# Conservation of momentum mine car problem

1. Dec 4, 2013

### genevievelily

1. The problem statement, all variables and given/known data

A mine car of mass 440kg rolls at a speed of 0.50m/s on a frictionless horizontal track. A chunk of coal of mass 150kg has a speed of 0.80m/s as it leaves a chute above the car. The angle of the chute is 25 degrees from the horizontal. After the coal has come to a rest in the car, what is the velocity of the car and coal system.

2. Relevant equations

mv1 + mv1 = v(m1+m2)

3. The attempt at a solution

Originally I thought all momentum was conserved but then that wouldn't make sense because if it did the cart would not continue rolling straight forward. I got the correct answer if only taking x component into account. I just don't understand why you do this and how come the y component is not included in calculating momentum.

The solution says:

Momentum is conserved in the horizontal direction during the "collision." Let the coal be object 1 and the car be object 2.

So my question in other words is how come its only conserved in horizontal direction?

Thanks.

2. Dec 4, 2013

### Panphobia

There is an x and y component to the coal's momentum, but when the coal collides with the mine car the only velocity you want is the horizontal. So all you need is the the x component of the coal momentum in the chute. So momentum is always conserved, but the y momentum doesn't matter to the x momentum, but you are looking for the x component.

3. Dec 4, 2013

### PeroK

It is conserved in the y-direction (downwards) but the momentum in this direction is transmitted through the car to the Earth. So, technically, the Earth is perturbed ever so slightly. In practice, though, this moment is lost to the system you are observing.

Another example is a bouncing ball. Each time the ball bounces the momentum of the ball changes direction. So, by conservation of momentum, the Earth must also be "bouncing" by a tiny amount.

4. Dec 4, 2013

### genevievelily

got it thanks!