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

Answer:

D=0.085520 m

Explanation:

Step 1:

For first block

Work-wood=ΔK

F*D=V-final - V- initial

F*(0.086m)=0-$\frac{1*m*v^{2} }{2}$

F*(0.086m)=-$\frac{1*0.007kg*v^{2} }{2}$

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=$\frac{1*m*v-f^{2} }{2}$-$\frac{1*m*v^{2} }{2}$

(-0.04069v^2)*D=$\frac{1*1.214*(5.76606*10^-3*v)^{2} }{2}$-$\frac{1*0.007*v^{2} }{2}$

D=0.085520m