A rope is passed over a pulley of radius R . Two people of…
A rope is passed over a pulley of radius $R$. Two people of equal mass $m_A = m_B = m$ hang from its ends, A on one side and B on the other, at rest. What happens to B if A starts climbing the rope at a constant speed $v$ relative to the rope? Neglect the masses of the rope and pulley, and friction.
Explanation of Method
Two people on a rope over a pulley.
Step 1: Same force on both
A massless rope over a massless, frictionless pulley has the same tension $T$ all along. Each person is acted on by $T$ upward and $mg$ downward: $m\,a_A = T - mg$.
\[ m\,a_B = T - mg \;\Longrightarrow\; a_A = a_B. \]
Both start at rest, so at every moment they have the same velocity relative to the ground.
Step 2: How fast
Let both move up at $u$. The rope on B's side then moves up with B at $u$, so the rope on A's side moves down at $u$. A's speed relative to the rope is
\[ v = u - (-u) = 2u \;\Longrightarrow\; u = \frac{v}{2}. \]
\[\begin{gathered}\text{B rises too, as fast as A:} \\ u = \tfrac{v}{2} \text{ upward} \end{gathered}\]
Result
\[\boxed{\,\begin{gathered}\text{B rises too, as fast as A:} \\ u = \tfrac{v}{2} \text{ upward} \end{gathered}\,}\]
Remark
About the original solution. It concluded that B goes down with speed $v$, using "conservation of momentum". That does not apply: the pulley's axle and gravity are external forces, and the upward pull on the system changes when A starts climbing. What does hold is the equal tension on both sides, which gives equal accelerations: A and B rise together and stay level with each other. The radius $R$ does not matter. If the rope is elastic, it stretches a little while A starts, but the tension is still equal on both sides, so the conclusion is the same.