# physics

A 0.060-kg handball is thrown straight toward a wall with a speed of 9 m/s. It rebounds straight backward at a speed of 6.7 m/s
(a) What impulse is exerted on the wall?
magnitude
0.942 kg·m/s

(b) If the ball is in contact with the wall for 3.0 ms, what average force is exerted on the wall by the ball?
314 N

(c) The rebounding ball is caught by a player who brings it to rest. During the process, her hand moves back 0.40 m. What is the impulse received by the player?
magnitude 0.402 N·s

(d) What average force was exerted on the player by the ball?

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1. For d) you need to use equation: delta p (or impulse)= Average Force * delta t, however, you know neither delta t nor the average force, so you have to find delta t using kinematics.
Vf^2 - Vo^2= 2a*deltax.
0-(6.7 m/2)^2=2*a*.4m
a=-56.1125 m/s^2
delta v= a * delta t
-6.7 m/s= -56.1125 m/s^2 *delta t
delta t= .1194 s
Plug this in the equation at the top:
.402 Ns= Average Force * .1194 s
Average Force= 3.36 N
Hope this helps!

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