# Cart-Cart Collision

#### r34racer01

A cart (m1 = 140 kg) is moving to the right along a track at v1i = 24 m/s when it hits a stationary cart (m2 = 300 kg) and rebounds with a speed of v1f = 7 m/s in the opposite direction.

a) With what speed does the 300 kg cart move after the collision?

An observer moves in the same direction as the incoming cart with a speed of 10 m/s.

Using the convention that the positive direction is to the right, what are the following velocities with respect to this observer:
b) v1i,ob =
c) v2i,ob =
d) v1f,ob =
e) v2f,ob =

f) What is the total momentum of the system before the collision as seen by this moving observer?

g) What is the total momentum of the system after the collision as seen by this same observer?

So I thought that I could use 1/2mv^2 to get the momentum of cart 1 before impact, and I got 1/2(140)(24^2)= 40320. And then use 40320=300(v) to solve for be, but that didn't work. Can anyone tell me what I should be doing?

1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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#### cristo

Staff Emeritus
So I thought that I could use 1/2mv^2 to get the momentum of cart 1 before impact, and I got 1/2(140)(24^2)= 40320. And then use 40320=300(v) to solve for be, but that didn't work. Can anyone tell me what I should be doing?
The 140kg cart is moving also after the collision.

#### r34racer01

The 140kg cart is moving also after the collision.
Yes its moving 7m/s after the collision, but I don't get how I can use that to get the 300kg's speed after collision.

#### r34racer01

Never mind I realized we just nee to use conservation of momentum, so you just do (24*140)-(-7*140)/300 = 14.46 = V2f. And I was able to solve for everything else so thanks.

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