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 the right with speed u and collides completely inelastic 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)

We choose the +x-direction to be towards the right. The spring is at its natural length when $$x=X$$, so the potential energy there is zero. Every collision takes place exactly at this point. A small particle always comes from the right, therefore its velocity just before every collision is $$\;-\;u$$ (towards the left).

Initially the oscillator of mass $$M=10\ \text{kg}$$ is at rest, so its velocity is $$0$$ and its mechanical energy is also zero.

First collision  ($$n=0\to1$$)

Conservation of linear momentum gives

$$ m(-u)+M(0)=(M+m)\,v_1 . $$

Hence

$$ v_1=-\dfrac{m\,u}{M+m}=-\dfrac{5\cdot1}{10+5}=-\dfrac13\;\text{m\,s}^{-1}. $$

The kinetic energy of the combined body immediately after the impact is completely stored in the spring when it reaches its extreme position, so

$$ \dfrac12(M+m)\,v_1^{2}=\dfrac12\,k\,A_1^{2}. $$

Stating the relation between the speed at equilibrium and the amplitude, we have

$$ A_1=|v_1|\sqrt{\dfrac{M+m}{k}} =\dfrac{m\,u}{M+m}\,\sqrt{\dfrac{M+m}{k}} =\dfrac{m\,u}{\sqrt{k(M+m)}} . $$

With the numerical data $$k=1$$, $$m=5$$, $$M=10$$ this gives

$$ A_1=\dfrac{5}{\sqrt{15}}\;{\rm m}\approx1.29\;{\rm m}. $$

The system now moves leftward to the negative extreme, comes back, and returns to the equilibrium point with the same speed but in the opposite (+x) direction:

$$ v_1^{\text{(return)}}=+|v_1|=+\dfrac13\;\text{m\,s}^{-1}. $$

Second collision  ($$n=1\to2$$)

The second particle (again with velocity $$-u$$) meets the oscillator when its velocity is $$+v_1$$. Applying conservation of momentum,

$$ m(-u)+(M+m)v_1=(M+2m)\,v_2 . $$

Because $$m\,u=(M+m)|v_1|$$ (seen from the first collision), the two momenta are equal and opposite, so the right-hand side is zero:

$$ v_2=0 . $$

With zero velocity, the spring stores no energy; therefore

$$ A_2=0 . $$

Beginning of the pattern

After every even collision the oscillator is at rest and its amplitude is zero. Consequently, before every following (odd) collision it is again at rest at the equilibrium position. Thus each odd collision is identical in character to the very first one, except that the total mass has increased.

General odd collision

Let the total mass just before the $$(2k+1)$$-th collision be

$$ M_{2k}=M+2k\,m . $$

The oscillator is at rest, the incoming particle still has momentum $$-m\,u$$, so momentum conservation gives

$$ m(-u)+M_{2k}(0)=(M_{2k}+m)\,v_{2k+1}\;\;\Longrightarrow\;\; v_{2k+1}=-\dfrac{m\,u}{M_{2k}+m}. $$

The corresponding amplitude is therefore

$$ A_{2k+1}=|v_{2k+1}|\sqrt{\dfrac{M_{2k}+m}{k}} =\dfrac{m\,u}{\sqrt{k\,[M_{2k}+m]}} . $$

Amplitude after 13 collisions

The $$13$$-th collision is odd ($$2k+1=13\;\Rightarrow\;k=6$$). Thus

$$ M_{12}=M+12\,m=10+12\cdot5=70\ \text{kg}, $$

and

$$ A_{13}=\dfrac{m\,u}{\sqrt{k\,[M_{12}+m]}} =\dfrac{5\cdot1}{\sqrt{1\,(70+5)}} =\dfrac{5}{\sqrt{75}} =\dfrac{5}{5\sqrt3} =\dfrac1{\sqrt3}\;{\rm m}. $$

Hence, the correct answer is Option B.