Answer to Question #157085 in Mechanics | Relativity for Peter Frank

Question

7g bullet when fired from a gun into a 1kg block of wood in a vise penetrate the block to a depth of 8cm .what if? This block of wood is placed on a frictionless horizontal surface and a second 7g bullet is fired from the gun into a block. To what depth will the bullet penetrate the block in this case?

Answer 1

Solution:

work is done = change in KE

$F*d=-\dfrac{mv^2}{2}$ → where v is the speed of the bullet

$F*0.08m=-\dfrac{0.007kg*v^2}{2}$

F = -0.04375v²

conserve momentum in free block:

mv = (M + 2m)*V → V is the post-impact block velocity

0.007v = 1.014V

V = 0.0069v

Again, friction work = change in KE:

$F*d=\dfrac{(M+2m)V^2}{2}-\dfrac{mv^2}{2}$

$-0.04375v^2*d=\dfrac{1.014kg*(0.0069v)^2}{2}-\dfrac{0.007kg*v^2}{2}$

divide through by v²; the rest solves to

d = 0.0794 m = 7.94 cm