This problem is ok in concept, but the numbers are not going to work out. The difference in mass between the bullet and block are so great (7 grams vs 1000 grams) that the answer will still come out to about 7.20 cm. It works like this:
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There are two imporant concepts involved here:
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1. work is done on the bullet to stop it, and the work done is proportional to the depth of penetration.
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2. momentum is conserved in the second situation, when the block is free to slide.
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So the bullet has initial speed "v" and initial KE (1/2) m v2 . The work done to stop the bullet is equal to its change in kinetic energy. Since in the first situation it has final KE of zero, the work done is equal to the initial kinetic energy. In other words:
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Situation 1: bullet penetrates to depth d1 = 7.20 cm work done = (1/2) m v2 = K1
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Now for situation 2... the bullet actually does not stop! It embeds in the block and the two continue at some final speed after the collision. We have to use conservation of momentum to find the final speed:
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initial momentum of bullet = final momentum of bullet & block
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m v = (m + M) vf
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This means the final speed of the bullet (while it is in the block) is vf = mv / (m +M)
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And the final KE of the bullet is: (1/2) m vf2 = (1/2) m [ mv / (m+M) ]2
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Or... final KE = (1/2) m [m2 / (m +M)2 ] v2 = K1 [m2 / (m+M)2 ]
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So in the second situation, the work done is the difference between initial and final kinetic energy of the bullet:
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work done = K1 - K1 [m2 / (m+M)2 ] and this is proportional to the depth the bullet penetrates. So...
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Situation 2: bullet penetrated to depth d2 = ?? Work done = K1 - K1 [m2 / (m+M)2 ]
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NOW... in each case, the work done is proportional to the depth. So we can write:
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depth for #2 / depth for #1 = work done for #2 / work done for #1
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d
2 / d
1 = ( K
1 - K
1 [m
2 / (m+M)
2 ] ) / K
1
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Simplify by eliminating the K...
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d2 = d1 ( 1 - [m2 / (m+M)2 ] ) =
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= 7.20 * ( 1 - [ 7.002 / (7.00 + 1007)2 ] ) = 7.20 * ( 1 - 0.000047656 ) =
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= 7.19966 cm ˜ 7.20 cm
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The problem is that the final speed of the bullet is practically zero in the second situation... due to conservation of momentum, the bullet's final speed is only 0.7% of its initial speed. So its final KE is only 0.0049% of its initial KE. This means the work done it both situations is practically the same, so the depth is practically the same.
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If the mass of the bullet and block were closer to each other, the answer would come out much different.