# Linear momentum problem (ballistic pendulum)

## Homework Statement

A ballistic pendulum is a device that may be used to measure the muzzle speed of a bullet. It is composed of a wooden block suspended from a horizontal support by cords attached at each end. A bullet is shot into the block, and as a result of the perfectly inelastic impact, the block swings upward. Consider a bullet (mass m) with velocity v as it enters the block (mass M). The length of the cords supporting the block each have length L. The maximum height to which the block swings upward after impact is denoted by y, and the maximum horizontal displacement is denoted by x.

a) In terms of m, M, g and y, determine the speed of the bullet.

mgh=1/2 m v2

## The Attempt at a Solution

(1/2)(m)v2 = (m + M)(g)(y)
v =sqroot { [(2)(g)(m + M)(y)]/m }

The real solution states that one must find the initial velocity of the bullet-block system, and then use this equation. Why can I not just use the initial velocity of the bullet? The real final answer is what I have, except the (m+M)/m is outside the sqroot.

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## Homework Statement

A ballistic pendulum is a device that may be used to measure the muzzle speed of a bullet. It is composed of a wooden block suspended from a horizontal support by cords attached at each end. A bullet is shot into the block, and as a result of the perfectly inelastic impact, the block swings upward. Consider a bullet (mass m) with velocity v as it enters the block (mass M). The length of the cords supporting the block each have length L. The maximum height to which the block swings upward after impact is denoted by y, and the maximum horizontal displacement is denoted by x.

a) In terms of m, M, g and y, determine the speed of the bullet.

mgh=1/2 m v2

## The Attempt at a Solution

(1/2)(m)v2 = (m + M)(g)(y)
v =sqroot { [(2)(g)(m + M)(y)]/m }

The real solution states that one must find the initial velocity of the bullet-block system, and then use this equation. Why can I not just use the initial velocity of the bullet? The real final answer is what I have, except the (m+M)/m is outside the sqroot.
What can you say about energy in an inelastic collision?

• Dennis Heerlein
What can you say about energy in an inelastic collision?
:( I stared at that for 30 minutes, thinking I was using conservation of momentum and not conservation of energy. Thank you for reading that block of text and helping me though! Much appreaciated