Two blocks of masses 2 kg and 3 kg are connected by string of 1 m. At any instant the velocities of blocks mass 2 kg 3 kg are 5 m/s in opposite direction and perpendicular to the length of string. (Assume gravity free space)

tension in the string is 300 N

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tension in the string is 120 N

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tension in the string is 75 N

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tension in the string is 200 N

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Solution

The correct option is B tension in the string is 120 N
First of all, find velocity of center of mass of system of bodies.
We know, velocity of center of mass of system of bodies is given by,

$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
Here, $m_{1} = 2 k g, m_{2} = 3 k g$

$v_{1} = 5 m / s$ $v_{2} = - 5 m / s$

so, $v_{c m} = \frac{(2 \times 5 + 3 \times (- 5))}{(2 + 3)} = - 1 m / s$

Now, velocity of $m_{1}$ with respect to centre of mass, $v_{c m} = 5 - (- 1) = 6 m / s$

Distance between two bodies is $r = 1 m$ .

Dis. b/w $m_{1}$ and position of centre of mass, $r_{1} = \frac{m_{2} \times r}{m_{1} + m_{2}}$

$= 3 / 5 \times 1 = 0.6 m$

Tension of string $= \frac{m v_{1 c m}^{2}}{r_{1}} = \frac{2 \times 6^{2}}{0.6} = 120 N$

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Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be: