Common velocity after truck starts pulling a car

AI Thread Summary
When a truck accelerates and begins to pull a car, the common velocity just after the car starts moving can be determined using the conservation of momentum. The discussion highlights that conservation of kinetic energy does not apply in this scenario due to the inelastic nature of the interaction between the truck and the car. The differing results from momentum and energy calculations stem from the fact that momentum is conserved while kinetic energy is not, as internal forces affect energy but not momentum. The key takeaway is that momentum conservation is the correct approach for determining the common velocity immediately after the car begins to move. Understanding these principles is essential for solving related physics problems effectively.
nikolafmf
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Homework Statement



Truck is connected to a car with a rope of given length. The truck starts accelerating with a given acceleration and at some moment starts pulling the car. What is their common velocity just after the car starts moving?

Homework Equations



I know how to calculate the velocity of the truck just before the car starts moving. Than I use the conservation of momentum or the conservation of energy.

The Attempt at a Solution



I get different results when I use momentum and when I use kinetic energy. Which one is conserved, which is not and why?
 
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Of course you get different results. In order for the car to instantly get up to the speed of the truck you have what is essentially a completely inelastic collision between the two.
 
Thank you. So conservation of momentum would be the right thing. I have also found that inner forces in the system are relevant for the change of the kinetic energy of a system, but not for the change of momentum. And I do have change of inner forces in this system.
 
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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