Q23 / 92 · NTA Abhyas (JEE Main) 2020 · Physics · Work Power Energy

The figure shows two blocks of mass $m$ each, connected by an ideal unstretched spring and then placed on a frictionless floor. If the blocks are given velocities $v_{0}$ and $2 v_{0}$ as shown, then the maximum extension in the spring is

A
$\sqrt{\frac{m v_{0}^{2}}{k}}$
B
$\sqrt{\frac{m v_{0}^{2}}{2 k}}$
Correct answer
C
$\sqrt{\frac{9 m v_{0}^{2}}{2 k}}$
D
None of these

Solution

$2 m v = m 2 v_{0} - m v_{0}$

$v = \frac{v_{0}}{2}$

$w_{s} = k_{f} - k_{i}$

$- \frac{k}{2} (x^{2} - 0^{2}) = \frac{1}{2} 2 m v^{2} - (\frac{1}{2} m v_{0}^{2} + \frac{1}{2} m {(2 v_{0})}^{2})$

$= m {(\frac{v_{0}}{2})}^{2} - \frac{5 m v_{0}^{2}}{2}$

$\Rightarrow - \frac{k x^{2}}{2} = - \frac{9 m v_{0}^{2}}{4}$

$x = \sqrt{\frac{9}{2} \frac{m v_{0}^{2}}{k}}$

Hence, the correct answer is option: 'C'.

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The figure shows two blocks of mass m each, connected by an ideal unstretched spring and then placed on a frictionless floor. If the blocks are given velocities v0 and 2v0 as shown, then the maximum extension in the spring is

Solution

More Work Power Energy Questions

The figure shows two blocks of mass $m$ each, connected by an ideal unstretched spring and then placed on a frictionless floor. If the blocks are given velocities $v_{0}$ and $2 v_{0}$ as shown, then the maximum extension in the spring is