# Using the law of conservation of kinetic energy

1. Dec 26, 2004

### Moe_the_Genius

HI,
I am wondering how to solve this physics question:

-We have an object A and we have an object B
-We place these objects at a distance from each other.
-Object A has mass m.
-Object B has mass 3m.
-Object A moves toward object B with velocity v.
-Object B moves toward object A with velocity -v.

Q; What is the velocity of each object after impact?
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Can you tell me what I am doing wrong here: (I don't know where I have made a mistake)

(v1) = velocity of Object A after impact.
(v2) = velocity of Object B after impact.

Total momentum = mv-3mv= -2mv = m(v1) + 3m(v2)

Dividing both sides by m we find that:
-2v = (v1) + 3(v2)

and solving for v1 we find,

(v1) = -2v-3(v2) (1)

Using the law of conservation of kinetic energy we also get:

(.5)mv^2 + (.5)(3m)v^2 = (.5)m(v1)^2 + (.5)(3m)(v2)^2

Now divide by .5 and then by m to come up with:

4v^2 = (v1)^2 + 3(v2)^2 (2)

Inserting v1 from (1) into (2):

4v^2 = (-2v-3(v2))^2 + 3(v2)^2
4v^2 = 4v^2 + 12v(v2) + 9(v2)^2 + 3(v2)^2
-12v(v2) = 12 (v2)^2

and finally (dividing by (v2) and 12),
(v2)=-v

plugging (v2)=-v into (2) to come up with:
(v1)= v

It doesn't even make sense or look right and I have been trying to determine where I have made a mistake for days!
Thanks for any help...

2. Dec 26, 2004

### krab

You only have one of the solutions, namely the one where they fail to interact. The other solution is v2=0, v1=-2v.

3. Dec 26, 2004

### Moe_the_Genius

Wow, you really caught my mistake!
I understand now, instead of dividing by (v2) I should have brought -12v(v2) to the other side and factored to find (v2). Thanks a bunch.