A light spring is permanently connected between two trolleys of masses $M$ and $m$ which can move over a smooth horizontal table, along a straight line parallel to the length of the spring as shown in figure. The trolleys are brought nearer to compress the spring and then released. In the subsequent motion.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Initially they move in opposite directions with speeds inversely proportional to their masses.

Angular momentum of system about $O$ is increasing initially.

linear momentum and energy of the system must be conserved.

their center of mass remains stationary.

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

No net external force, so linear momentum is conserved, hence $m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}} = 0 i . e v_{1} and v_{2}$ are in opposite direction and $|\frac{v_{1}}{v_{2}}| = \frac{m_{2}}{m_{1}}$ . Spring force is conservative in nature no external force hence number external torque so angular momentum is still conserved.