A 3.42-kg steel ball strikes a massive wall at 10.0 m/s at an angle of

θ = 60.0°
with the plane of the wall. It bounces off the wall with the same speed and angle. If the ball is in contact with the wall for 0.170 s, what is the average force exerted by the wall on the ball?
magnitude N
direction

1 answer

LOL - That same old impulse thing again.

find change of momentum
= final m v - initial mv
(doing only component perpendicular to wall, v = 10 sin 60 inbound
v = - 10 sin 60 outbound
change in v = 2 * 10 sin 60

Force = change in momentum/time
It only acted perpendicular to the wall. We know that because the velocity component parallel to the wall did not change.