Q53 / 75 · JEE Main 2020 (03 Sep Shift 2) · Physics

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become ƒA. The value of ƒ is:

A
$\frac{1}{\sqrt{2}}$
B
$1$
C
$\frac{1}{2}$
Correct answer
D
$\sqrt{2}$

Solution

Using mometum conservation

$m_{1} v_{1} = m_{2} v_{2} m A ω = \frac{m}{2} A_{2} ω_{2} m A \sqrt{\frac{k}{m}} = \frac{m}{2} A_{2} \sqrt{\frac{k}{m / 2}} A_{2} = \sqrt{2} A \Rightarrow f = \sqrt{2}$

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

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