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Impulsive Forces in Collision

Three blocks ...

Question

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings as shown on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ , the tension $T_{2}$ will be

20

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40

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10

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32

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Solution

The correct option is D $32$ N
$Given :$ $T_{3} = 40 N$ , $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$
$Step 1: Acceleration of complete system$
Total mass of system, $m = m_{1} + m_{2} + m_{3} = 10 + 6 + 4 = 20 k g$
Acceleration of system $= a$
Applying Newton's law, $Σ F = m a$
$\begin{matrix} ∴ a = \frac{T_{3}}{m} = \frac{40 N}{20 K g} = 2 m / s^{2} \end{matrix}$

$Step 2: Calculation of$ $T_{2}$
From F.B.D OF ( $m_{1} + m_{2}$ ) Force $T_{2}$ is pulling $1$ and $2$ blocks.
If we consider $m_{1}$ and $m_{2}$ as a system then $T_{1}$ will get cancel as $T_{1}$ acts as internal force.
(As shown in figure)
Therefore from Newton's second law-
$Σ F = m a$
$\begin{matrix} T_{2} = (m_{1} + m_{2}) a = (10 + 6) \times 2 = 32 N \end{matrix}$

Hence, Option $(D)$ is correct answer.

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