# Homework Help: Hopelessly Lost

1. Dec 12, 2004

### Afro_Akuma

Having missed most of this material, I am finding myself at odds wih the last question of a homework assignment:

A bullet of mass 4.0 g moving horizontally with a velocity of 500 m/s strikes a wooden block, initially at rest, on a rough wooden surface. The bullet passes through the block in a negligible time interval, emerging with a velocity of 100 m/s and causing the block to slide 40 cm along the surface before coming to rest. With what velocity does the block move just after the bullet exits?

Can anybody suggest a good way to set up this problem?

2. Dec 12, 2004

### quasar987

Are you given the coefficient of friction of the surface?

3. Dec 12, 2004

### ms. confused

I don't think you need the coefficient of friction to solve a problem like this. This is a type of collision question. Since there are two types of collisions (elastic and inelastic), there are two ways of setting up the problem. We'll just treat yours as an elastic collision since the bullet and the block don't remain stuck together. So, set it up like this:

m1v1 + m2v2 = m1v1' +m2v2'
m= mass in kg
v= velocity-->(v2' is the velocity of the block after the bullet goes through, which is what you want to find)

Try plugging some of the numbers into this formula and solve for v2'.

4. Dec 12, 2004

### ms. confused

O.K. wait a sec...do you know the mass of the block?

5. Dec 12, 2004

### quasar987

I figured we didn't have the mas of the wooden block, otherwise the "it moves 40 cm part" would be useless and pi = pf would be all there is to the problem.
My idea was to use the fact that in moving of 40 cm, the friction does a work equal to the variation of kinetic energy. If we know the coefficient of friction mu, then

$$W = 40cm*m_2g \mu = \frac{1}{2}m_2 v_2'^2$$

and we can isolate v_2' as a fucntion of mu.

Last edited: Dec 13, 2004
6. Dec 12, 2004

### ms. confused

OK then we do need mu. Woops...my mistake. I wasn't sure how to take the "40cm" part into account. There isn't another way to do it without mu then, is there?

7. Dec 14, 2004

### Diane_

The collision can't be treated as elastic, since the bullet does deform the block. That takes energy, which will have to come out of the initial kinetic energy of the bullet. Kinetic energy therefore cannot be conserved, so the collision cannot be elastic.

It does seem to me that more information is needed here. Clearly, the amount of friction is going to make a difference. Since the bullet passes through the block "in negligible time", friction will not matter greatly in the amount of energy transferred to the block. The distance the block will slide, then, will depend only on the force of friction. If we knew the mass of the block, we could determine from that the velocity of the block: the distance it slides would then tell us the force of friction and, therefore, the coefficient of friction. Alternately, if we knew the coefficient of friction, we could determine the force it exerts. Given that, we could find the velocity and, hence, the mass of the block. Without one or the other, though, it seems to me the problem is insoluable as stated.