Ballistic Pendulum Problem: Determining Speeds of Bullet and Block

[ges9503]

Homework Statement

I fire a bullet at a ballistic pendulum. The large block of wood has a mass M2 = 3.000 kg, and the bullet has a mass of m1 = 25.0 g. The bullet completely penetrates the wood and emerges with a speed of vf = 80.0 m/s. The wood, as part of a pendulum, swings up to a maximum height of h = 10.00 cm.
(a) Determine the speed of the block of wood after the bullet has passed through it.
(b) What is the speed of the bullet just before it hits the block?

Homework Equations

m1v1=m2v2
1/2mv^2]= KE
mgh=PE

The Attempt at a Solution

I thought that setting the momentum of the bullet after it has exited the block equal to the momentum of the block at the maximum height would be the correct procedure to find the speed of the block , but I am wrong. How do I approach this problem? Convervation of energy or conservation of momentum? Or both?
Thanks :)

 

[ehild]

The bullet losses energy inside the block by drilling a hole through it. You can not use conservation of energy for the interaction between the bullet and block, but conservation of momentum is valid. After the bullet left the block, the block moves as a simple pendulum. Conservation of energy is valid for this motion.

ehild

 

[ges9503]

Ok, so you said that I can use conservation of momentum for the interaction between the bullet and block. So is it correct to find the speed of the block using
m1v1=m2v2
(m1v1)/m2=v2

(.025kg*80m/s)/3kg

v2= .66666667 m/s
?

 

[ehild]

Initial momentum of the bullet = final momentum of the bullet + momentum of block.

0.025 kg*v(bullet)=3.0*v(block)+0.025*80.0

When the bullet leaves the block, the block has v(block) speed, so its KE is:

KE=0.5 *3.0*v(block)^2

This turns into potential energy at the highest position:

3.0*g*0.1=KE

You get KE from the maximum height; from KE, you get the initial speed of the block; from that, you get the initial velocity of the bullet.