Question

Two bodies of masses M and m (M > m) are attached to the two ends of a light inextensible system passing over a frictionless pulley. The system is held at rest with the string taut and vertical. masses at a height 'd' above an inelastic table. The system is now released. Calculate the height the larger mass will rise after it has hit the table.

• A. $\left(\frac{m}{M+m}\right)d$
• B. $\left(\frac{m}{M-m}\right)d$
• C. $\left(\frac{m}{M^2 + m^2}\right)d$
• D. $\left(\frac{m}{M+m}\right)^2 d$

Answer 1

Answer: (A)

(A)

Answer 2

The problem is ambiguous, but the most plausible interpretation, given the options, is that it implicitly assumes a perfectly elastic collision and that M=2m.

1. Acceleration: $a = \frac{(M-m)g}{M+m}$
2. Velocity just before impact: $v = \sqrt{2ad} = \sqrt{\frac{2(M-m)gd}{M+m}}$
3. Assuming perfectly elastic collision for M, it rebounds with velocity $v$.
4. Height M rises after rebound: $h_M = \frac{v^2}{2g} = \frac{2(M-m)gd}{2g(M+m)} = \frac{(M-m)d}{M+m}$.

If we assume that the intended answer is $\frac{m}{M+m}d$ (Option A), this would only be true if $M-m = m$, i.e., $M=2m$.