1. Lecture series for Conceptual Physics 8 th Ed.

2. Momentum p 82 momentum =mass x velocity p =m x v Impulse p83 impulse =force x time I =F x t Together: F x t = m x  v Impulse = change in momentum

3. Case 1 : Increasing Momentum p85 Increasing the force…hitting it harder. and / or Increasing the time of contact. The force on the golf ball builds up to a huge amount then decreases. We’ll use the average force. F t F tF t

4. Case 2: Decreasing Momentum Over a Long Time p85 Long means gentle, gradual, softly. A large momentumcan be reduced by a small forcein a long time. A large momentum can by reduced by a large force in a short time. Not so gently.

5. m  v = F t Case 3: Decreasing Momentum Over a Short Time

6. A flower pot falling on your head is bad enough. But, if the pot bounces, it’ll really hurt! First, there’s the impulse required to stop the pot. Then, there’s the impulse in the other direction to throw it back. When something bounces, it hits twice as hard. Figure 5-8 the Pelton water wheel.

7. Conservation of Momentum p 88 The momentum before a collision equals the momentum after a collision. BEFORE: p = m v = m x 0 = 0 AFTER: m (-v ) + m v = 0 Momentum is conserved…it is the same before and after.

8. Collisions p 90 p before = p after These are elastic collisions

9. More on Collisions Notice that these box cars stick together. These are inelastic collisions. p before = p after (m x 10 m/s) before = (2m x v) after v after = 5 m/s Before During After

10. More inelastic collisions p91 Before:m v + m (-v) = 0 After:m 0 + m 0 = 0 Before: m v A + m v B = k After:v (m + m) = k Fig 5-12 p91 m v + m (-v) = m 0 + m 0 m v A + m v B = v (m + m)

11. 1 m/s m = 5kg 4 m/s m = 1 kg Mv + m(-v) = (M+m)v (5kg)(1m/s) + (1kg)(-4m/s) = (5kg + 1kg)v v = 1/6 = 0.17 m/s What’s the velocity of the fish after lunch? We’ll do the more complicated collisions on the board. The End