# Inelastic Collision with Friction

1. Oct 16, 2004

### Gallium

I forget how to solve one dimensional motion problems with friction, which is holding me up in this problem:

a 5.00 g bullet is fired horizontally into a 1.2 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between the block and surface is .2. The bullet remains embedded in the block and the block moves 0.23 m before stopping. What is the initial speed of the bullet?

bullet------->block----.230 M-[ ]

looking at it as a momentum problem:

M(bullet)*V(bullet) = M(Bullet+Block)*V(bullet+block)

you need to solve for v(bullet+block) which can be determined by the following problem:

A block of mass 1.25 kg on a surface with kinetic friction of .2 is acted on upon a force and move .230 M before coming to rest on a horizontal surface. Find the initial speed of the bullet.

m = 1.25
v_0 = x
v_x = 0
mu_k = .2

2. Oct 16, 2004

### arildno

Well, in the second phase, you have constant deceleration equal to 0.2g

3. Oct 16, 2004

### Gallium

I understand that.

What I dont understand is the logic to find the initial velocity.

I know what this answer is, but I need to apply the exact same logic to a seperate problem. So what I am really asking for is someone to solve this problem and show the steps.

Gallium

4. Oct 16, 2004

### arildno

Well, you have:
$$v_{f}^{2}-v_{0}^{2}=2*(-0.2g)d$$
Right?
($$v_{f}=0$$, d is 0.230)
This enables you to solve for $$v_{0}$$

5. Oct 16, 2004

### Gallium

that doesnt yield the correct answer.

6. Oct 16, 2004

### Pyrrhus

Because that's the speed for the bullet+box system, not for the bullet, now use conservation of momentum to find the initial speed of the bullet.

7. Oct 16, 2004

### Gallium

.05 * V_1 = .9212*1.25

V_1 = 23.03 m/s

V_1 actually = 229 m/s

8. Oct 16, 2004

### Pyrrhus

1 Kg is 1000 grams not 100 grams.

5grams is 0.005 Kg.

9. Oct 16, 2004

### Gallium

i see, thank you. I found the correct logic.

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