a. A rail road car of mass M is moving without friction on a straight horizontal track with a velocity ct. A man of mass m lands on it normally from a helicopter. What will be the new velocity of the car? ltbgt b. If now the man begins to run on it with speed um with respect to car in a direction opposite to motion of the car, what will be the new velocity of the car?

Options

A. `v_1 = (mu)/((M + m))`
B. `v_1 = (m u)/((M + m))`
C. `v_1 = (mu)/((M - m))`
D. `v_1 = (Mu)/((M - m))`

Answer

C

Explanation

In the direction of motion of the train `barF_(ex) = 0 implies Mu = (m + M) V_1 implies V_1 = (Mu)/(m + M)`.