# Homework Help: How do I show that each mass endures an equal and opposite change in momentum

1. Oct 25, 2007

### pebbles

in this condition: Mass 1 = 2.5 kg and v =25 m/s
Mass 2 = 4.0 kg and v = - 10 m/s and how it produces conservation of momentum??? and, find the percentage of energy lost in the collision. which cart will lose more energy and why?

heeellpppppppppp!!!!

i know that the inital and final momentums are equal, but i don't know the rest!!

2. Oct 25, 2007

### dynamicsolo

I take it that since the problem asks about energy lost, this is an inelastic collision. If it's a perfectly inelastic collision, the definition will tell you what you need to do in setting up the conservation of linear momentum equation. If it isn't perfectly inelastic, then we need an additional piece of information, because conservation of linear momentum by itself won't tell us what the speeds of the masses after the collision will be.

Do the problem say what kind of inelastic collision you have?

3. Oct 25, 2007

### pebbles

it's a completely inelastic collision question.

4. Oct 25, 2007

### dynamicsolo

That means that the two masses have stuck together and have the same velocity after the collision. If you find the sum of the momenta of the two masses before the collision, you will also have the total momentum of the merged masses afterwards. This will tell you the velocity of the merged masses.

Now find the sum of the kinetic energies of the two masses before the collision and compare that with the kinetic energy of the merged masses afterwards. That will tell you how much kinetic energy was "lost" (transformed); the fractional loss will be the ratio

kinetic energy lost / original total kinetic energy

and the percentage of loss will be this fraction times 100%.

5. Oct 25, 2007

### pebbles

ok, so

kinetic energy initial = (1/2)mv initial ^2
kinetic energy final = (1/2) mv initial ^2 + (1/2)mv final^2

---> 1-KE final/KE initial---> 1-((v initial + v final)^2 + v final^2)/v initial ^2

....yes...???

6. Oct 25, 2007

### pebbles

then times 100 of course for percentage.

7. Oct 25, 2007

### dynamicsolo

All right, first of all, each mass has its own speed before the collision and the two together have a single speed afterwards. So you would have

kinetic energy initial = (1/2) · m_1 · (v_1_initial ^2) + (1/2) · m_2 · (v_2_initial ^2)
and
kinetic energy final = (1/2) · m_1 · (v_final ^2) + (1/2) · m_2 · (v_final ^2) .

You still need to use conservation of linear momentum to find that final velocity.

The first part is fine, but the rest won't simplify that way in this problem because the two masses are different. (So no nice cancellations here...)

8. Oct 25, 2007

### pebbles

conservation of linear momentum= change in momentum/change in time?

9. Oct 25, 2007

### dynamicsolo

Conservation of linear momentum means that the total linear momentum in a system does not change. That would mean that change in momentum/change in time = 0 .

So you want to set up an equation for this system which says that

linear momentum of mass 1 (initial) + linear momentum of mass 2 (initial) =
linear momentum of mass 1 (final) + linear momentum of mass 2 (final) .

If there are any unknowns, use the appropriate variable for each.
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