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

In summary, the conversation discusses the use of equations and principles for solving a physics problem involving a collision and friction. After applying the conservation of mechanical energy principle, it is determined that the work-energy conservation theorem must also be used to account for the frictional force. The final equation includes the coefficient of kinetic friction and correctly factors in the mass of both objects involved in the collision.
  • #1
roam
1,271
12

Homework Statement



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

Homework Equations



[itex]v_f = \frac{m v_1 + M v_2}{m+M}[/itex]

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):

[itex]v_B = \frac{m v_{1A}}{m+M}[/itex]

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

[itex]K_B = 1/2 (m + M) v_B^2[/itex]

Substituting vB we get

[itex]K_B = \frac{m^2v_{1A}^2}{2(m+M)}[/itex]

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

[itex]K_B+U_B = K_C + U_C[/itex]

[itex]\frac{m^2v_{1A}^2}{2(m+M)} + 0 = 0+ (m+M) gh[/itex]

[itex]v_{1A} = \left( \frac{M+m}{m} \right) \sqrt{2gh}[/itex]

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? :confused:
Any help is appreciated.
 
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  • #2
We then apply the conservation of mechanical energy principal to the system to get:

[itex]K_B+U_B = K_C + U_C[/itex]

[itex]\frac{m^2v_{1A}^2}{2(m+M)} + 0 = 0+ (m+M) gh[/itex]

[itex]v_{1A} = \left( \frac{M+m}{m} \right) \sqrt{2gh}[/itex]

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 μk (coefficient of friction) into my equation?

Whats the definition of work by a force?
 
  • #3
Infinitum said:
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:

[itex]W_{net}= K_f -K_i[/itex]

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

[itex](F-f_k)d = (F- \mu_k mg) d = \frac{m^2 v_o^2}{2(m+M)}[/itex]

If we solve for vo, how come we don't end up with the correct expression? (I tried that) :confused: Is there any other way to factor μk into the equation?
 
  • #4
roam said:
[tex](F-f_k)d = (F- \mu_k mg) d = \frac{m^2 v_o^2}{2(m+M)}[/tex]

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

You're almost there...Though, where did that 'F' come in from? :confused: (there is no F!)
 
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  • #5
Infinitum said:
You're almost there...Though, where did that 'F' come in from? :confused: (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)...
 
  • #6
roam said:
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.
 
  • #7
Okay but there is another problem

[itex]\mu_k mgd = \frac{m^2v_o^2}{2(m+M)}[/itex]

[itex]v_o^2 = 2 \mu_k g d \frac{m+M}{m}[/itex]

[itex]v_o = \sqrt{2 \mu_k gd} \sqrt{\frac{m+M}{m}}[/itex]

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

[itex]v_o = \sqrt{2 \mu_k gd} \frac{m+M}{m}[/itex]

So, what's wrong? :confused: How can I get rid of this square root?
 
  • #8
roam said:
Okay but there is another problem

[itex]\mu_k mgd = \frac{m^2v_o^2}{2(m+M)}[/itex]

Here's your mistake. Is the frictional force only acting on the 'm'?
 
  • #9
Infinitum said:
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! :)
 
  • #10
Glad to help :smile:
 

1. What is bullet muzzle velocity?

Bullet muzzle velocity refers to the speed at which a bullet exits the barrel of a firearm. It is typically measured in feet per second (fps) or meters per second (m/s).

2. How is bullet muzzle velocity calculated?

Bullet muzzle velocity is calculated by measuring the distance the bullet travels in a given amount of time. This can be done using specialized equipment or by using a chronograph, which measures the speed of the bullet as it passes through two sensors.

3. What factors affect bullet muzzle velocity?

Several factors can affect bullet muzzle velocity, including the type and weight of the bullet, the type of firearm and its barrel length, the type of propellant used, and environmental factors such as air temperature and altitude.

4. Why is bullet muzzle velocity important?

Bullet muzzle velocity is important for several reasons. It can affect the accuracy and trajectory of a bullet, as well as its impact and penetration power. It is also important for safety considerations, as a high muzzle velocity can increase the risk of ricochets and over-penetration.

5. What is the average bullet muzzle velocity?

The average bullet muzzle velocity can vary greatly depending on the specific firearm and ammunition being used. However, for a standard 9mm handgun, the average muzzle velocity is around 1,000 fps, while for a .223 rifle it can be around 3,000 fps.

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