A 220-g ball strikes a wall with a speed of 7.0 m/s and rebounds with only 37% of its kinetic energy. What is the speed of the ball immediately after rebounding?

Question

What is the magnitude of the impulse of the ball on the wall?
If the ball was in contact with the wall for 9.9 ms, what is the magnitude of the average force exerted by the wall on the ball during this time interval?

Answer 1

KE2=0.37KE1
(m(v2)^2)/2=0.37(m(v1)^2)/2.
The speed of the ball immediately after rebounding is
v2=v1(sqroot(0.37)) =0.6•7=4.2 m/s.

The magnitude of the impulse of the ball on the wall:
delta(p) = p2 - p1= mv2 - (- mv1) = mv2 + mv1=
m(v2+ v1) = 0.22•(4.2 + 7) = 2.464 kg•m/s.

The magnitude of the average force
exerted by the wall on the ball during this time interval:
F•delta(t) = delta(p),
F= delta(p)/ delta(t) = 2.464/9.9=0.249 N.

Answer 2

Thanks. the impulse is correct but the magnitude of the average force is not.