Question

A bullet of mass m is fired at speed v0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction μ. Write an expression for how far the block slides (s) in terms of m, M, Vo , μ, and g.

Answer 1

Answer:

$s = \frac{1}{2\mu g}\left (\frac{mv_{0}}{m + M}  \right )^{2}$

Explanation:

mass of bullet = m

mass of block = M

initial speed of bullet = vo

coefficient of friction = μ

Let the horizontal distance traveled before stopping is s.

Let v be the velocity of the bullet and block system after collision.

Use conservation of momentum

m x vo = (M+m) x v

$v = \frac{mv_{0}}{m + M}$

Now use third equation of motion

v² = u² - 2 a s

So,

$0 = \left (\frac{mv_{0}}{m + M}  \right )^{2}-2\mu gs$

$s = \frac{1}{2\mu g}\left (\frac{mv_{0}}{m + M}  \right )^{2}$