Calculating Average Force on a Bouncing Steel Ball

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To calculate the average force exerted on a 9 kg steel ball bouncing off a wall, the impulse-momentum theorem is applied. The ball strikes the wall at 10 m/s at a 49.8-degree angle and bounces off with the same speed and angle. The change in the x component of momentum is determined, and this value is divided by the contact time of 0.0911 seconds to find the average force. The calculations confirm that understanding momentum and impulse is crucial for solving such problems. The discussion emphasizes the importance of these concepts in physics.
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A 9 kg steel ball strikes a wall with a speed of 10 m/s at an angle of 49.8 degrees 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 .0911 sec, what is the magnitude of the average force exerted on the ball by the wall?

KE = 1/2 m v^2 p = mv

I wasn't sure how to tackle this one, I made a few attempts but to no avail. Can anyone learn me up a bit.
 
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Look up the impulse-momentum theorem.
 
Thanks, that helped alot. I found the change in the x component of the momentum then divided by the time.
 
That sounds good to me. :smile:
 

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