Q 11-05-036NEETJEE MainMedium
A $0.5$ kg block moving at $4$ m/s on a smooth floor hits a spring of force constant $200$ N/m. The maximum compression of the spring is
Answer: (C) $0.2$ m
$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2 \;\Rightarrow\; x = v\sqrt{\frac{m}{k}} = 4\sqrt{\frac{0.5}{200}} = 4 \times 0.05 = 0.2\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics