A man of mass m is standing on a stationary flat car of mass M. The car can move without friction along horizontal rails. The man starts walking with velocity v relative to the car. Work done by him

Options

A. is greater than `1/2 mv^2` if the walks along rails.
B. is less than `(1)/(2) mv^(2)` if the walks along rails.
C. is equal to `(1)/(2) mv^(2)` if he walks normal to rails.
D. can never be less than `(1)/(2) mv^(2)`.

Answer

B, C

Explanation

If the man starts walking with velocity v (relative to the car) along the rails, the car recoils. Let the velocity with which the car recoils be V.
Then, resultant velocity of the man will be equal to (v- V). Since the car moves without friction, there is no external horizontal force on the system of man and car. Hence, applying conservation of momentum, we get
`MV= m(v-V)` or `V= (mv)/( m+M)`
Work done by the man is used to provide kinetic energy to the car and to the man himself.
Hence work done by him
`W= (1)/(2) m (v-V)^(2) + (1)/(2) m V^(2) = (1)/(2) ((m M)/( m+ M) ) v^(2)`
But [m M/m+ M] is less than m and M both, therefore W is less than `(1)/(2) mv^(2)`. Hence, option (a) is wrong and (b) is correct.
If the man moves normal to the rails, then the car will not move. Hence, in that case, man moves along with velocity v. Hence, work done by him is used to provide kinetic energy to his own body which is equal to `(1)/(2)mv^(2)`. Hence, option ( c) is correct.
Since option (b) is correct, option (d) is wrong.