Posted by Katie on Monday, February 27, 2012 at 1:33pm.
A 220g 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?
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?

physics  Elena, Monday, February 27, 2012 at 5:33pm
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.

physics  Katie, Tuesday, February 28, 2012 at 10:19am
Thanks. the impulse is correct but the magnitude of the average force is not.
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