Momentum and Kinetic Energy problem

AI Thread Summary
In the discussion about the momentum and kinetic energy of two carts on a track, it is established that the final momentum of both carts will be equal due to the conservation of momentum principle, with cart A having a greater velocity because of its lower mass. The kinetic energy comparison is more complex, as cart A, despite having a higher velocity, may have less kinetic energy than cart B due to its mass being lower. The conversation emphasizes that while potential energy is not applicable in horizontal movement scenarios, the energy from the spring in cart A is relevant. Participants clarify that the problem is more of a conceptual exercise rather than purely mathematical. Overall, the key takeaway is the relationship between mass, velocity, and energy in this context.
BuddyGoodness
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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.
 
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Welcome to PF!

Hi BuddyGoodness! Welcome to PF! :smile:
BuddyGoodness said:
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. :frown:

Call the speeds vA and vB, 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 :wink:)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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