Two railroad freight cars with masses 130 Mg and 140 Mg approach with equal spee

Question

Two railroad freight cars with masses 130 Mg and 140 Mg approach with equal speeds of 0.300 m/s. They collide, the lighter car rebounding opposite its original direction at 0.240 m/s. Find the velocity of the heavier car after the collision. Assume the original direction of the lighter car is positive. Express your answer to two significant figures and include the appropriate units. What fraction of the original kinetic I energy was lost in this inelastic collision? Express your answer using two significant figures.

Explanation / Answer

you need to set up a sign convention. "Assume the direction of the lighter car is +"
v1i = +0.300 m/s
v2i = -0.300 m/s
v1f = -0.240 m/s
v2f = +0.20m/s (according to your answer in 1st part)

m1 = 130Mg

m2 = 140Mg

KEi = 1/2*m1*v1i^2 + 1/2*m2*v2i^2
KEf = 1/2*m1*v1f^2 + 1/2*m2*v2f^2

Fraction = (KEi - KEf)/KEi
fraction = 1 - KEf/KEi
fraction = 1 - (m1*v1f^2 + m2*v2f^2) / (m1*v1i^2 + m2*v2i^2)

fraction = 1 - (130*(0.240)^2 + 140*(0.20)^2)) / (130*(0.300)^2 + 140*(0.300)^2) = 0.4613

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