Question

Assuming the surfaces to be frictionless. The smaller block m is moving horizontally with acceleration a and vertically downwards with acceleration 2a. Then the acceleration of smaller block m with respect to ground

• A. $\frac{2\sqrt{5}mg}{(5m+M)}$
• B. $\frac{2mg}{(5m+M)}$
• C. $7\sqrt{5}g$
• D. none of these

Answer 1

The correct option is A $\frac{2\sqrt{5}mg}{(5m+M)}$

Free body diagram for m

For m,

$mg-T=m2a$ ........(i)

N = ma .......(ii)

Free body diagram for M

For M

2T - N = Ma ...........(iii)

On solving, $a=\frac{2mg}{(M+5m)}$

$\therefore$ Net acceleration of m,

$a_m=\sqrt{4a^2+a^2}=\sqrt{5}a=\frac{2\sqrt{5}mg}{(5m+M)}$