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July 12, 2014

Homework Help: Physics

Posted by Sarah on Wednesday, November 24, 2010 at 12:28am.

A bullet of mass 45 g is shot at a speed of 220 m/s into a 5.0-kg bag hanging from a string from the ceiling. The bag absorbs the bullet and begins to swing. To what maximum vertical height will it rise? Hint: Energy is not conserved as the bullet enters the bag but is conserved after the bullet comes to rest in the bag and the bag is swinging upward.

Is this right?

mv + Mv= v(m+M)
(.045)(200) + (5)(0)= v(.045+5)
9= v(5.045)
v= 1.7839444m/s (this will be the velocity of both at the bottom of the arc)

Step 2, use the conservation of energy law:

(.5)mv2=mgh
(.5)(5.045)( 1.7839444)2=(5.045)(9.8)h
(.5)(5.045)(3.182457622)=(5.045)(9.8)h
8.027749351=49.441h
h=.1623702868

Do I have the right idea? thanks for your help.

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