A locomotive of mass 21X10^3 kg travelling at 2.0m/s runs into an identical stat
ID: 1704530 • Letter: A
Question
A locomotive of mass 21X10^3 kg travelling at 2.0m/s runs into an identical stationary locomotive.
a) if the collision is perfectly inelastic, and the two trains lock together, what is their speed after the collision?
b) if the collision is perfectly inelastic, what fraction of the energy is lost after the collision?
c) if the collision is perfectly elastic, find the motion of the locomotives after the collision.
d) if the collision is perfectly elastic and lasts for 1.5s, find the average force exerted on the stationary locomotive during the collision.
Explanation / Answer
(a) Conservation of momentum (momentum = mass x speed): let v be the final speed
Total momentum before collison = total momentum after collison
(21000 x 2.0) + (21000 x 0) = (21000 + 21000)v
v = 1.0 m/s
(b) Kinectic energy = 0.5 x mass x (speed)2
Total energy before collison = 0.5 x 21000 x (2.0)2 = 42000 J
Total energy after collision = 0.5 x (21000 + 21000) x (1.0)2 = 21000 J
Energy lost = 42000 - 21000 = 21000 J
Fraction of energy lost = 21000/42000 = 0.5
(c) If the collision is perfectly elastic, then both momentum and kinetic energy are conserved.
The first locomotive will come to a complete stop (i.e speed will drop to 0 m/s), while the second locomotive will move off at 2.0 m/s (i.e. with the original speed of the first train).
So the total momentum is the same before and after the collison. The total kinetic energy is also the same before and after the collision.
(d) Force exerted = change in momentum of first locomotive/time
= ((21000 x 2.0) - (21000 x 0))/1.5 = 28000 N
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