Physics Collision

Question

A .075 kg bullet strikes a 4.500 kg block of wood. The bullet exits the opposite side of the block with a speed of 180 m/s. The block slides for 3.1 m on a horizontal surface before coming to rest. The coefficient of sliding friction is .30. What was the original speed of the bullet?

Answer 1

I will set this up for you, but YOU do the calculations. First, I assume that the block started to slide at the moment the bullet finished passing through it. Let us define some variables:

m = mass of bullet (given)

u = initial speed of bullet = ? (what we want to solve for)

v = speed at which bullet exits the block (given)

M = mass of block (given)

w = speed at which block begins to slide = ? (we will need to find its value in order to get u)

μ = coefficient of friction (given)

d = distance that block slides before coming to rest (given)

a = acceleration of block while moving under friction = ? (we will use this to get u)

Use conservation of momentum at moment that the bullet exits the block.

mu = mv + Mw

Get w from kinematics:

0² - w² = 2ad

a = (-μMg) / M = -μg

-w² = 2(-μg)d

w = √(2μgd)

Now plug this expression for w into the equation where we used conservation of momentum, and then solve for u.

mu = mv + M√(2μgd)

u = [mv + M√(2μgd)] / m

You know m, v, M, μ, g, and d. Plug in the numbers and get u in m/s. I will leave that to you.