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

AI Thread Summary
A 3 kg steel ball strikes a wall at 10 m/s and bounces off at the same speed and angle, with contact lasting 0.2 seconds. The impulse-momentum principle is applied to calculate the average force exerted by the wall, resulting in a force of -260 N. The confusion arises regarding the signs used in calculating the change in momentum, specifically the initial and final velocities along the x-axis. The initial velocity is positive, while the final velocity after the bounce is negative, justifying the negative sign in the momentum equation. Understanding these sign conventions is crucial for correctly solving the problem.
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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