# Homework Help: Bullet striking block, maximum height obtained?

1. Aug 25, 2007

### JFonseka

1. The problem statement, all variables and given/known data

A 10-g bullet moving 1000 m/s strikes and passes through a 2.0-kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To what maximum height will the block rise above its initial position?

2. Relevant equations

K = (mv^2)/2
W = F x d
F = ma

3. The attempt at a solution

I first calculated the initial kinetic energy of the bullet which is 5000 J, and the final as it exits the block, which is 800J, therefore am I right in assuming 4200 J has been given to the block?

I tried using E = mgh for the block, then I realized I did not know how quickly the block accelerated upwards, and to find that I would need F = ma, but I don't know the force either, do i need to find those two to finish this question off? Or is there another way to go about it.

2. Aug 25, 2007

### learningphysics

Can you describe the picture, or post it? I'm guessing the block is on an incline, and the bullet is moving horizontally and strikes the block?

This isn't an elastic collision... so energy is converted to heat here... so you can't use conservation of kinetic energy during the collision.

3. Aug 25, 2007

### JFonseka

No the block is not on an incline, and I think heat energy produced here is negligible.

The bullet moves upwards, and the block is sitting on a ledge, the ledge has a small gap in which the bullet goes through and strikes the block

4. Aug 25, 2007

### JFonseka

Pictures doesnt fit, edited

5. Aug 25, 2007

### learningphysics

Do they give the height of the block? And also, is the 1000m/s right before it hits the block?

You should use conservation of momentum during the collision to calculate the velocity of the block right after the collision.

Then you can use conservation of energy (kinetic energy and grav. potential energy) to see the height it goes up to.

6. Aug 25, 2007

### learningphysics

That's actually very good.

7. Aug 25, 2007

### JFonseka

No they don't give the height of the block, but thanks, I'll try momentum, forgot about that!

Cheers

8. Aug 25, 2007

### JFonseka

Lol the drawing doesn't hold once it's posted, damn formatting

9. Aug 25, 2007

### JFonseka

Yep, so I did mv + mv = mv + mv

Which means, 2(0) + 0.010(1000) = Momentum before collision = 10 kg/ms^-1

2(v) + 0.010(400) = Momentum after collision, leading to v = 3 m/s

K.E. of the block therefore is .5x2x9 = 9J

h = 9/19.62 = 0.46m, which is one of the answers, so I'm guessing it's right, thanks!

10. Aug 25, 2007

### learningphysics

cool. you're welcome.