1. Collisions

2. Review Momentum is a quantity of motion. p = mv A change in momentum is called impulse. Impulse =  p = m  v Impulses are caused by forces applied over time.  p = Ft

3. Collisions Collision – an isolated event in which 2 or more bodies exert strong forces over short periods of time against each other. Momentum is conserved in all collisions (assuming no outside force). “sticky” collision – objects stick together after colliding and move as one. “non-sticky” collision – objects bounce apart after colliding.

4. Collisions A train car with a mass of 3.00x10 4 kg traveling north at 1.5 m/s collides and couples with a 3.20x10 4 -kg train car going south at 0.80 m/s. What is the velocity of the coupled cars after the collision?

5. Collisions Sketching the situation can help. North 1.5 m/s -0.80 m/s mass = 3.20x10 4 kgmass = 3.00x10 4 kg p 1 = m 1 v 1 p 2 = m 2 v 2 p 1 = (3.00x10 4 kg)(1.5 m/s)p 2 = (3.20x10 4 kg)(-0.80 m/s) p 1 = 4.5x10 4 kg*m/sp 2 = -2.6x10 4 kg*m/s p total = (4.5x10 4 kg*m/s) + (-2.6x10 4 kg*m/s) = 1.9x10 4 kg*m/s

6. Collisions North m total = 6.20x10 4 kg p total = 1.9x10 4 kg*m/s p = mv v = p/m v = (1.9x10 4 kg*m/s) / (6.20x10 4 kg) = 0.31 m/s north

7. Collisions A 20.0-kg ball is moving north at 4.00 m/s toward a 5.00-kg ball that is stationary. After they collide, the 20.0-kg ball is moving at 2.40 m/s north. What is the 5.00-kg ball’s velocity after the collision? Neglect friction.

8. Collisions Before Collision m 2 = 5.00 kg North m 1 = 20.0 kg 4.00 m/s p 1 = m 1 v 1 p 2 = m 2 v 2 p 1 = (20.0 kg)(4.00 m/s)p 2 = (5.00 kg)(0 m/s) p 1 = 80.0 kg*m/sp 2 = 0 kg*m/s p total = (80.0 kg*m/s) + (0 kg*m/s) = 80.0 kg*m/s

9. Collisions After Collision m 2 = 5.00 kg North m 1 = 20.0 kg 2.40 m/s ??? p total = 80.0 kg*m/s p 1 = (20.0 kg)(2.40 m/s) p 1 = 48.0 kg*m/s p 2 = (80.0 kg*m/s) – (48.0 kg*m/s) = 32.0 kg*m/s v 2 = p 2 / m 2 v 2 = (32.0 kg*m/s) / (5.00 kg) = 6.40 m/s

10. Collisions A 0.0085-kg bullet strikes a stationary wooden block with a mass of 40.0 kg on a frictionless surface. After the impact, the block-bullet combo is moving at 0.20 m/s. What was the bullet’s original velocity?

11. Collisions Before Collision 40.0 kg m 1 = 0.0085 kg p 1 = m 1 v 1 Don’t know! p 2 = m 2 v 2 p 2 = (40.0 kg)(0 m/s) p 2 = 0 kg*m/s ??? p total = Don’t know yet!

12. Collisions After Collision m total = 40.0085 kg = 40.0 kg 0.20 m/s p total = m total v = (40.0 kg)(0.20 m/s) = 8.0 kg*m/s

13. Collisions Before Collision 40.0 kg m 1 = 0.0085 kg p total = 8.0 kg*m/s ??? v 1 = p 1 / m 1 v 1 = (8.0 kg*m/s) / (0.0085 kg) v 1 = 940 m/s

14. Recoil (Collision in Reverse) Whenever one object pushes another object away, momentum is still conserved. Think of it like a collision in reverse.

15. Recoil A 60.-kg student on ice skates stands at rest on a frictionless frozen pond holding a 10.-kg brick. He throws the brick east with a speed of 18 m/s. What is the resulting velocity of the student?

16. Recoil Before Throwing m 1 = 60. kgm 2 = 10. kg p total = 0 kg*m/s

17. Recoil p total = 0 kg*m/s After Throwing m 1 = 60. kgm 2 = 10. kg 18 m/s??? p 2 = m 2 v 2 p 2 = (10. kg)(18 m/s) p 2 = 180 kg*m/s p 1 + p 2 = p total p 1 + (180 kg*m/s) = 0 kg*m/s p 1 = -180 kg*m/s v 1 = p 1 / m 1 = (-180 kg*m/s) / (60. kg) v 1 = -3.0 m/s