JEE Main · 2005 · 08 May 2005 · Offline · Work, Power & Energy

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The block of mass M moving on the frictionless horizontal surface collides with a spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

(A) $MK L$ ✓ Correct
(B) $\frac{K L ^{2}}{2 M}$
(C)zero
(D) $\frac{ML ^{2}}{K}$

Answer: (A)

Solution

At maximum compression

L

the block is momentarily at rest and all its kinetic energy has passed into the spring,

\frac{1}{2} K L^{2} = \frac{p ^{2}}{2 M}

, where

p

is the momentum the block carried on arrival. Solving,

p^{2} = M K L^{2}

, so the maximum momentum is

p = L M K

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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