Conservation of energy or conservation of momentum

AI Thread Summary
In an inelastic collision, such as a bullet hitting a block on wheels, momentum is conserved while kinetic energy is not. The discrepancy in results arises because the conservation of momentum equation yields a different velocity after the collision compared to the conservation of kinetic energy equation. Since mechanical energy is lost in inelastic collisions, using kinetic energy conservation leads to incorrect results. The correct approach for calculating the final velocity of the system is to apply the conservation of momentum principle. Understanding this distinction is crucial for solving problems involving inelastic collisions.
salami
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OK this is about a bullet hitting a block on wheels (inelastic collision), no friction or energy losses in the problem.

I'm trying to find Velocity of the system after collision

Why is it that if I use conservation of momentum I get a different answer of V than if I used conservation of Kinetic energy.

Momentum
m1*v1=(m1+m2)*v2

K energy
0.5*m1*v1^2=0.5*(m1+m2)*v2^2

You can make up your own numbers...I just don't understand the physics!
 
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It's an inelastic collision--energy is not conserved. Mechanical energy will be lost.
 
The reason you get different numbers is that mechanical energy is NOT conserved in an inelastic collision, while momentum is.
 
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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