A train of mass M is traveling with uniform velocity on a level line. The last carriage, whose mass is m . becomes uncoupled at an instant. The driver discovers it after traveling a distance / and stops the engine. The resistance to the motion is unifom and proportional to the weight. The pull of the engine is uniform. When both parts of the train come to rest, the distance between them is

Step-by-step solution

Suppose last carriage gets detached at t=0 . Suppose resistance to motion = mg Distance travelled by last carriage before it stops = 1 2 m v^2 m g = v₀^2 2 g While the train travels a distance of l, its velocity will increase because of extra pull generated by the engine. The net force acting on remaining train aligned & F= m g & a= m g (M-m) aligned velocity r = v₀^2+ 2 m g l (M-m) After driver puts off the engine, the distance travelled by train aligned l^ & = 1 2 (M-m) v^2 (M-m) g = v^2 2 g = v₀^2+ 2 m g l M-m 2