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

 

[asteriatic]

The mass of the second car (M2) can be calculated using the conservation of momentum equation: M1V1 + M2V2 = M1V`1 + M2V`2. We know that the initial velocity of the second car is 0 m/s and the final velocity is V`2, so the equation becomes M1V1 = M1V`1 + M2V`2. Plugging in the values, we get (1 kg)(2 m/s) = (-1 kg)(1 m/s) + M2V`2. Solving for M2, we get M2 = (1 kg)(3 m/s) / V`2. Since we know that the final velocity of the second car is in the original direction of the first car, V`2 = 3 m/s. Therefore, the mass of the second car is M2 = (1 kg)(3 m/s) / 3 m/s = 1 kg.

The speed of the second car can be calculated using the conservation of momentum equation as well. Since we know that the initial momentum of the second car is 0 kg m/s and the final momentum is (1 kg)(3 m/s), the equation becomes 0 = (1 kg)(3 m/s) + M2V`2. Solving for V`2, we get V`2 = -3 m/s. However, since the second car is moving in the original direction of the first car, the speed is positive and we can take the absolute value, giving us a final speed of 3 m/s for the second car.

In summary, the mass of the second car is 1 kg and the speed is 3 m/s. It is important to note that the conservation of momentum principle states that the total momentum of a system before and after a collision remains constant, as long as there are no external forces acting on the system. This means that the individual momentums of the cars may change, but the total momentum remains the same. In this case, the first car experiences a change in momentum from 2 kg m/s to -1 kg m/s, while the second car experiences a change from 0 kg m/s to 3 kg m/s. This demonstrates the conservation of momentum in an elastic collision.