# Why is energy not conserved for an impulsive force? [duplicate]

Why is energy conservation not valid in the case of an impulsive force?

Let me give an example of a problem that concerns my question.

Suppose a horizontally flying bullet of mass m gets stuck in a body of mass M.

So I came up with 2 equations:

1. $$(m)(v_i) = (m+M)(v_f)$$
2. $$1/2(m)(v_i)^2 = 1/2(m+M)(v_f)^2$$

So why is equation 2 incorrect?

Any help is appreciated!

• In case of elastic collision, kinetic energy is conserved (as change in potential energy, $\Delta U=0)$. But in case of inelastic collision, the structure of colliding bodies change forvever at the time of collision, thus $(\Delta U\neq 0)$ and thus kinetic energy is not conserved.
– Iti
May 24, 2021 at 11:24
• Does this answer your question? Conservation of $\vec{p}$ in Ballistic pendulum May 24, 2021 at 11:26
• @Bhavay No, the link which you gave is regarding the momentum conservation part which I didn't ask. And that question is a little different from mine. May 24, 2021 at 11:47

Energy is always conserved, that is one of the strongest principles in physics. However, in some cases depending of its practicality, we talk just about the mechanical energy, which is the one not conserved in your example. In your case it is said to be an inelastical collision.

They key fact is that, what first were two distinct objects, transform into just one. When the bullet collides with the other body, it penetrates into it, deforming its volume and tearing come molecular/structural bonds apart, giving energy to some atoms of the body and increasing its temperature. This temperature, at the same time, transforms into radiation (emision of light) that also take away energy.

As you can see, there is a lot of things going on in just a simple inelastic collision. If you still want to apply energy conservation, you would need to know all this changes in total energy and write

$$\frac{1}{2}m v_i^2=\frac{1}{2}(M+m) v_f^2 + E_{friction} + E_{chem} + E_{rad}...$$

Or you could just group all the unknown energies into $$\Delta E_{lost}$$ and say $$\frac{1}{2}m v_i^2=\frac{1}{2}(M+m) v_f^2 + \Delta E_{lost}$$

With that little correction you could still apply the energy conservation formula, but obviously the energy lost is not usually known, and therefore you could not use the formula in that case.

Here energy as a whole is conserved but mechanical energy is not conserved, the energy that is used up in penetrating the the body of mass M is not recovered in the form of Mechanical energy but is used up for deformation of the body, changing temperature of the body, sound, etc. And these energies cannot be retrieved back in the form of Mechanical energy at the time of reformation.

It does not depend on the fact if the Force is Impulsive or not but on the nature of collision i.e. if it is elastic or inelastic collision. Mechanical Energy would be conserved in elastic collision and does not depend if the force is Impulsive or not.

• Understood! But are there other cases where mechanical energy is not conserved? (except this one) could you name a few? May 24, 2021 at 11:53
• Practically any collision is not perfectly elastic but rather ideal cases and practically not possible due to the fact that there is some kind of deformation or sound, friction etc in a collision. Collision between two electrons is elastic, as the two bodies do not hit each other but rather interact via electrostatic forces. May 24, 2021 at 12:03
• Yes, and electrostatic forces are conservative forces too! Thanks a lot! May 24, 2021 at 12:31

Some kinetic energy of the bullet is converted to potential energy of deformation of the object it hit, heat and sound; thus the kinetic energy is not conserved

Kinetic energy is only conserved in an elastic collision. Since the bullet gets stuck in the mass $$M$$, this collision is not elastic - in fact it is perfectly inelastic - and energy is lost due to friction etc.

Note that the key factor here is the nature of the collision, not whether the force is impulsive. Even if the bullet slowed down over a significant distance and so exerted a non-impulsive force on mass $$M$$, kinetic energy would still be lost in this scenario.