The block of mass $M$ moving on the frictionless horizontal surface collides with a spring of spring…

The block of mass $M$ moving on the frictionless horizontal surface collides with a spring of spring constant $\mathrm{K}$ and compresses it by length $\mathrm{L}$. The maximum momentum of the block after collision is
  1. $\sqrt{\mathrm{MK}} \mathrm{L}$
  2. $\frac{\mathrm{KL}^2}{2 \mathrm{M}}$
  3. zero
  4. $\frac{M L^2}{\mathrm{~K}}$

Solution

$\frac{1}{2} \mathrm{KL}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \quad \therefore \mathrm{P}=\sqrt{\mathrm{MK}} \mathrm{L}$

Asked in: JEE Main 2005

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