Posted by Josh on Wednesday, May 18, 2011 at 11:42am.
In a frictionless mount, momentum would have been conserved.
Vfinal for both would be given by
0.007 Vbullet = (1.007)Vfinal
Vfinal = 0.00695 Vbullet
The bullet kinetic energy loss would be
0.99305^2 = 0.986 of the value lost in the clamped block.
Assume penetration depth is proportional to kinetic energy loss. This would be true if the stopping force were constant.
The penetration depth would then be 7.89 cm.
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