Solving Momentum Problem: 10 g Bullet & 10 kg Wood Block

[mattward70]

I am having difficulty understanding a basic idea about momentum.
I already learned several different types of momentum problems but this one is confusing me. Thanks for your help in advance.

"A 10 g bullet is fired into a 10 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. What is the speed of the bullet?"

I have tried to read past threads on this for the sake of repitition but i couldn't find any like this. We haven't covered Energy yet, and most of the past threads either dealt with friction (which I assume is neglegent, because it is not mentioned, and our teacher wouldn't require us to find a chart on the coefficient of wood to wood) or the problems dealt with energy, which I don't know how to do yet.

I understand that this is an inelastic collision, but I don't understand where to start. What confuses me is that you know the box slides 5 cm but you don't know how long it took or its acceleration... It just seems like there are too many unknowns to solve the problem. No, I didnt wait until the last minute to do the homework assignment, and yes i will be going to my teachers office hours tomorrow, but its bugging me and I'd like to know this technique before i cal sleep tonight so any help will be appreciated...thanks

 

[arildno]

Welcome to PF!
This is really confusing!
No energy? No friction?
I'm sorry, I can't see how you've been given adequate info here.

Some force MUST DO WORK on the system; otherwise, the system won't ever stop moving (due to Newton's 1.law of motion)

 

[pmrazavi]

The surface MUST have friction, else it would continue going on with the final velocity...I guess you should look up the friction coeffiecent of wood, then solve the problem...

m1v1=(m1+m2)Vf
(.001)V1=(10.001)Vf

You are looking for V1..to do that first you should find Vf:

you should find the final velocity by this equation

delta K = W , (there is no way to find work without friction coe...)

 

[mattward70]

Alright guys, I suppose I have to hunt down that coe... thanks a lot :)

 

[pmrazavi]

by the way..you can solve it without using energy stuff: use vf^2-Vi^2=2a(Xf-Xi)
You got the delta X, find "a" by F=ma, finalV=0 so you can easily find Vi..

 

[arildno]

by the way..you can solve it without using energy stuff: use vf^2-Vi^2=2a(Xf-Xi)
You got the delta X, find "a" by F=ma, finalV=0 so you can easily find Vi..

As it happens, this is nothing BUT energy stuff; but I agree: this must have been what they were after!

Great clarification.

 

[Andrew Mason]

As it happens, this is nothing BUT energy stuff; but I agree: this must have been what they were after!

Great clarification.

I don't see how you can solve it thiis way. a is the deceleration of the block from its initial speed after impact (v_{0_bullet} x .01/10.01 ). The rate of acceleration has to do with friction. In order to determine a, you have to know F, the force of friction on the block/bullet. If F = K(M+m), then F/(m+M) = K = a. So you would need to know K, the co-efficient of friction for this system in order to solve for a.

Alternatively, using $\triangle d = \frac{1}{2}at^2$ you need to know t - the time it took to slow down to solve for a.

There is not enough information to solve this problem. Which stands to reason, because the standard way to measure the speed of a bullet is to use a ballistic pendulum: the same system but with the block as the bob of a pendulum.

If the distance 5 cm represents the height that a ballistic pendulum reached, one could determine the speed of the bullet quite easily using:
$V_{bullet}=\frac{(m+M)}{m}\sqrt{2gh}$

Using these figures, V_bullet = 1000 m/sec,

AM