### JEE Main On 16 April 2018 Question 15

##### Question: 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 comes from right with speed u and collides completely in elastically with and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after collisions is: $ (M=10,m=5,u=1,k=1). $ [JEE Main 16-4-2018]

#### Options:

A) $ \frac{1}{2} $

B) $ \frac{1}{\sqrt{3}} $

C) $ \frac{2}{3} $

D) $ \sqrt{\frac{3}{5}} $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

Initial momentum of mass ’m’ = mu =5

Final momentum of system $ =(M+m)v=mu=5 $

For second collision, mass (m=5, u = 1)

coming from right strikes with system of mass 15,

both momentum have opposite direction.

$ \therefore $ net momentum = zero

Similarly for12th collision momentum is zero.

For 13th collision, total mass $ =10+12\times 5=70 $

Using conservation of momentum $ 70\times 0+5\times 1=(70+5)v’ $

$ v’=\frac{1}{5} $ Total mass $ =10+13\times 5=75 $

Finally KE of system $ =\frac{1}{2}mv^{2}=\frac{1}{2}\times 75\times [ \frac{1}{15} ][ \frac{1}{15} ] $ $ \frac{1}{2}k,A^{2}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15} $

$ =\frac{1}{7}\times (1)A^{2}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15} $ $ A^{2}=\frac{1}{3} $ $ A=\frac{1}{\sqrt{3}} $