A train of mass M was moving with a uniform velocity on a horizontal plane. The last bogie of the train suddenly got detached. The driver could detect the detachment after he had travelled through a distance L, where he stopped the engine. If resistance against motion of train is directly proportional to the mass, show that, when both the detached bogie and the rest of the train come to rest, the distance between them ML/(M – m), where m is the mass of the detached bogie. Assume that the engine maintains a constant pull.

Open in App

Solution

Dear Student,

Let the initial velocity of the train is V

$l e t t h e p u l l b e F r e s i s \tan c e a g a i n s t m o t i o n = F = k M, \sin c e i t i s m o v i n g w i t h c o n s \tan t s p e e d . r e s i s \tan c e f o r b o g i e F = k m w h e r e i s t h e c o n s \tan t a c c n o f b o g i e = - k 0 = V^{2} - 2 k S S = \frac{V^{2}}{2 k} i s t h e d i s \tan c e t r a v e l l e d b y t h e b o g i e a f t e r d e t a c h m e n t . a c c n o f t h e r e s t o f t h e t r a i n c a n b e f o u n d f o r d i s \tan c e L a = \frac{k m}{(M - m)} f i n a l v e l o c i t y t h r o u g h t h e V'^{2} = V^{2} + \frac{2 k m L}{(M - m)} n o w d i s \tan c e t r a v e l b y t h e t r a i n a f t e r e n g i n e b e s t o p p e d . d e a c c n = - k 0 = V'^{2} - 2 k S' S' = \frac{V^{2} + \frac{2 k m L}{(M - m)}}{2 k} t o t a l d i s \tan c e t r a v e l b y t r a i n = \frac{V^{2} + \frac{2 k m L}{(M - m)}}{2 k} + L = \frac{V^{2}}{2 k} + \frac{m L}{(M - m)} + L t o t a l d i s \tan c e t r a v e l b y b o g i e = \frac{V^{2}}{2 k} d i s \tan c e b e t w e e n t h e t r a i n a n d b o d i e = \frac{V^{2}}{2 k} + \frac{m L}{(M - m)} + L - \frac{V^{2}}{2 k} = \frac{m L}{(M - m)} + L = \frac{M L}{(M - m)} R e g a r d s$

Suggest Corrections

Similar questions

A train consists of 12 bogies, each bogie being 18 metre long. The train crosses a tree in 18 seconds. Due to some reasons, two bogies were detached. In how much time will the train now cross the tree?

Q. A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The ratio of distance covered by the boggy and distance covered by the train when the boggy has stopped is

Q. A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

v_{A}

and

v_{B}

towards each other on smooth parallel tracks. At an instant a boy of mass m from bogie A and a boy same mass from bogie B exchange their position by jumping in a direction normal to the track, then bogie A stops while B keeps moving in the same direction with new velocity

v_{B}

. The initial velocities of bogie A and B are given by :

Q. A train of mass M is driven with acceleration 'a' along a straight level track against a constant external resistance R. When the velocity of the train is v, the rate at which the engine of the train is doing work will be

Join BYJU'S Learning Program