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Most Important Selected Qs for JEE AdvancedPhysicsCenter of Mass, Momentum and Collision

The friction coefficient between the horizontal surface and each of the block shown in the figure is 0.2 . The collision between the blocks is perfectly elastic. Find the separation (in cm ) between them when they come to rest. Take g =10 ~m / s ^2 .

Correct answer

5

Step-by-step solution

Velocity of first block before collision, aligned & v₁^2=1^2-2(2) 0.16 & =1-0.64 & v₁=0.6 ~m / s aligned By conservation of momentum, 2 0.6=2 v ₁^ +4 v ₂^ also v₂^ -v₁^ =v₁ for elastic collision It gives aligned & v₂^ =0.4 ~m / s & v₁^ =-0.2 ~m / s aligned Now distance moved after collision aligned & s ₁= (0.4)^2 2 2 & ~s ₂= (0.2)^2 2 2 & s = s ₁+ s ₂=0.05 ~m =5 ~cm aligned

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