Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a light inextensible string which passes over a fixed pulley. Initially, mass $m_{1}$ moves with a velocity $v_{0}$ when the string is not taut. Neglect friction at all contact surfaces. Find the velocity of the blocks just after the string is taut, if $m_{1} = 2 kg, m_{2} = 3 kg$ and $v_{0} = 5 m/s$

2 m/s

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3 m/s

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5 m/s

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0 m/s

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Solution

The correct option is A $2 m/s$
Tension developed in both parts of the string will be same just after the string become taut.
Hence, impulsive forces acting on both blocks will be the same
$∴_{1} =_{2} = \to J . . (i)$

Because of string constraint, speed of both blocks will be equal. Let it be $v$ just after the string becomes taut.

Applying impulse-momentum theorem (taking positive direction rightward):
$J = m Δ v$
For block $m_{1}$ :
$- J = m_{(} v_{f} - v_{i}) = 2 [(+ v) - (+ v_{0})] = 2 (v - v_{0})$
$\Rightarrow - J = 2 (v - 5) . . (i i)$

For block $m_{2}$ :
$- J = m_{2} (v_{f} - v_{i}) = 3 (- v - 0) = - 3 v . . . (i i i)$

Equating Eq. $(i i)$ and $(i i i)$ :
$2 v - 10 = - 3 v$
$∴ v = \frac{10}{5} = 2 m/s$

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