Q126 / 160 · TS EAMCET 2023 (13 May Shift 1) · Physics
A block of mass M moving on a frictionless horizontal surface collides with a spring of spring constant K, as shown in the figure. If the spring compresses by a length $\mathrm{L}$, then the maximum momentum of the block after the collision is.
- AZero
- B$\frac{\mathrm{ML}^2}{\mathrm{~K}}$
- Correct answerC$\mathrm{L} \sqrt{\mathrm{MK}}$
- D$\frac{\mathrm{KL}^2}{2 \mathrm{M}}$
Solution
Initial kinetic energy, $\mathrm{KE}=\frac{1}{2} \mathrm{Mv}^2$
Final potential energy, $P E=\frac{1}{2} k^2$ from the conservation of energy K.E $=$ P.E
$$
\begin{aligned}
& \frac{1}{2} \mathrm{Mv}^2=\frac{1}{2} \mathrm{Kx}^2 \quad \text { Here compression, } \mathrm{x}=\mathrm{L} \\
& \mathrm{M} v^2=K \mathrm{~L}^2 \\
& \mathrm{v}=\sqrt{\frac{K}{M}} \mathrm{~L}
\end{aligned}
$$
The maximum momentum of block, $\mathrm{p}=\mathrm{mv}$
$$
=m \sqrt{\frac{k}{m}} \mathrm{~L}=L \sqrt{M K}
$$
Hence, the correct answer is option: 'C'.