QJEEM18Ph133d: JEE Main 2018 Physics - Class 11 JEEM18Ph133d Solution: An oscillator of mass M is at rest in its equilibrium position…

Potential of the given oscillator is

$V = {1 \over 2}k{(x - k)^2}$

Given: M = 10; m = 5, u = 1; k = 1

Initial momentum of the particle of mass m

= mu = m $\times$ 5 = 5m

Momentum of (oscillator + particle) after collision = (M + m)

Velocity of oscillator after collision = v

So, momentum of system = (M + m)v

From conservation of linear momentum, we have

(M + m) = mu = 5 $\times$ 1 = 5

For second collision, oscillator and particle have momentum in opposite direction.

Net or total momentum is zero.

Likewise after 4^th, 6^th, 8^th, 10^th, 12^th collision the momentum is zero. After 12^th collision, Mass of oscillator and 12 particles will be (10 + 12 $\times$ 5) = 70

Now, from conservation of linear momentum, for 13^th collision, we have

$70 \times 0 + 5 \times 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}$

Total mass after 13^th collision = (10 + 13 $\times$ 5) = 75

Kinetic energy of system $= {1 \over 2}mv{'^2}$

$\Rightarrow KE = {1 \over 2} \times 75 \times {1 \over {15}} \times {1 \over {15}}$

$\Rightarrow {1 \over 2}k{A^2} = {1 \over 2} \times {{75} \over {225}} = {1 \over 6}$

$\Rightarrow {1 \over 2} \times 1 \times {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}$

$\Rightarrow A = {1 \over {\sqrt 3 }}$