Question

A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. If the car recoils with a speed v backward on the rails, with what velocity is the man approaching the engine?

Answer 1

When a man in a car moves, Using the third law of motion, we can say that there is going to be a reaction to this motion. The car will move in such a way that the motion of the man is nullified or the center of mass doesn’t move much.

Complete answer: As per the conditions given in the question, let us suppose ${V_{mE}}$ be the velocity of a man with respect to the earth. ${V_{mC}}$ (v ’) be the velocity of the man with respect to the car and ${V_{cE}}$ (v) be the velocity of the car with respect to the earth.
Now, the Velocity of a man with respect to the earth is equal to the velocity of a man with respect to a car minus the velocity of the car with respect to earth.
${V_{mE}} = {V_{mC}} - {V_{cE}}$
${V_{mE}} = v' - v$
Now, as the net force applied is zero, therefore, the momentum is conserved, which implies that
${V_{cM}} = 0$
Now, by using law of conservation of momentum, we get,
$\dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}} = 0$
$M\left( { - v} \right) + m\left( {v' - v} \right) = 0$
$Mv = m\left( {v' - v} \right)$
$v' = \dfrac{{(m + M)v}}{m}$
Therefore, It is inferred that, the man is approaching the engine with velocity of $v' = \dfrac{{(m + M)v}}{m}$.

Note: It is crucial to note the sign conventions. The motion of the car is opposite to the motion of the man. This means that their velocities will be negative with respect to each other. One usually skips this point and lands up incorrect solutions.