More than just a momentum problem

[macgirl06]

When a bullet traveling at 721 m/s strikes a block of wood originally at rest on a frictionless surface, the bullet emerges from the other side of the block of wood traveling at 349 m/s. If the mass of the bullet is 5.38 g and the mass of the block is 744 g, what is the speed of the block after the collision?


I don't know what to do, I tried doing m1v1 + m2v2 = m1v1' + m2v2', and I couldn't get the answer. Any help would be greatly appreciated.

Homework Statement

 

[Doc Al]

I don't know what to do, I tried doing m1v1 + m2v2 = m1v1' + m2v2', and I couldn't get the answer.

But this is "just" a conservation of momentum problem, so that should work. Show exactly what you did. What values did you use?

 

[macgirl06]

I did the following:

(0.0058kg)(721 m/s) - (0.0058kg)(349m/s) \ 0.744 kg

and that didnt work

and thnks for the fast reply

 

[Doc Al]

I did the following:

(0.0058kg)(721 m/s) - (0.0058kg)(349m/s) \ 0.744 kg

That looks perfectly OK to me:
Speed of block = [(0.0058kg)(721 m/s) - (0.0058kg)(349m/s)]/(0.744 kg)

Check your arithmetic; perhaps you made an error there.

 

[macgirl06]

I have tried that many many times, are you sure there arent any tricks to this question or something you have overlooked?

 

[Doc Al]

This is as straightforward a momentum conservation problem as you are likely to find. What answer did you get and why do you think it's wrong?

 

[macgirl06]

I got it, it was just a calculational error. Thanks for the clarification!