Momentum and Kinetic Energy problem

[BuddyGoodness]

Homework Statement

Two carts are put back-to-back on a track. Cart A has a spring-loaded piston; cart B, which has twice the inertial mass of cart A, is entirely passive. When the piston is released, cart A pushes against cart B, and the carts move apart. How do the magnitudes of their final momenta p and kinetic energies K compare?

1. pA > pB, KA > KB.
2. pA > pB, KA = KB.
3. pA > pB, KA < KB.
4. pA = pB, KA > KB.
5. pA = pB, KA = KB.
6. pA = pB, KA < KB.
7. pA < pB, KA > KB.
8. pA < pB, KA = KB.
9. pA < pB, KA < KB.

Homework Equations

I think this is more of a thought exercise rather than a mathematical problem. Relevant equations are P - mass x velocity and KE = 1/2mass(v)²

The Attempt at a Solution

I would think that final momentum would be the same cause the spring would push them both back with the greater mass cart being at a lesser velocity momentum of A = -Momentum of B = conservation of momentum. This is basically like an elastic collision problem though I am just considering the problem after they have bounced off each other.


As for kinetic energy I don't really know. No mention is made of potential energy but if the total energy is conserved then cart A's velocity would be greater so in the equation for potential energy (mass x gravity x displacement) cart A would have more displacement so it's kinetic energy would be lesser than Cart B. Am I right in my thinking here?


Also as a side question: am I right in thinking potential energy is ok to use in horizontal movement? I have been replacing the height in the equation mgh with displacement in horizontal movement problems.

 

[tiny-tim]

Welcome to PF!

Hi BuddyGoodness! Welcome to PF!

I think this is more of a thought exercise rather than a mathematical problem. Relevant equations are P - mass x velocity and KE = 1/2mass(v)²

Nooo … physics is equations, and this is no exception.

Call the speeds v_A and v_B, and the energy of the spring E.

Then use the usual equations.

Also as a side question: am I right in thinking potential energy is ok to use in horizontal movement? I have been replacing the height in the equation mgh with displacement in horizontal movement problems.

If you mean the mgh PE, then NO!

The spring energy is entirely different …

but you don't need to know how to calculate it …

as I said, just call it E (all you need to know is that E > 0 )