Question

A 7.00 g bullet, when fired from a gun into a 1.20 kg block of wood held in a vise, penetrates the block to a depth of 8.60 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?

Answer 1

The depth a bullet penetrates into a block on a frictionless surface should be the same as when the block is held in a vise. This assumes there are no changes in the bullet's conditions or the material properties of the wood.

Step-by-step explanation:

The question is about predicting the depth a bullet will penetrate into a block of wood placed on a frictionless surface compared to its depth when the block is held in a vise. This scenario involves concepts such as momentum conservation, energy conservation, and possibly material resistance to penetration. Considering that the conditions of the bullet (mass and speed) and the block (mass) are the same in both cases, and ignoring any material changes in the wood, the depth of penetration should be the same in both scenarios. This is because the bullet will lose the same amount of kinetic energy in coming to a halt inside the wood. The only difference here is that when the block is on a frictionless surface, the block will be free to move, and so it will gain some kinetic energy, which means the bullet-block system will conserve momentum. However, since the question asks for the penetration depth, not the final velocities, and assuming the bullet penetrates to the same depth, the energy loss due to penetration should still be the same.

Answer 2

D=0.085520 m

Step-by-step explanation:

Step 1:

For first block

Work-wood=ΔK

F*D=V-final - V- initial

F*(0.086m)=0-
`(1*m*v^(2) )/(2)`

F*(0.086m)=-
`(1*0.007kg*v^(2) )/(2)`

F=-0.04069v^2

Step 2:

For 2nd block

Conservation of momentum

Momentum of 2nd block= Total momentum

mv=(M+m+m)v-f

0.007v=(1.2+0.007+0.007)v-f

v-f=5.76606*10^-3v

Work-wood=ΔK

Force on 1st block= Force on 2nd block

V-final = V-f which is calculated using conservation of momentum.

F*D=V-final - V- initial

(-0.04069v^2)*D=
`(1*m*v-f^(2) )/(2)`-
`(1*m*v^(2) )/(2)`

(-0.04069v^2)*D=
`(1*1.214*(5.76606*10^-3*v)^(2) )/(2)`-
`(1*0.007*v^(2) )/(2)`

D=0.085520m