Physics · 2018
JEE Main · 16 April 2018, Shift 1 · Q5
An oscillator of mass M is at rest in its equilibrium position in a potential V =1/2 k (x- X )^2. A particle of mass m comes from right with speed u…
An oscillator of mass M is at rest in its equilibrium position in a potential $\displaystyle \mathrm{V}=\frac{1}{2} \mathrm{k}(x-\mathrm{X})^2$. 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 $\displaystyle 13$ collisions is : $\displaystyle (\mathrm{M}=10, \mathrm{~m}=5, \mathrm{u}=1, \mathrm{k}=1)$
Official answer
(1)
$\displaystyle \frac{1}{\sqrt{3}}$
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JEE Main 2018 Physics question, with the answer from NTA’s final answer key. Where our answers come from.