Momentum Problem: Final Momenta, Kinetic Energies Compared

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In the discussion about the momentum and kinetic energy of two carts, it is established that momentum is conserved, meaning the total initial momentum of zero results in equal and opposite final momenta for the carts. Cart B, having twice the mass of Cart A, will have a lower velocity, leading to a situation where the final momenta are equal in magnitude but opposite in direction. The kinetic energy of each cart is derived from their respective velocities and masses, with the kinetic energy of Cart A potentially being greater due to its higher velocity. The relationship between kinetic energy and momentum is highlighted, suggesting that kinetic energy can be expressed in terms of momentum. Overall, the discussion emphasizes the conservation of momentum and the interplay between mass, velocity, and kinetic energy.
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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, it pushes against cart B, and the carts move apart. How do the magnitudes of the final momenta p and the kinetic energies K compare?

Please tell me if my approach is correct, because I'm not very clear on momentum:
Well if the piston is pushing against Cart B, then its momentum would be bigger than Cart A because it has both a greater velocity and a greater mass and momentum= velocity * mass. The kinetic energies- wouldn't the KE of cart A be bigger because it has more PE?
 
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Maiia said:
Please tell me if my approach is correct, because I'm not very clear on momentum:

Hi Maiia! :smile:

If there are no external forces, then momentum is always conserved.
 
that means that momentum inital = momentum final and since momentum inital is zero than momentum final is zero which means that final momentum for both is zero?
hmm but if i use KE= 1/2mv^2 then the KE of cart B would be bigger...?
 
Maiia said:
that means that momentum inital = momentum final and since momentum inital is zero than momentum final is zero which means that final momentum for both is zero?

Yes … the total momentum is zero, so one velocity will be twice the other (which one? :wink:)
hmm but if i use KE= 1/2mv^2 then the KE of cart B would be bigger...?

Hint: KE = momentum times what? :smile:
 
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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