An oscillator of mass M is at rest in its equilibrium position in a… | JEE Main 2018 (16 Apr, Shift 1)

Q84JEE Main 2018 · 16 APR · SHIFT 1Difficult

Spring Systems

An oscillator of mass M is at rest in its equilibrium position in a potential

V = $\frac{1}{2}$ k(x $-$ X)2. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: (M = 10, m = 5, u = 1, k = 1)

Show answer and solution

Answer: (a)

Each collision happens at the equilibrium position, so momentum is conserved and the speed there gives $A = \frac{v}{\omega}$ .
1st collision: momentum $mu$ (leftward). The oscillator returns moving right with momentum $mu$ ; the 2nd particle brings $-mu$ , total $0$ . So after every even collision it stops, after every odd one momentum is $mu$ .
After 13 collisions: mass $= M + 13m = 75$ , $v = \frac{mu}{75} = \frac{5}{75} = \frac{1}{15}$ .
$\omega = \sqrt{\frac{k}{75}} = \frac{1}{\sqrt{75}}$ , $A = \frac{v}{\omega} = \frac{\sqrt{75}}{15} = \frac{1}{\sqrt{3}}$
So the answer is (a).

Report a mistake

Spring Systems question from JEE Main 2018 (16 Apr, Shift 1)

More questions on Spring Systems