What is the Average Force Exerted by the Wall on a Bouncing Ball?

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SUMMARY

The average force exerted by a wall on a bouncing 3 kg steel ball, which strikes at a speed of 10 m/s at a 60° angle, is calculated to be -260 N. The impulse-momentum theorem is applied, where the change in momentum is determined by the velocities before and after the collision. The ball's contact time with the wall is 0.2 seconds, leading to the force calculation using the formula F = Δp/Δt. The negative sign indicates the direction of the force opposing the ball's initial motion.

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roam
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Homework Statement



A 3 kg steel ball strikes a wall with a speed of 10 m/s at an angle of 60° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.2 s, what is the average force exerted by the wall on the ball?

[PLAIN]http://img708.imageshack.us/img708/5262/15128648.gif

The Attempt at a Solution



Here is the solution to this problem:

I= \Delta p = F \Delta t (I is the impulse)

\Delta p_y =m(v_{fy}-v_{iy})=m(vcos60)-mv cos 60=0

\Delta p_x = m(-v sin60-vsin60)=-2mvsin60

=-52

F=\frac{\Delta P}{\Delta t}= \frac{-52}{0.2}=-260

Now, where did they get the first minus sign from in "-v sin60-vsin60" (in the third line)?? Isn't the ball initially moving towards the positive x direction? :confused:
 
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i don't see anything wrong.

vxf = -v sin60 (-ve x)
vxi = v sin60 (ve x)

vxf - vxi = -2vsin60
 
roam said:
Now, where did they get the first minus sign from in "-v sin60-vsin60" (in the third line)?? Isn't the ball initially moving towards the positive x direction?
Yes, the ball is initially moving in the +x direction. But that first minus sign comes from the final velocity along the x-axis. After it bounces off the wall it's moving in the negative direction.
 

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