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
Text Solution
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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
$\frac{1}{2} M v^{2} = \frac{1}{2} K L^{2}$ $\begin{matrix} \bullet \\ \bullet & \bullet \end{matrix}$ $v = \sqrt{\frac{K}{M}} L$
Where v is the velocity of block by which it collides with spring. So, its maximum momentum
$\mathrm{P} = \mathrm{Mv} = \mathrm{M} \sqrt{\frac{\mathrm{K}}{\mathrm{M}}} \mathrm{L}$ = $\sqrt{MKL}$
After collision the block will rebound with same linear momentum.
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