Two railroad freight cars with masses 130 {\ m Mg} and 140 {\ m Mg} approach wit

Question

Two railroad freight cars with masses 130 { m Mg} and 140 { m Mg} approach with equal speeds of 0.360 { m m/s}. They collide, the lighter car rebounding opposite its original direction at 0.300 { m m/s}.

(a) 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.

(b)What fraction of the original kinetic energy was lost in this inelastic collision?
Express your answer using two significant figures.

Explanation / Answer

The mass of first car is m1 = 130 kg
The mass of the second car m2 = 140 kg
The speed of approach is u1 = -u2 = 0.36 m/s
The final speed of first car is v1 = -0.3 m/s

a)
According to conservation of linear momentum
m1u1+m2u2 = m1v1+m2v2
130*0.36+140*-0.36=130*-0.3+140*v2
v2 = 0.25 m/s

b)
The initial KE of the system is
KEi = 0.5*(m1+m2)*u^2
KEi = 0.5*(130+140)*0.36^2
KEi = 17.496 J

The final KE of the system is
KEf = 0.5*m1v1^2 + 0.5*m2v2^2
KEf = 0.5*130*0.3^2 + 0.5*140*0.25^2
KEf = 10.225 J

The los  of KE due to the inelastic colllision is

KE = KEi - KEf

KE = 17.496 J - 10.225 J

KE = 7.271 J

The fraction of the original kinetic energy was lost in this inelastic collision is

KE/KEi = 7.271/10.225

KE/KEi = 0.71

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