A man weighining 80 kg is standing in a trolly weighing 320 kg the trolly is resting on frictionless horizontal rails.If the man startswalking on the trolly with a speed of 1m/s and after 4sec his placement is belived to the ground

Here, the external force on the system of man+trolley is zero along horizotnal direction.

There fore conservation of LINEAR MOMENTUM is applicable.

(M+m)(0) = (M)(V) + (m)(V-u).   M, m -> mass of the trolley & man. V,u -> vel of trolley wrt ground &man wrt the trolley.

=> V = mu/(m+M) = 0.2m/s.

vel of the man wrt ground = V-u = 0.2 -1 = -0.8m/s.

The position of the man wrt the initial position is (-0.8 * 4) = -3.2m.

here , it is assumed that the man is moving towars left, so we got -ve displacemnt of the man.

the previous answer was rite accordin to me . but i dint understand why u took the velocity after interaction as m(v-u)??????