Question

Two blocks A and B of mass m and 2m connected by a light spring of spring constant lie at rest on a fixed smooth horizontal plane are given velocities initially of magnitudes 2u and u as shown in figure. In the subsequent motion, the only horizontal force acting on blocks is due to spring. Match the conditions in column I with the instants of time they occur as given in column II.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A)The length of spring is least at time}& \text{(p)}\frac{\pi }{2}\sqrt{\frac{2m}{3k}}\\ \text{(B) The length of spring is maximum at time}& \text{(q)}\pi \sqrt{\frac{2m}{3k}}\\ \text{(C) The acceleration of both blocks is}& \text{(r)}\pi \sqrt{\frac{3m}{2k}}\\ \text{zero simultaneously at time}& \\ \text{(D) velocity of center of mass at time t = 0 is}& \text{(s) zero}\end{array}$

Choose the correct option

• A. A-p ; B-r ; C-q; D-s
• B. A-p; B-q; C-r; D-s
• C. A-q; B-p; C-s; D-q
• D. A-p; B-s; C-q; D-s

Answer 1

The correct option is A

A-p ; B-r ; C-q; D-s

${}^{v}CM=\frac{2u\times m-u\times 2m}{3m}=0$

$\omega =\sqrt{\frac{k}{\mu }}=\sqrt{\frac{k.3m}{m\times 2m}}=\sqrt{\frac{3k}{2m}}$

(A) Spring is maximum compressed when phase of block B reaches position 1, and travels an angle

$\frac{\pi }{2}\Rightarrow t=\frac{\pi }{2\omega }=\frac{\pi }{2}\sqrt{\frac{2m}{3k}}$

(B) Spring is maximum elongated when phase of block B reaches position 3 and travels angle

$\frac{3\pi }{2}\Rightarrow t=\frac{3\pi }{2\omega }=\frac{3\pi }{2}\sqrt{\frac{2m}{3k}}=\pi \sqrt{\frac{3m}{2k}}$

(C) Acceleration for both is zero when both pass their respective mean positions i.e. phase of B reaches position 2 and hence travels $\pi$ angle.

$\Rightarrow t=\frac{\pi }{\omega }=\pi \sqrt{\frac{2m}{3k}}$.