Question

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

• A. $0$
• B. $\frac{M{L}^2}{K}$
• C. $L\sqrt{MK}$
• D. $\frac{K{L}^2}{2M}$

Answer 1

The correct option is C

$L\sqrt{MK}$

Explanation of correct answer:

Option C -

The diagrammatic representation of the given case is,

Momentum will be maximum when kinetic energy is maximum and in this case, total elastic potential energy is converted to kinetic energy.

According to the law of conservation of energy, the total energy of a system remains the same. Thus, the loss in potential energy is equal to the gain in kinetic energy.
$$\begin{gathered} \frac{1}{2}K{L}^2=\frac{1}{2}M{v}^2 \\ \Rightarrow K{L}^2=\frac{{\left(Mv\right)}^2}{M} \\ \Rightarrow MK{L}^2=p^2\left(\because p=Mv\right) \\ \therefore p=L\sqrt{MK} \end{gathered}$$

where $K$ is spring constant, $L$ is length compressed, $M$ is mass of block, $v$ is velocity and $p$ is momentum.​

Therefore, the maximum momentum of the body after the collision is $L\sqrt{MK}$

Hence, option C is correct.