Solving for Initial Speed of a 5.5g Bullet Impacting Wood Block

[CoolBlueR]

I need help with this. I don't know what equation to use.

A 5.5g bullet is fired into a block of wood w/a mass of 22.6g. the wood block is initially at rest on a 1.5m tall post. after the collision, the wood block and bullet land 2.5m from the base of the post. how do you find the initial speed of the bullet?

I think that you would use the momentum equation, but I don't understand where the height and displacement come in.

 

[greybird]

I think this is how you would do it:

Find the time it takes the block to hit the floor:

s =ut + 0.5at^2
1.5 = 0.5*9.8t^2
t = 0.55 seconds

Then use this to work out what acceleration would be needed to move the block 2.5m in that time.

2.5 = 0.5a*0.55^2
a = 16.33 m/s

Then work out the velocity of the block when it hits the ground.

v=u+at
v=8.98 m/s

Use this in the momentum forumulas.

m1v1 + m2v2 = momentum after

8.98*0.0281 = 0.25
(final velocity*total mass of both the block and the bullet)

Then work back with the momentum forumula.

0.0055v + 0.0226*0 = 0.25
v = 45.8 m/s

Therefore the bullet must have hit the block of wood at a speed of 45.8 m/s

 

[chroot]

1) CoolBlueR, do not post your question more than once. The appropriate place to post homework questions is here, in the homework help forum.

2) greybird, do not give away answers. This does not help the student. You should instead guide the student with thoughtful advice and hints about what to do next, filling in the steps for him/her only when he/she cannot do it independently.

- Warren