Question

3. A small 0.2 kg bouncy ball with a speed of 10 m/s to the left approaches head-on alarge door at rest. The ball bounces straight back with a speed of 8 m/s.a. What is the change in momentum of the ball?b. What is the impulse exerted on the ball?C.What is the impulse exerted on the bat?

Answer 1

m = mass = 0.2 kg

v1 = speed = 10 m/s

Taking left as positive direction:

• p1 = momentum = m*v1 = 0.2 x 10 = 2 kgm/s

Bounce back:

• p2 = m*v2 = 0.2 x -8 = -1.6 kgm/s

Change of momentum = P2 - P 1 = -1.6 - 2 = -3.6 kgm/s (a)

(b) impulse exerted on the ball is equal to the change of momentum in the ball

-3.6 kgm/s

(c) impulse exerted on the bat is the opposite way.

3.6 kgm/s