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Velocity of COM

Three blocks ...

Question

Three blocks of masses $M_{1}, M_{2}$ and $M$ are arranged as shown in the figure. All the surfaces are frictionless and string is inextensible. Pulleys are light. A constant force $F$ is applied on block of mass $M_{1}$ . Pulleys and string are light. Part of the string connecting both pulleys is vertical and part of the string connecting pulleys with masses $M_{1} and M_{2}$ are horizontal.

(P) Acceleration of mass $M_{1}$ (1) $\frac{F}{M_{2}}$

(Q) Acceleration of mass $M_{2}$ (2) $\frac{F}{M_{1} + M_{2}}$

(R) Acceleration of mass $M$ (3) zero

(S) Tension in the string (4) $\frac{M_{2} F}{M_{1} + M_{2}}$

P - 1, Q - 2, R - 2, S - 3

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P - 2, Q - 2, R - 3, S - 4

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P - 2, Q - 4, R - 3, S - 1

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P - 2, Q - 2, R - 3, S - 3

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Solution

The correct option is B $P - 2, Q - 2, R - 3, S - 4$

Let the tension generated in string be $T$ and accceleration of block $M_{1} be a_{1}$ .
So, from string constraint the acceleration of block $M_{2}$ will be $a_{1}$
also, let acceleration of block $M$ be $a_{2}$
Now, drawing FBD of all the blocks we have

Since the motion of blocks are in horizontal, considering horizontal forces only we have
for block $M_{1} = F - T = M_{1} a_{1} . . . . . (i)$
for block $M_{2} = T = M_{2} a_{1} . . . (i i)$
from equation $(i)$ and $(i i)$
$F = (M_{1} + M_{2}) a_{1} \Rightarrow a_{1} = \frac{F}{M_{1} + M_{2}}$
and $T = M_{2} a_{1}$
$\Rightarrow T = \frac{M_{2} F}{M_{1} + M_{2}}$

for block $M :$
$T - T = M a_{2}$
$\Rightarrow a_{2} = 0$
$∴$ option $(b)$ is correct answer.

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(P) Acceleration of mass $M_{1}$	(1) $\frac{F}{M_{2}}$
(Q) Acceleration of mass $M_{2}$	(2) $\frac{F}{M_{1} + M_{2}}$
(R) Acceleration of mass $M$	(3) zero
(S) Tension in the string	(4) $\frac{M_{2} F}{M_{1} + M_{2}}$