Two blocks $M_{1} a n d M_{2}$ having equal mass are to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in figure. Initially $M_{2}$ is at rest and $M_{1}$ is moving toward $M_{2}$ with speed v and collides head-on with $M_{2}$

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

If the surface on which blocks are moving has friction, then collision cannot be elastic

While spring is fully compressed the system momentum is not conserved though final momentum is equal to initial momentum

If spring is massless, the final state of the M₁is state of rest

While spring is fully compressed all the KE of M₁ is stored as PE of spring

answer is C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

While spring is fully compressed, the entire KE of M₁ is not stored as PE of spring as M₂ may move. If spring is massless, than as $M_{1} = M_{2},$ velocities of $M_{1} a n d M_{2}$ are interchanged on collision. $M_{1}$ comes to rest, instead of M₂Choice (3) is correct. If surface on which blocks are moving friction loss of energy is involved. Collision cannot be elastic. Choice (4 is correct