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 $m$ 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: $(M=10,m=5,u=1,k=1)$.

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{2}{3}$
• D. $\sqrt{\frac{3}{5}}$

Answer 1

Answer: (B)

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

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 for ${12}^{th}$ collision momentum is zero.

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

Using conservation of momentum

$70\times 0+5\times 1=(70+5)v^{\prime }$

$v^{\prime }=\frac{1}{15}$

Total mass $=10+13\times 5=75$

Finald KE of system $=\frac{1}{2}m{v}^2=\frac{1}{2}\times 75\times \left[\frac{1}{15}\right]\left[\frac{1}{15}\right]$

$\frac{1}{2}k$ $A^2=\frac{1}{2}$ $75\times \frac{1}{15}\times \frac{1}{15}$

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

$A^2=\frac{1}{3}$

$A=\frac{1}{\sqrt{3}}$