A train consisting of 8 cars of mass m each is moving down a straight track with velocity ??. The train collides and couples with a stationary coal car of mass ?? = 3?. a. What is the final speed of the train? b. What would the mass of the coal car have to be in order to reduce the speed of the train by 50%?
(a) Solve this problem using conservation of momentum.
Initial momentum of the system = Pi = m*v0
Since the train couples with the stationary coal car.
Hence, final mass, m' = m+Mc = m+3m = 4m
Suppose, v' is the final velocity of the train.
So, final momentum, Pf = m'*v' = 4m*v'
Apply conservation of momentum -
m*v0 = 4m*v'
=> v' = v0 / 4 = 0.25*v0 (Answer)
(b) Suppose in this case, mass of the coal car = Mc'
As per the condition -
m*v0 = (m+Mc')*0.50*v0
=> m+Mc' = 2m
=> Mc' = 2m - m = m
So, the mass of the coal car should be equal to the mass of the train.
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