Question

The pulley (figure) is fixed, light and smooth. The string is light and mass m < mass M. At t = 0, M and m are at rest, when the system is released. At time t = $t_1$ the mass 'M' is displaced by 'd'. The speed of the big block M at time $t_1$ with respect to the pulley is:

• A. $\sqrt{\frac{2(M+m)gd}{M}}$
• B. $\sqrt{\frac{2Mgd}{M+m}}$
• C. $\sqrt{\frac{2(M-m)dg}{M+m}}$
• D. $\sqrt{\frac{2(M-m)dg}{M}}$

Answer 1

The correct option is C $\sqrt{\frac{2(M-m)dg}{M+m}}$

Let $a$ be the acceleration of the system as well as of the block of mass M.
for mass $M$,

$Ma=Mg-T....\left(1\right)$
for mass $m$,

$ma=T-mg.....\left(2\right)$
$\left(1\right)+\left(2\right),a=\frac{(M-m)g}{(M+m)}$
now using the formula, $v^2-u^2=2as$
here, $u=0,s=d$ so , $v^2=2ad=\frac{2(M-m)gd}{(M+m)}$
$\therefore v=\sqrt{\frac{2(M-m)gd}{(M+m)}}$
Ans:(C)