Posted by **Anonymous** on Tuesday, October 7, 2008 at 5:31pm.

A 1 kg steel ball strikes a wall with a speed of 9.99 m/s at an angle of 55.1 ◦ with the normal to the wall. It bounces off with the same speed and angle.

If the ball is in contact with the wall for 0.292 s, what is the magnitude of the average force exerted on the ball by the wall? Answer in units of N.

- physics -
**bobpursley**, Tuesday, October 7, 2008 at 6:18pm
Ok, what is the change in momentum. Since the momentum parallel to the wall is the same, it does not change. So the momentum perpendicular (along the normal) is changed, it actually reverses, so one knows the change there, twice the initial momentum.

Momentum normal=mvcos55.1

Double that, is the change in momentum. then the Impluse must equal that.

Force*time=change in momentum.

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**carly**, Tuesday, January 20, 2015 at 7:37pm
I can confirm that the answer above is correct.

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**Ben**, Tuesday, December 15, 2015 at 6:46pm
The impulse equals that, and then you divide the impulse by change in time (.292 s) in order to obtain your answer, the average force exerted on the ball.

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