# Writing energy conservation equation for a collision? by \!/_PeacePlusOne

Question by …: Writing energy conservation equation for a collision?
For this problem, cart A is given an initial velocity towards a stationary cart B. There are pads at the end of each cart. The pads allow the carts to stick together after the collision.

Write down the energy conservation equation for this collision (Remember to take into account the energy dissipated).

right now i just have potential E(initial) + kinetic E(initial) = potential E(final) + kinetic E(final). i have no idea how to take into account the energy dissipated. any ideas?

What kind of potential energy are you tracking?

In the final state, some energy is in another form (dissipated):

potential E(initial) + kinetic E(initial) = potential E(final) + kinetic E(final) + dissipated E

where the total energy initially must equal the total energy finally.

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Edit: oldprof does, of course, mean “If this were perfectly ELASTIC, qe = 0 would be true.”

## 3 thoughts on “Writing energy conservation equation for a collision?”

1. physic says:

symmetry→conservation law
translation in time → energy
” ” space → momentum

2. Madhukar says:

Let Ma and Mb = masses of the carts A and B respectively.
Let u = velocity of the cart A before the collision
and v = common velocity of both the carts after the collision

By the law of conservation of linear momentum,
Ma * u + 0 = (Ma + Mb) * v … ( 1 )

This being the case of inelastic collision, kinetic energy is lost due to collision.
=> The loss of kinetic energy
= Initial kinetic energy of both the carts before the collision – Final kinetic energy of both the carts together after the collision
=> Loss of K.E.
= (1/2) Ma u^2 – (1/2) (Ma + Mb) v^2
Plugging v = [Ma / (Ma + Mb)] * u from ( 1 ),
=> Loss of K.E.
= (1/2) Ma u^2 – (1/2) (Ma + Mb) * [Ma / (Ma + Mb)]^2 * u^2
= (1/2) Ma u^2 [1 – Ma/ (Ma + Mb)]
= (1/2) MaMb * u^2 / (Ma + Mb).

3. oldprof says:

At the abstract level, TE = KE = 1/2 mV^2 is the total energy before the collision. Afterward it’s TE = ke + qe = 1/2 (m + M)v^2 + qe where qe is the deformation and heat energy lost.

We assume PE = 0 as there is no stored energy before or after. If restitution were assume, there would be some potential energy pe in that after the collision. That would be like a compressed spring ready to pop open. But we assumed permanent deformation and no restitution.

Then the COE gives us TE = KE = ke + qe = TE. And rearranged, it gives us KE – ke = qe, and 1/2 mV^2 – 1/2 (m + M)v^2 = qe is in fact that energy dissipation you were looking for. V > v is the speed of A and v < V is the speed of A and B smooshed together. Note. If this were perfectly inelastic, qe = 0 would be true.