Question

A block of mass 10 kg moving with a velocity of 1 m s -1 collides with a spring of force constant 1000 N m-1 . If the spring is initially at rest, calculate the maximum compression of the spring

Answer 1

$$\begin{gathered} \mathrm{PE}\mathrm{of}\mathrm{spring}=\frac{1}{2}{\mathrm{kx}}^2 \\ \mathrm{KE}\mathrm{of}\mathrm{block}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{From}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{conservation}\mathrm{of}\mathrm{energy}, \\ \mathrm{PE}\mathrm{of}\mathrm{spring}=\mathrm{KE}\mathrm{of}\mathrm{block} \\ \frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{mv}}^2 \\ {\mathrm{kx}}^2={\mathrm{mv}}^2 \\ {\mathrm{x}}^2=\frac{{\mathrm{mv}}^2}{\mathrm{k}} \\ \mathrm{x}=\sqrt{\frac{{\mathrm{mv}}^2}{\mathrm{k}}} \\ =\sqrt{\frac{10\times 1^2}{1000}} \\ =\sqrt{\frac{1}{100}} \\ =\frac{1}{10}\mathrm{m} \\ =\frac{1}{10}\times 100\mathrm{cm} \\ \mathbf{=}\mathbf{10}\mathbf{}\mathbf{cm} \end{gathered}$$