Conservation of momentum problem with a bullet and stationary block

[seanmcgowan]

Homework Statement

A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface. The initial speed of the bullet is 896m/s, and the speed of the bullet after it exists the block is 435 m/s. At what speed does the block move after the bullet passes through it?

Homework Equations

I think I have the corect momentum, but the answer for the acceleration of the block seems a bit too easy. Is it right, or am I way off?

The Attempt at a Solution

First I drew a free body digram. The diagram has the bullet on the left, with its velocity and mass (Mass=.006 Kg Velocity=896 m/s), and the stationary block of wood with its mass and velocity (Mass=1.25 Kg Velocity=0 m/s) .

Since this is a collision problem, I figured I should find the momentum using the formula Mass*Velocity=Momentum. So, for the bullet... .006 Kg * 896m/s= 5.376 Kg*m/s (initial momentum) and... .006 Kg * 435 m/s= 2.610 (final momentum). For the block of wood... 1.25 Kg * 0 m/s= 0 Kg*m/s (Initial momentum) and... 1.25 Kg * 461 m/s (the difference of the bullet's Initial and Final velocities)= 576 Kg * m/s

I realize this is momentum not the block's velocity. I'm not sure what else to do because the only other logical answer is 896-435= 461 m/s and that just seems way to easy for a physics problem. Or am I over thinking this?

 

[Doc Al]

Since this is a collision problem, I figured I should find the momentum using the formula Mass*Velocity=Momentum. So, for the bullet... .006 Kg * 896m/s= 5.376 Kg*m/s (initial momentum) and... .006 Kg * 435 m/s= 2.610 (final momentum).

This is good.

For the block of wood... 1.25 Kg * 0 m/s= 0 Kg*m/s (Initial momentum) and... 1.25 Kg * 461 m/s (the difference of the bullet's Initial and Final velocities)= 576 Kg * m/s

That last step isn't correct. Why would you multiply the change in the bullet's velocity by the mass of the block?

Instead, set up a conservation of momentum equation:
Intial momentum (of bullet and block) = Final momentum (of bullet and block)

Initially, you just have the bullet moving; After the collision, the bullet and the block move. Set up the equation and solve for the speed of the block.

 

[seanmcgowan]

So your saying that what I shoud do is this:

5.367+0=2.610+x
5.367=2.610+x
2.757 Kg*m/s=x

Where X equals the blocks momentum.

2.206(Kg*m/s)/1.25(Kg)=
2.2056m/s.

Using significant figures it would be 2.206.

Or am I still wrong?

 

[Doc Al]

So your saying that what I shoud do is this:

5.367+0=2.610+x
5.367=2.610+x
2.757 Kg*m/s=x

Where X equals the blocks momentum.

2.206(Kg*m/s)/1.25(Kg)=
2.2056m/s.

Using significant figures it would be 2.206.

Or am I still wrong?

Your method is exactly correct. Two small problems:
(1) Check the digits that I highlighted; you have them transposed. (Fix that and redo the calculation.)
(2) Your final answer should be rounded to 3 significant figures, since your data is only given to 3 digits.