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Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Problem set 9 number 17
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Problem Set 9 # 17 states:
A bullet of mass 46 grams is shot into a 6.7 kg mass, which quickly and completely absorbs the bullet. The mass is originally at rest, and is attached at the equilibrium position to a spring whose force constant is 1948 Newtons/meter. The mass is observed to move to a maximum displacement of .114 meters from the equilibrium position. Find the velocity of the mass immediately after absorbing the bullet, and the velocity of the bullet immediately before impact. Assume that no dissipative forces act on the mass after the collision.
Solution
At the extreme position, 46 meters from equilibrium, the potential energy of the mass is .5 kx ^ 2 = 8.51 Joules. Since only the spring forces were acting on the mass between that time and the time the extreme position was attained, no work was done on the system between the instant after the collision and the time at which the maximum displacement was attained. Thus no energy was added to the system during this time.
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I don't understand where the 46 meters comes in? Should we be using .114 meters?
If so, I'm not sure I understand how we arrive at 8.51 Jules.
I get .5 kx ^ 2 = .5 * 1948N/m * (.114m)^2 = 12.7 J.
Am I doing that correctly?
Thanks,
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Remember that these problems use a random number generator, which occasionally malfunctions. So the numbers in these problems are sometimes unreliable.
The procedures are all correct, as far as I know, but the numbers always need to be taken with a grain of salt.
In this case the 46 from the 46 grams got mixed up with the displacement from equilibrium.
It looks like there was an additional error in the number used for the energy.
Your solution is correct.
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