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.
Answers
Complete question:
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
(a) is greater than 1/2 mv² if he walks along rails.
(b) is less than 1/2 mv² is he walks along rails.
(c) is equal to 1/2 mv² is he walks normal to rails.
(d) can never be less than 1/2 mv²
Answer:
The work done by him (b) is less than 1/2 mv² is he walks along rails and (c) is equal to 1/2 mv² is he walks normal to rails.
Explanation:
When the man starts walking along the rail, the man gains V velocity.
On applying conservation of momentum, we get,
MV = m(v - V)
∴ V = mv/(m + M)
The kinetic energy is provided to man and the car, thus, the kinetic energy is:
W = 1/2 m(v - V)² + 1/2mV²
W = 1/2 (mM/(m + M)) v²
The value of (mM/(m + M)) is less than m and M. Thus, work done is less than 1/2 mv². So, option (b) is correct.
When the man moves normal to the rail with velocity v, car will not move.
The kinetic energy of the man is equal to 1/2 mv². Thus, work done is is equal to 1/2 mv². So, option (c) is correct.
Answer:
If he walks along the rails he pushes the car back on gaining velocity V so some so the work done will be less than its gained kinetic energy but, when he walks normal to the rail his total work has converted into kinetic energy so work done will be
2 1 mv 2
.