# Bullet shot into a block #2

1. Homework Statement

A bullet of mass m is fired into a block of mass M initially at rest on a frictionless table of height h. The bullet remains m in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

2. Homework Equations

I used (m +M)gh for the potential energy and set that equal to the kinetic energy ½(m + M)v₂² and found that v₂ = √(2gh).

And since it was a completely inelastic collision (right?) I used mv₁ = (m + M)v₂ → v₁ = (m + M)v₂ * 1/m

Is this right?

3. The Attempt at a Solution

initial velocity = √(2gh)*((m + M)/m)

## Answers and Replies

Homework Helper
What about the distance d?

What about the distance d?

That's the thing, I din't know what do with it. That's why I'm here for help.

mathman44
First solve for the speed of the bullet+block by finding time, (you already have distance):

$$d=v*t$$

From kinematic equation $$h = 1/2 at^2$$, solve for t: $$t=\sqrt(2h/g)$$

Now we know the speed of the bullet+block is just the distance "d" divided by time, so $$v_2 = d * \sqrt(g/2h)$$.

Using momentum equations, just as you did, $$v_1 = ((m+M)/m) * v_2 = ((m+M)/m) * [d * \sqrt(g/2h)]$$