# Physics

posted by on .

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.

• Physics - ,

In your first equation, there are two different velocity variables: before and after. You should not use the same symbol for both. You have the right idea applying conservation of momemtum, but you used the wrong bullet velocity. It is 220, not 200.

You are also correct applying conservation of energy to the upward swing