Conservation of mechanical energy

Question

Is the mechanical energy after the collision conserved or not conserved? Explain.

Answer 1

Cart A has Potential energy of 2MgH and when it gets to the bottom of the hill, all potential energy has converted to kinetic energy so:

2MgH = 1/2Mv²

4gH = v²

v = 2√(gH)

Cart A then has momentum of Mv = 2M√(gh)

Cart B has Potential energy of 2MgH and when it gets to the bottom of the hill, all potential energy has converted to kinetic energy so:

2MgH = 1/2(2M)v²

2gH = v²

v = √(2gH) but since it is in the opposite direction, we can say v = -√(2gH)

Cart B then has momentum (mv) = -2M√(2gh)

Momentum (P) is conserved so:

P_{cart A} + P_{cart B} = P_{stuck carts}

2M√(gh) - 2M√(2gh) = P_{stuck carts}

2M√(gh) - 2M√2√(gh) = P_{stuck carts}

2M√(gh)(1 - √2) = P_{stuck carts} = 3Mv_{stuck carts}

v_{stuck carts} = [2M√(gh)(1 - √2)]/(3M)

v_{stuck carts} = [2√(gh)(1 - √2)]/3

Then the kinetic energy (1/2mv²) would be (1/2)(3M)[2√(gh)(1 - √2)/3]² = (3/2)M(4gh(3 - 2√2))/9 = (2/3)MgH(3 - 2√2) ≈ 0.114MgH

Initial Total Mechanical Energy = 4MgH while final is 0.114MgH therefore, energy is NOT conserved