Question

Depth to which bullet penetrates into a block

Answer 1

look at example


a 7.00g 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 8.00 cm. What if? this block of wood is 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? 

If you assume that force decelerating the bullet is 
constant Fo, then depth of penetration D is proportional 
to energy loss: 
ΔE = Fo x D 

In the former case energy loss in collision is equal to 
initial kinetic energy of the bullet: 
Fo x D1 = ΔE1 = K = 1/2 mv² 

In the latter case of frictionless surface, linear momentum 
of bullet + block is conserved. Energy of center of mass is 
Ecm = 1/2 p²/(M+m) = 1/2 (mv)²/(M+m) = 1/2 mv² * m/(M + m) 
Again, depth of penetration in this case is proportional 
to the loss of energy: 

Fo x D2 = ΔE2 = K - Ecm = 
= 1/2 mv² - 1/2 mv² * m/(M + m) = 
= 1/2 mv² * M/(M + m) 

Therefore 
D2/D1 = M/(M + m) 
D1 = D1 * M/(M + m) = 8cm * 1000/1007 = 7.94 cm 

New depth of penetration, as it turns out is almost the same.