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.
A 7.00 g bullet, when fired from a gun into a 1.00 kg block of wood held in a vice,
penetrates the block to a depth of 8.00 cm. If the block had instead been at rest on a
frictionless horizontal surface, to what depth would the bullet have penetrated?
1 answer
1 answer