Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is:

Option 1 -

mv / √K

Option 2 -

mv / √2K

Option 3 -

√(m/2K) * v

Option 4 -

v * √(m/2K)

4 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

For an elastic collision where C comes to rest, and the compression in the spring is maximum, the velocities of A and B are equal (v). Using the conservation of mechanical energy:
(1/2)mv? ² = 2 * (1/2)mv² + (1/2)kx²
This gives the maximum compression x as:
x = v? * √* (m / 2k)*

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