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

An 8.0g bullet is fired into and becomes embedded in a 2.5kg block of wood at the end of a pendulum. The block of wood swing to a height of 6.0cm. (Watch Units)

a) What is the PE of the block & bullet at the top?

b) What is the KE at the bottom? (COE)

c) What is the velocity of the block & bullet right after the impact?

d) What is the pmomentum at the bottom?

e) What was the initial speed of the bullet? (COM)

2. Relevant equations

KE= PE

KE= 1/2 mv^{2}

PE= mgh

P=mv

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v'

3. The attempt at a solution

a)PE = (.008kg + 2.5kg)(9.8 m/s^{2})(.06m) = 1.47 J (doesnt seem like a reasonable number to me)

b)KE = .5(.008kg + 2.5kg)(v)^{2}

c) mgh=1/2mv^{2}

2gh=v^{2}

v=(square root)2gh

(square root) (2)(9.8m/s^{2})(.06m)=1.08 m/s

d)Does "bottom" mean not moving? if not, then

P=mv P=(.008kg+2.5kg)(1.08m/s)=2.71 kgm/s

e)(.008kg)(v_{1}) + (2.5kg)(v_{2}) = (.008kg + 2.5kg)v'

.008kg(v_{1}) + 2.5kg(v_{2}) = 2.508kg(v')

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# Homework Help: Balistic Pendulum

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