Energy and conservation of momentum confusion

[Jimmy87]

If you consider a bullet firing from a gun then you have conservation of momentum and a Newton's third law pair (according to what i have read on the internet anyway). They both experience the same force if they are a third law pair but, generally, what is it that determines which object receives more energy? For instance, the bullet has a huge kinetic energy after the gun has fired whereas the gun has very little. So where does energy conservation fit into this example and what is the relationship between energy and force (if there is one)? They are both receiving the same force yet experience different energies?

 

[WannabeNewton]

The relationship between kinetic energy and force is the work energy theorem. It states that $\frac{1}{2}mv^{2}(x) - \frac{1}{2}mv^{2}(x_0) = \int _{x_0}^{x}F(x')dx'$.

 

[nasu]

So where does energy conservation fit into this example and what is the relationship between energy and force (if there is one)?

It doesn't.
The kinetic energy is not conserved in this process.
It is like the reverse of a plastic (non-elastic) collision.
The initial KE is zero, the final KE is not.

The distribution of KE is determined by the masses of the two objects.
And maybe the external factors.

 

[technician]

The relationship between kinetic energy and force is the work energy theorem. It states that $\frac{1}{2}mv^{2}(x) - \frac{1}{2}mv^{2}(x_0) = \int _{x_0}^{x}F(x')dx'$.

This is of no help in this problem!
You are correct to realize that the bullet and the gun are a "Newtons 3rd law pair"
However conservation of energy is not enough to sort out the details, conservation of energy requires consideration of all the energies involved in the interaction.

 

[WannabeNewton]

This is of no help in this problem!
You are correct to realize that the bullet and the gun are a "Newtons 3rd law pair"
However conservation of energy is not enough to sort out the details, conservation of energy requires consideration of all the energies involved in the interaction.

What are you even talking about? He asked for a relationship between force and kinetic energy. The work energy theorem doesn't require conservation of energy to hold. The determination of the final velocities of the two objects involved in the process only requires conservation of momentum and by itself illuminates why the significantly more massive object doesn't have a recoil velocity comparable to the exit velocity of the smaller mass; he asked for something extra.

 

[nasu]

For instance, the bullet has a huge kinetic energy after the gun has fired whereas the gun has very little.

Are you sure about this? It is not obvious.
The bullet has a high speed and the gun has a much smaller recoil speed.
But the mass of the bullet is much smaller than that of the gun.
Their kinetic energies my be quite similar.

 

[james26]

Jimmy, you are right, an equal force acts on both. Both the bullet and the gun will receive equal momentum (mv), so if you know the masses of each and the velocity after firing of one, you can work out the velocity of the other. You can then calculate the kinetic energy (mv^2)/2 of each. You will find that the kinetic energy of the bullet is greater because of the squared factor.

When the gun fires, chemical energy in the exposive is converted to kinetic energy.

 

[lightarrow]

If you consider a bullet firing from a gun then you have conservation of momentum and a Newton's third law pair (according to what i have read on the internet anyway). They both experience the same force if they are a third law pair but, generally, what is it that determines which object receives more energy? For instance, the bullet has a huge kinetic energy after the gun has fired whereas the gun has very little.

Yes.

M = gun's mass; V = gun's velocity after shot
m = bullet's mass; v = bullet's velocity after shot

momentum conservation:

MV + mv = 0 → v = -(M/m) V

from here you see that |v| > |V|.

Now let's see kinetic energy T:

T_gun = 1/2 M V²
T_bullet = 1/2 m v² = 1/2 m (-M/m)² V² = 1/2 (M²/m) V²

T_bullet/T_gun = M/m.

So, for example, if m = 20g and M = 5kg, T_bullet is 250 times T_gun.

 

[Jimmy87]

Wow that's a lot of very useful information, thanks to all!