# Collision of falling and thrown upward balls

1. Mar 16, 2014

### nikolafmf

Hi,

I am wondering if total momentum of the system is conserved in this case. Our system is consisted of two balls: one is falling vertically down to ground, another is thrown vertically up, so they collide in air. Is total momentum of the two balls conserved after the collision?

Now I know that if no external force is applied in the direction of momentum, is is conserved. But here there is gravitational force in the direction of the initial momentum. So I am not sure if total momentum before and after collision is conserved. Any help will be very appreciated.

Sincerely,
Nikola

2. Mar 16, 2014

### Staff: Mentor

Yes. (I could type in this part of my response before I even read the rest of your post )

The total momentum of the system is conserved, because the system consists of *three* objects - the two balls and the earth. The momentum of the two balls, in isolation, is not conserved - it continuously increases in the downwards direction, for exactly the same reason that the momentum of a single ball, released from the top of a tall building, will increase in the downwards direction.

3. Mar 16, 2014

### nikolafmf

Just to be clear, I meant if the momentum of the two balls is conserved just before and just after the collision. Well, this "just" has to be clarified. The collision takes some time to happen, so momentum will not be conserved exactly. Question that bothers me here is, can we take that momentum (of two balls just before and just after the collision) is conserved, if we are going to calculate the speeds after the collision and we don't want to make large error (not larger than, say, 1%)?

4. Mar 16, 2014

### Staff: Mentor

If you just consider the momentum of the two balls, yes. Usually, this effect is negligible. And even if it is not: gravity will give a constant rate of momentum change ((m1+m2)*g), independent of the collision process.

It is impossible to express this error relative to the total velocity without looking at the actual process.

5. Mar 17, 2014

### sophiecentaur

The easiest way to approach this is to do your initial calculation in the reference frame of the Centre of Mass of the two balls. That will give you the approach and parting velocities of the balls. Then, if you want to know what the collision will look like in the Earth's frame, you can just introduce the velocity of the CM and find its resulting trajectory, under g, adding this value to the velocities of the two balls.

Don't even consider that you can do without Conservation of Momentum!
I don't understand that your problem with errors will be worse or better, depending on the calculation method. (Assuming a given accuracy of measurements)