# Momentum Questions

## Homework Statement

Two heavy frictionless carts are at rest. They are held together by a loop of string. A
light spring is compressed between them. When the string is burned, the spring expands
from 2 cm to 3 cm, and the carts move apart from one another. Both hit the bumpers
fixed to the table at the same instant, but cart A moved 0.45 metre while cart B moved
0.87 metre.
What is the ratio of :
A. The speed of A to the of B after the interaction?
B. Their masses?
C. The impulses applied to the carts?
D. The accelerations of the carts while the spring pushes them apart?

## Homework Equations

This is an Explosion, therefore momentum before is going to be 0.
0 = m1v1 + m2v2.
The impulse will also equal 0
Ft = 0.

## The Attempt at a Solution

Well I wish I knew where to being, right now the only that I can think about is the displacement of the respective carts, and that they hit at an instant. But I cannot figure out much on the ratio and the solution. Any help would be appreciated.

Thank You.

Alright I can help you with the masses: Thinking back to the units that make up momentum p=(kg times m)/s So this can be broken down further to be p=(mass times distance)/time. Time is the same for both so. m1d1=m2d2. m1/m2=.87/.45 so I hope that helps.

I would say the impulses are the same based on the fact that they are moving at different velocities... and have different masses. Impluse if based on change in momentum so I believe the an equal impulse is given to each cart in an opposite direction (so they do cancel out). So there are parts B and C. :)

Thank you for the quick reply. Well to figure out the velocity ratio, this is what I derived.

Because of you I was able to figure out the velocity ratio.

va/vb = da/t * t/db (*Note: the velocities are just equal to their respective d/t equations, where time will cancel out, therefore you are just left with the ratio of displacements.)

va/vb = .45m/.87m = .51

I just want to know how exactly I could prove the impulse, and acceleration, but I will keep working at it.

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