1. Aim: How can we explain momentum and impulse? Do Now: Which is easier to do: Stop a skateboard traveling at 5 m/s or stop a car traveling at 5 m/s? Why?

2. Which is easier to do: Stop a bullet fired from a gun or stop a bullet that is thrown at you? Why?

3. Momentum (p) “Mass in motion” Momentum = mass * velocity Or p = mv Vector quantity What about the units? Units are kg·m/s

4. What is the momentum of a 60 kg halfback moving 9 m/s, eastward? p = mv p = (60 kg)(9 m/s) p = 540 kg·m/s eastward

5. If a 1 kg ball bounces off of a wall with the same velocity as shown below, is there a change in momentum? Yes because there has been a change in direction Momentum is a vector quantity v = 10 m/s

6. So how do we define a change in momentum? Δp = p f – p i Δp = mv f – mv i Δp = m(v f – v i ) Δp = mΔv

7. Calculate the change in momentum of the bouncing ball Δp = mΔv Δp = (1 kg)(-10 m/s – 10 m/s) Δp = (1 kg)(-20 m/s) Δp = -20 kg·m/s The negative sign indicates a change in direction

8. Why do you follow through when: Swinging a baseball bat Swinging a tennis racket Swinging a golf club Kicking a football

9. Impulse (J) A force must act on an object for a time in order to change its velocity Impulse (J) = Force * time Or J = Ft Vector quantity What about the units? The units are N·s

10. Calculate the impulse on a baseball being hit by a baseball bat with a force of 1200 N over 0.02 s J = Ft J = (1200 N)(0.02 s) J = 24 N·s

11. Egg Demo Why doesn’t the egg break when it hits the bed sheet?

12. We know: Doesn’t J = Ft? Doesn’t Δp = mΔv? So J = Δp Impulse is a change in momentum!!

13. mΔv is a constant mass has not changed initial and final velocities have not changed Time to slow down increased Therefore force has to decrease Hence, the egg does not break!

15. Real-World Applications Baseball and tennis player’s ‘following through’ Boxer’s ‘riding the punch’ Airbags Padded dashboards

16. What if force and time are constant but the mass changes? Astroblaster Demo If mass decreases, velocity must increase!

17. A car with m=725 kg is moving at 32 m/s to the east. The driver applies the brakes for 2 s. An average force of 5.0 x 10 3 N is exerted on the car. What is the change in momentum? Δp = Ft Δp = (-5 x 10 3 N)(2 s) Δp = -1 x 10 4 N What is the impulse on the car? J = Δp J = -1 x 10 4 N What is the car’s final velocity? Δp = mΔv -1 x 10 4 N = (725 kg)(v f – 32 m/s) -1 x 10 4 = 725v f – 23,200 v f = 18.2 m/s

18. An impulse of 30.0 N·s is applied to a 5.00 kg mass. If the mass had a speed of 100 m/s before the impulse, what is its speed after the impulse? J = mΔv 30 N·s = (5 kg)(v f – 100 m/s) 30 = 5v f – 500 v f = 106 m/s

19. A car with a mass of 1.0 x 10 3 kg is moving with a speed of 1.4 x 10 2 m/s. What is the impulse required to bring the car to rest? J = mΔv J = (1 x 10 3 kg)(0 m/s – 1.4 x 10 2 m/s) J = -1.4 x 10 5 N·s