# Homework Help: Conservation of Momentum problem needs quick reply

1. Apr 24, 2005

### Byrne

Okay, here's the problem:

A 24 g bullet is fired horizontally, embedding itself in a 10 kg block initally at rest on a horizontal ice surface. The block slides the ice, coming to rest in 2.0 s at a distance of 60 cm from its original position. Assuming that the frictional force stopping the block was constant, calculate the velocity of the bullet.

Okay, I use the m1v1 + m2v2 = m1v1' + m2v2' formula to determine the v12'. However, my answer is half of what the answer in the back of the book is (My answer is 125 m/s while the book's is 250 m/s). Now, I think my problem might have something to do with the frictional force, but I dont really know where to begin to factor in the friction because the actual frictional force nor the frictional constant is given... please help.

2. Apr 24, 2005

### dextercioby

U realize that this is a plastic collision & u can find the initial velocity of the system bullet through consevation of momentum...U can find the acceration due to friction using Galilei's formula.Then,knowing that this acceleration is constant,u can find the velocity of the system block-bullet and then the velocity of the bullet before impact (<--conservation of linear momentum).

Daniel.

3. Apr 24, 2005

### Byrne

Well, only one out of three of the things you mentioned sound familiar to me. I mean, I am taking a high school physics course and we haven't learned anything about plastic collision or Galilei's formula so using those terms doesnt really help me.

4. Apr 24, 2005

### Berislav

No, it's $$mv=(m+M)v'$$

Since the bullet embeds itself inside the block.

Also, you don't need the constant of friction, since by knowing how far the block traveled and for how long you can figure out it's initial speed since the deaccelearation is constant (frictional force is constant).

EDIT:
Dextercioby was faster.

5. Apr 24, 2005

### dextercioby

Galileo's formula is valid for constant acceleration/deceleration movement and it reads

$$v_{final}^{2}=v_{initial}^{2}+2 a t$$

For deceleration,a<0...

Plastic collision,u have to be kidding,right...?

Daniel.

6. Apr 24, 2005

### Byrne

Okay, well the Galilei's forumula is obviously familiar, but as we learned it, its just one of the five acceration formulas. I guess my teacher doesnt care too much for the formal names of formulas. And I'm not kidding about plastic collision.

Thanks to both Berislav and dextercoiby for the help!

7. Apr 24, 2005

### Byrne

I'm sorry guys, I've been trying to plug in these numbers for the last half an hour and cannot get the right answer.

You gave me Galileo's formula to use, but I still dont understand how to use it. I'm plugging in the info I do have and am not getting the answer... (200 m/s)

8. Apr 24, 2005

### Berislav

$$v'=\frac{2s}{t}$$
$$v'=0.6 m/s$$

$$v=\frac{(m+M)v'}{m}$$
$$v=250.6 m/s$$