(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 2.3 kg wood block hangs from the bottom of a 1.3 kg, 1.3 m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 35 degrees.

2. Relevant equations

Conservation of Energy.

Conservation of Momentum

3. The attempt at a solution

(1/2)(m)(v^2) = mgh

Drawing a triangle, h is 1.3 * cos 35 = 1.06m

So (1/2)(v^2) = (9.8)(1.06)

v = 4.568 m/s

Conservation of momentum:

(mB)(vB) = (mB + mP)(v)

(.012)vB = (2.3+.012)(4.568)

vB = 10.561 / .012 = 880.21 m/s

Not sure where I went wrong.

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# Homework Help: Ballistic Pendulum

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