Let's consider the example of a gun firing a bullet. By Newton's third law we know that an identical force is applied by the gun towards the bullet and vice versa, resulting in both bullet and gun increasing their kinetic energy. So, the chemical energy of the bullet is transformed in kinetic energy of both bodies? Is it true to assume that in any similar case in which a body exerts a force to another body, the energy required to generate that force is converted in work for both bodies? And if we consider another case in which the gun is attached to the ground and remains still so it doesn't increase its kinetic energy, then does the bullet receive more energy and have a higher acceleration?
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2Not a gun specialist but I think it's usually the "chemical energy of the gunpowder". The bullet is usually a rigid body with no chemical energy of its own. – Gaurang Tandon Dec 08 '17 at 11:56
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4Possible duplicate of Newton's third law does not apply to the kinetic energy formula? – sammy gerbil Dec 08 '17 at 12:00
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In the second part, the answer is no as it will increase in the kinetic energy of the earth. There could be some change due to the decrease in relative velocity of bullet and the gun chamber that could manifest itself through a change in chemical reaction. – Rishabh Jain Dec 08 '17 at 12:50
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Yes, if your gun were attached to the ground the bullet would seem to receive a different amount of kinetic energy from the blast than the chemical energy stored in the bullet. But this is fine because energy is not conserved between reference frames. Total energy is just conserved within a single frame. Let's start with the gun and bullet in space, and assume that a total energy of $E_c$ (chemical energy) is introduced into the system and look at the different reference frames to see how the energy changes and when it's not concerved:
Frame 1: Center of Mass
Let's say the gun and bullet are stationary in front of you in space. You're reference frame is now fixed at the center of mass. The gun fires with $E_c$ and the gun and bullet get some respective amounts of energy:
$$E_{k_2}=\frac{m_b v_b^2}{2}+\frac{m_g v_g^2}{2}=\Delta E_k = E_c$$
Frame 2: Following the gun
Here's the tricky part. If you started off staring at the stationary gun before the explosion, then let your reference frame start following the gun as it moves after the explosion (the gun seems stationary in this frame) then you've changed reference frames from a stationary frame before the blast to a moving frame after the blast. In this case the notion that your chemical energy is converted completely into kinetic might not hold: $E_c=\Delta E_k$. But it will hold if we don't change frames.
Let's say you want the gun to be stationary after the shot is fired without changing reference frames. For this to happen your reference frame would have to be moving the future speed of the gun $v_g$ prior to the explosion. So you would be barreling towards your gun and bullet at $v_g$. The energy you see in the system before the explosion will be: $$E_{k_1}=\frac{(m_g + m_b)v_g^2}{2}$$ then the explosion goes off and you see the kinetic energy change to: $$E_{k_2}=\frac{m_b (v_b + v_g)^2}{2}$$
Which is basically saying that the gun isn't moving relative to you and the bullet is moving away at the speed of the bullet and the gun from Frame 1. Ok so let's see if $\Delta E_k=E_c$ which we expect to be the case:
$$\Delta E_k=E_{k_2}-E_{k_1}=\frac{m_b (v_b + v_g)^2}{2}-\frac{(m_g + m_b)v_g^2}{2}=\frac{m_b v_b^2 + 2 m_b v_b v_g + m_b v_g^2 - m_b v_g^2 - m_g v_g^2}{2}=\frac{m_b v_b^2 + 2 m_b v_b v_g - m_g v_g^2}{2}$$
Promising but we're not there yet, we need another piece to keep going, conservation of momentum. Since momentum is conserved we can look at the momentum before the explosion (both the gun and bullet are moving at $v_g$ relative to the frame) and after (when the gun seems stationary and the bullet is traveling at $(v_g + v_b)$):
$$p = (m_b + m_g) v_g = m_b (v_b + v_g) $$ $$ -> m_g v_g = m_b v_b$$
Using this equation in $\Delta E_k$ above, we get:
$$\Delta E_k=\frac{m_b v_b^2 + m_g v_g^2}{2}= E_c$$
which is exactly the energy of the explosion and is equal to the energy change in Frame 1.
So what happens when you're on earth and standing next to the gun. Your reference frame changes. You are no longer following the center of mass of the gun and bullet. Friction with the earth moves you along. So you can't say anything about how the chemical energy of the bullet before and after the shot are conserved or compare it with other frames from an energy perspective. Momentum however is conserved even between reference frames, but that's a topic for another question.
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