Block of mass M moving on frictionless horizontal surface collides with spring of spring cons

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5.Work, Energy, Power and Collision

medium

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

Zero

$\frac{{M{L^2}}}{K}$

$\sqrt {MK} \,L$

$\frac{{K{L^2}}}{{2M}}$

(AIEEE-2005)

Solution

(c)When block of mass $M$ collides with the spring its kinetic energy gets converted into elastic potential energy of the spring.
From the law of conservation of energy
$\frac{1}{2}M{v^2} = \frac{1}{2}K{L^2}$ $\therefore $ $v = \sqrt {\frac{K}{M}} L$
Where $ v$ is the velocity of block by which it collides with spring. So, its maximum momentum
$P = Mv = M\sqrt {\frac{K}{M}} \,L$ = $\sqrt {MK} \,L$
After collision the block will rebound with same linear momentum.

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Solution

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