Linear momentum and collisions

  • Thread starter roam
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  • #1
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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 [Broken]

The Attempt at a Solution



Here is the solution to this problem:

[tex]I= \Delta p = F \Delta t[/tex] (I is the impulse)

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

[tex]\Delta p_x = m(-v sin60-vsin60)=-2mvsin60[/tex]

[tex]=-52[/tex]

[tex]F=\frac{\Delta P}{\Delta t}= \frac{-52}{0.2}=-260[/tex]

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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Answers and Replies

  • #2
195
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i dont see anything wrong.

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

vxf - vxi = -2vsin60
 
  • #3
Doc Al
Mentor
45,093
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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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