A railroad cart with a mass of m1 = 12.4 t is at rest at the top of an h = 10.6

Question

A railroad cart with a mass of m1 = 12.4 t is at rest at the top of an h = 10.6 m high hump yard hill.

After it is pushed very slowly over the edge, it starts to roll down. At the bottom it hits another cart originally at rest with a mass of m2 = 19.2 t. The bumper mechanism locks the two carts together. What is the final common speed of the two carts? (Neglect losses due to rolling friction of the carts. The letter t stands for metric ton in the SI system.)

Please show all work and use sig-figures

Thank you for all your help.

Explanation / Answer

mass m1 = 12.4 ton = 12.4*103 kg mass m2 = 19.2 ton = 19.2*103 kg height of the hill h = 10.6 m .......................................................................... from law of conservation of energy,             m1gh + 0 = (1/2)m1v12    hence, velocity of the first cart before the collision with second cart is               v1 = 2gh     ......(1) after , collision from law of conservation of momentum,           m1v1+0 = (m1 + m2)V substitute the equation (1), in above equation, we get          m1(2gh) = (m1 + m2)V therefore, common speed of the two carts is           V = m1(2gh) / (m1 + m2)    ...... (2) substitute the given data in equation(2), we get          V = (12.4*103 kg)[(2)(9.8 m/s2)(10.6 m)] / (12.4*103 kg + 19.2*103 kg)              = 5.65 m/s    hence, velocity of the first cart before the collision with second cart is               v1 = 2gh     ......(1) after , collision from law of conservation of momentum,           m1v1+0 = (m1 + m2)V substitute the equation (1), in above equation, we get          m1(2gh) = (m1 + m2)V therefore, common speed of the two carts is           V = m1(2gh) / (m1 + m2)    ...... (2) substitute the given data in equation(2), we get          V = (12.4*103 kg)[(2)(9.8 m/s2)(10.6 m)] / (12.4*103 kg + 19.2*103 kg)              = 5.65 m/s from law of conservation of momentum,           m1v1+0 = (m1 + m2)V substitute the equation (1), in above equation, we get          m1(2gh) = (m1 + m2)V therefore, common speed of the two carts is           V = m1(2gh) / (m1 + m2)    ...... (2) substitute the given data in equation(2), we get          V = (12.4*103 kg)[(2)(9.8 m/s2)(10.6 m)] / (12.4*103 kg + 19.2*103 kg)              = 5.65 m/s

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