b = + 3.5 m/s v 1a = + 1.6 m/s p before = + 24.5 kg m/s = p after p after = m 1 v 1a + m 2 v 2a + 24.5 kg m/s = ( 7.0 kg )( + 1.6 m/s ) + ( 2.6 kg ) v 2a - 11.2 kg m/s + 13.3 kg m/s = ( 2.6 kg ) v 2a 2.6 kg v 2a = + 5.1 m/s + 24.5 kg m/s = 11.2 kg m/s + ( 2.6 kg ) v 2a

18. 7. A steel ball of mass 2.00 kg and moving at +16.0 m/s collides elastically with a steel ball of mass 3.00 kg moving in the opposite direction with a speed of 20.0 m/s (a velocity of -20 m/s). Find the velocity of each ball after the collision.