Elastic Collisions and Conservation of Momentum

[tehjinxman]

Homework Statement

A 1 Kg car moving at 2m/s collides elastically with a stationary car. The first car rebounds opposite to the original direction at 1m/s and the second car moves off in the original direction of the first car.
A) What is the mass of the 2nd car
B) What is the speed of the 2nd car

Homework Equations

P = M*V
M1V1 + M2V2 = M1V`1 + M2V`2

The Attempt at a Solution

I have been beating my head against this problem for hours now. It seems like it should be incredibly easy, and simple to figure out, but I just can't make it click in my head.
So I have:
Car1-- Initial momentum = 2*1 = 2, Final momentum = -1*1 = -1
Car2-- Initial momentum = 0*M2 = 0, Final momentum = M2*v`2

Since momentum is conserved --> Car1_i + Car2_i = Car1_f + Car2_f --> 2+0 = -1 + 3
Total final momentum of Car2 is 3 in the positive direction.

(1*2) + (M2*0) = (-1*1) + (M2V`2)

Ive spent i don't know how many hours trying to use these two equations to find some equality with M2. I mean the momentum of Car 2 after the collision, clearly needs to be +3, but there are an infinite combination of numbers that multiply to 3. I feel as though i am missing something extremely simple.

 

[ehild]

The collision is elastic, energy is conserved. Set up a second equation for conservation of kinetic energy.

ehild

 

[tehjinxman]

So Ke1_i = .5(1)(2^2) =2
Ke2_i = 0
KeTotal = 2

Ke1_f = .5(1)(-1^2) = 0.5
So Ke2_f must = 1.5
KeTotal = 2

Ke1_f+Ke2_f = 2
.5 + (mv^2)/2 = 2
mv^2 = 3

Then compare MV = 3 and MV^2 = 3? So MV^2 = MV
Which means V must be 1m/s and M must be 3 Kg?

 

[ehild]

It is correct now !

ehild