# The law of conservation of momentum

1. Oct 8, 2015

### Drizzy

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

a trailer with a mass of 150 kg happen to roll away with the speed 20 m/s. 450 kg cement fall staight down into the trailer! which speeds does the trailer get when the cement fall into it?

2. Relevant equations

I know how to solve it but I don't know why it works. Is it because this counts as a collision?

3. The attempt at a solution

150 * 20 = V(150+450)

2. Oct 8, 2015

### Daeho Ro

Where the situation is collision or merge, it satisfies
$$p_i = p_f.$$
That's it.

3. Oct 8, 2015

### Staff: Mentor

You can think of it as a collision. Or just think of "trailer + cement" as a system whose momentum (at least horizontally) is conserved.

4. Oct 8, 2015

### Drizzy

okay thanx :)

5. Oct 8, 2015

### stockzahn

Momentum is described by a vector, because velocity is a vector. You can bilance the momenta before and after the cement fell on the trailer in two perpendicular directions (x / horizontal & y / vertical) - in both directions momentum has to be conserved.

x: mT ⋅vT1x + mC ⋅vC1x = mT ⋅vT2x + mC ⋅vC2x = 150 kg ⋅ 20 m/s + 450 kg ⋅ 0 m/s = 150 kg ⋅ vT2x + 450 kg ⋅ vC2x with T2x = vC2x

The sum of the momenta of both objects in x-direction remains the same.

y: mT ⋅vT1y + mC ⋅vC1y = mT ⋅vT2y + mC ⋅vC2y = 150 kg ⋅ 0 m/s + 450 kg ⋅ vC1y > 150 kg ⋅ 0 + 450 kg ⋅ 0 m/s

As the momenta of the two objects can't be the same, a force must have affected them - the earth. Taking into account the momentum of earth (assuming it was standing still, when the cement hit the trailer):

y: mT ⋅vT1y + mC ⋅vC1y + mE ⋅vE1y = mT ⋅vT2y + mC ⋅vC2y + mE ⋅vE2y = 150 kg ⋅ 0 m/s + 450 kg ⋅ vC1y + 5.6 ⋅1024 kg ⋅0 m/s = 150 kg ⋅ vT2y + 450 kg ⋅ vCy2 + 5.6 ⋅1024 kg ⋅ vE2y with vT2y = vC2y = vE2y

and you can calculate what's the velocity of the earth (+ trailer + cement), due to this collision.