# Bullet striking a block on a frictionless surface

1. Mar 24, 2014

### Dustinsfl

1. The problem statement, all variables and given/known data
A 10g bullet with velocity 1000m/s strikes a 100g block at rest. What is their combined velocity? Can you work the problem with conservation of momentum? Energy?

2. Relevant equations

3. The attempt at a solution
For conservation of momentum, we have
$$10g\cdot 1000m/s + 100g\cdot 0m/s = 110gv_f\Rightarrow v_f = \frac{1000}{11}m/s.$$
How is this done with the conservation of energy? The KE right after impact is $55g\cdot v_f^2 = ???$
Am I using the 0 PE level of the block? If that is the case, I have $(m+M)gh=110g\cdot 9.8m/s^2\cdot h$. What is $h$? I have two unknowns then.

2. Mar 24, 2014

### SteamKing

Staff Emeritus
Why are you bringing gravity into this? What happens if the block is sliding on a long flat surface after it is struck by the bullet?

What happens to the KE of the bullet after it strikes the block?

What happens to the KE of the block after it is struck by the bullet?

3. Mar 24, 2014

### Dustinsfl

It is an inelastic collision so the KE is $55gv_f^2$ directly after impact. I already stated this though.

4. Mar 25, 2014

### ehild

The mechanical energy (kinetic+potential energy) is not conserved. The potential energy does not change, the kinetic energy decreases, as part of the initial energy transforms to heat.

ehild

5. Mar 25, 2014

### Dustinsfl

How am I supposed to use this information here?

6. Mar 26, 2014

### ehild

It means you cannot solve the problem assuming conservation of energy. You have to calculate the total energy after the collision, in joules. The total energy is solely kinetic. What does "g" mean in your formula?

ehild