[engineeringGUY]

A 3.00 kg steel ball stricks a wall with a speed of 10.0 m/s at an angle of 60.0 degree's with the surface. It bounces off with the same speed and angle (fig P9.9). If the ball is in contact with the wall for 0.200 seconds what is the average force exerted by the wall on the ball??


Ball @ 60 degree's O \
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________________________| wall what is the average force exerted
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O ball @ 60 degree's

[urmod4u]

There is a lot of stuff to tell about this.
Assumed are frictionless motions.
However, I have problems with the "avarage".
Is that RMS value, or mean value?
At the beginning of the contact, the ball will push against the wall with zero force, going up to a maximum, and going down zero again. This happens within 200ms.
This force is not a constant force, it isn't even linear with the displacement (it would be linear if it were a spring).
As we know nothing about the elasticy of the ball and the wall, let's assume that the force is constant during the 200 ms. In this particular case, the RMS and mean value are the same.
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The wall is X direction, the normal on it is Y direction. At the moment of impact, the angle of the ball's direction with the wall 's surface is 60 degrees.
The X component of the speed is 10cos(60°)= 10*0.5.
After bouncing, it is unchanged.
The Y component of the speed is 10sin(60°)= 10*sqrt2/2.
After bouncing, it's the opposite direction, or the speed is: 10(-sqrt2/2).
The speed change (dv) is 10(sqrt2/2 - (-sqrt2/2)) = 10sqrt2.
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A speed change of 10sqrt2 during 200ms is an acceleration of 5*10sqrt2 = 50sqrt2. (a=dv/dt)
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An acceleration of 50sqrt2 for a mass of 3 Kg is F=m*a
or 150sqrt2 = 212.132 Newton. Here's your "avarage"...
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If the ball would have bounced under an angle of 60° with the *normal* to the wall (30° with the surface), then the result would have been much nicer.
Then the avarage force would have been 150 Newton.

[drwls]

There are a couple of mistakes in the previous answer. One is in the algebra and one is in the trig.

Just use the rule that the impulse, (average force) x (time), equals the momentum change. In this case, only the momentum component perpendicular to the wall changes. Therefore

2 M V sin 30 = F t = 0.2 F (since t = 0.2 s)
F = 5 M V sqrt 3 = 150 sqrt 3 = 260 N since sin 30 = (sqrt 3)/2

[chimei89]

In regards to previous answer
sin30 does NOT equal (sqrt3)/2.
Sin30=1/2
sin60=sqrt3/2