What is the final velocity of a bullet entering a system after a collision?

[roam]

Homework Statement

http://img403.imageshack.us/img403/1751/problem1r.jpg

Homework Equations

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

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 v_B 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 μ_k (coefficient of friction) into my equation? The potential energy is given by mgh, so how do I write it in terms of μ_k and d?
Any help is appreciated.

 

[Infinitum]

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

$K_B+U_B = K_C + U_C$

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

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

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)

This is very close to the answer, but how can I bring μ_k (coefficient of friction) into my equation?

Whats the definition of work by a force?

 

[roam]

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?

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 f_k=μ_kmg 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 v_o, how come we don't end up with the correct expression? (I tried that) Is there any other way to factor μ_k into the equation?

 

[Infinitum]

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

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

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

 

[roam]

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

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

 

[Steely Dan]

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

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

 

[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?

 

[Infinitum]

Okay but there is another problem

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

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

 

[roam]

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

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

 

[Infinitum]

Glad to help