Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Two blocks A and B of masses m and 2m, respectively, are connected by a massless spring of force constant k and are placed on a smooth horizontal plane. The spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is

PhysicsCenter of Mass Momentum and CollisionNEET
Options:
  • A 3k2mx
  • B 2k3mx
  • C 2kxm
  • D 3km2x
Solution:
2461 Upvotes Verified Answer
The correct answer is: 3k2mx


Relative velocity will be v1+v2

Using conservation of linear momentum,

mv1=2mv2

   v1=2v2

Using conservation of energy

12kx2=12mv12+122mv22

12kx2=12m2v22+mv22

12kx2=2mv22+mv22

3mv22=kx22

v22=kx26m

v2= k6m.x

Relative velocity =3v2

=3k6m.x

= 3k2m.x

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.