# Momentum of a bullet problem

## Homework Statement

A 7.45g bullet has vi = 353 m/s. It strikes the 0.725 block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 12.1 cm, what was the velocity of the bullet as it emerged from the block?

## Homework Equations

Energy conservation and momentum

## The Attempt at a Solution

First I solved for the velocity of the block when the bullet hit it. I used conservation of energy 1/2mv2 = mgh. I got the velocity as 2.41 m/s. Then I used conservation of momentum.
.00745(353)=.725(2.41) + .00745vf.
I solved for vf and got 118.5 m/s. The answer is supposed to be 203 m/s. I have tried combining the masses with the 2.41 velocity, but it still doesn't work out.
Can some one help me out? Thanks.

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cristo
Staff Emeritus

## Homework Statement

A 7.45g bullet has vi = 353 m/s. It strikes the 0.725 block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 12.1 cm, what was the velocity of the bullet as it emerged from the block?

## Homework Equations

Energy conservation and momentum

## The Attempt at a Solution

First I solved for the velocity of the block when the bullet hit it. I used conservation of energy 1/2mv2 = mgh. I got the velocity as 2.41 m/s. Then I used conservation of momentum.
.00745(353)=.725(2.41) + .00745vf.
I solved for vf and got 118.5 m/s. The answer is supposed to be 203 m/s. I have tried combining the masses with the 2.41 velocity, but it still doesn't work out.
Can some one help me out? Thanks.
The number in bold that you have calculated (2.41) is actually the square of the velocity (you forgot to take the square root) Take sqrt(2.41) and plug this into your momentum conservation equation, and you should get the correct answer.

Oh, good catch, thanks!