Homework Help: Bullet muzzle velocity

1. May 9, 2012

roam

1. The problem statement, all variables and given/known data

http://img403.imageshack.us/img403/1751/problem1r.jpg [Broken]

2. Relevant equations

$v_f = \frac{m v_1 + M v_2}{m+M}$

3. The attempt at a solution

Let "A" be the configuration immediately before the collision and "B" the configuration immediately after the collision, and "C" the very final state.

I have used the above equation and solved for the velocity when the bullet enters the system (immidiently after the collision):

$v_B = \frac{m v_{1A}}{m+M}$

The expression for the total kinetic energy of the system right after the collision is

$K_B = 1/2 (m + M) v_B^2$

Substituting vB we get

$K_B = \frac{m^2v_{1A}^2}{2(m+M)}$

We then apply the conservation of mechanical energy principal to the system to get:

$K_B+U_B = K_C + U_C$

$\frac{m^2v_{1A}^2}{2(m+M)} + 0 = 0+ (m+M) gh$

$v_{1A} = \left( \frac{M+m}{m} \right) \sqrt{2gh}$

This is very close to the answer, but how can I bring μk (coefficient of friction) into my equation? The potential energy is given by mgh, so how do I write it in terms of μk and d?
Any help is appreciated.

Last edited by a moderator: May 6, 2017
2. May 9, 2012

Infinitum

Just applying conservation of mechanical energy will not work since there's an extra force acting. So, apply the work-energy conservation theorem, you need to include the total work done by all forces, including that done by the frictional force(missing in your equation)

Whats the definition of work by a force?

3. May 9, 2012

roam

Okay I tried the work-kinetic energy theorem but it's still not working:

$W_{net}= K_f -K_i$

Where work is force times d and fkkmg is the magnitude of kinetic friction.

$(F-f_k)d = (F- \mu_k mg) d = \frac{m^2 v_o^2}{2(m+M)}$

If we solve for vo, how come we don't end up with the correct expression? (I tried that) Is there any other way to factor μk into the equation?

4. May 9, 2012

Infinitum

You're almost there....Though, where did that 'F' come in from!?! (there is no F!)

Last edited: May 9, 2012
5. May 9, 2012

roam

So how should I write the equation? "F" is the force and fk is the friction slowing it down (W=Fxd)...

6. May 9, 2012

Steely Dan

What force? The only force acting after the collision is the friction force. So you only need the frictional force part of the work.

7. May 10, 2012

roam

Okay but there is another problem

$\mu_k mgd = \frac{m^2v_o^2}{2(m+M)}$

$v_o^2 = 2 \mu_k g d \frac{m+M}{m}$

$v_o = \sqrt{2 \mu_k gd} \sqrt{\frac{m+M}{m}}$

But the term "m+M/m" should not be under the square root, since the correct answer must be:

$v_o = \sqrt{2 \mu_k gd} \frac{m+M}{m}$

So, what's wrong? How can I get rid of this square root?

8. May 10, 2012

Infinitum

Here's your mistake. Is the frictional force only acting on the 'm'?

9. May 10, 2012

roam

Thank you so much, I had totally forgot the mass of the block M... I got the correct answer now. Thanks! :)

10. May 10, 2012