Simple part of a hard problem involving momentum/kinetic energy conservation

  • #1
"Simple" part of a hard problem involving momentum/kinetic energy conservation

Two balls, of masses mA = 36 g and mB = 64 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60° angle with the vertical and released.
(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)
c) What will be the maximum height of each ball (above the collision point) after the elastic collision?

I found the velocity for ball A which I know to be correct -.48 m/s with the kinetic energy and momentum conservation equations, and also the ball B speed 1.23 which I'm not sure about. The part I'm having problems with is the heights. I've been told to use this equation: mgh=1/2mv^2, but this equation gives me .01 for Ball A which is not correct. What am I doing wrong?
 

Answers and Replies

  • #2
Doc Al
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You need to describe the problem in more detail. Where's the figure, for example?

How did you find the initial speed of mass A just before the collision?
 
  • #3
Here is the image
I hope...
http://www.webassign.net/giancoli/7-44fig.gif
I found that velocity by solving the momentum conservation equation mav=mav1 + mbv2 for v2 and plugging into the kinetic energy conservation equation 1/2mav^2=1/2mav1^2+1/2mbv2^2
This became a quadratic equation which I solved, and then plugged v1 back into the equation for v2 that I found.
 
  • #4
Doc Al
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We must be talking about different things. I asked how you found the speed of mass A just before the collision--which is the answer to part (a). That answer is not found using conservation of momentum or kinetic energy.

My point is that the same principle used to solve (a) can be used to solve (c).
 

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