To what depth will the bullet penetrate the block in this case?

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SUMMARY

The discussion focuses on the penetration depth of a bullet fired into a block of wood under two different conditions. Initially, a 7.00 g bullet penetrates a 1.00 kg block held in a vise to a depth of 7.60 cm. When the same block is placed on a frictionless surface, the bullet's penetration depth will be less due to the conservation of momentum and the associated loss of kinetic energy. The key takeaway is that the penetration depth is influenced by the block's ability to move, which alters the energy dynamics of the bullet's impact.

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frosti
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Homework Statement


A 7.00 g bullet, when fired from a gun into a 1.00 kg block of wood held in a vise, penetrates the block to a depth of 7.60 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7.00 g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case?



Homework Equations


m1v1 + m2v2 = (m1+m2)vf


The Attempt at a Solution


I have no clue how to solve this problem. I don't really know what to do with the depth of penetration by the bullet. Can anyone please help?
 
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frosti said:

Homework Statement


A 7.00 g bullet, when fired from a gun into a 1.00 kg block of wood held in a vise, penetrates the block to a depth of 7.60 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7.00 g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case?



Homework Equations


m1v1 + m2v2 = (m1+m2)vf


The Attempt at a Solution


I have no clue how to solve this problem. I don't really know what to do with the depth of penetration by the bullet. Can anyone please help?
There's a significant loss of KE in part 1...like all of it, 1/2mv_b^2. In part 2 , apply conservation of momentum, and determine the total loss of KE. It' ll be a bit less; how does that ratio difference relate to the new depth of penetration?
 

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