Question

A hammer of mass 500g moving at 50 ms-1 strikes a nail. The nail stops the hammer in a small time interval of 0.01/s. What is the change in momentum suffered by the hammer? Also calculate the force of the hammer on the nail. What will be the force of the nail on the hammer? Give reason for your answer.

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Mass}\mathrm{of}\mathrm{hammer},m=500\mathrm{g}=0.5\mathrm{kg} \\ \mathrm{Initial}\mathrm{velocity}\mathrm{of}\mathrm{hammer},u=50\mathrm{m}{\mathrm{s}}^{-1} \\ \mathrm{Final}\mathrm{velocity}\mathrm{of}\mathrm{hammer},v=0 \\ \mathrm{Now}, \\ \mathrm{Change}\mathrm{in}\mathrm{momentum}\mathrm{of}\mathrm{hammer}=\mathrm{Final}\mathrm{momentum}-\mathrm{initial}\mathrm{momentum} \\ =mv-mu \\ =0.5\left(0-50\right)=-25\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1} \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{force}\left(F\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ F=\frac{\mathrm{Change}\mathrm{in}\mathrm{momentum}}{\mathrm{time}\mathrm{taken}} \\ \Rightarrow F=\frac{-25}{0.01}=-2500\mathrm{N} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{newton}\text{'}\mathrm{s}\mathrm{third}\mathrm{law},\mathrm{force}\mathrm{on}\mathrm{a}\mathrm{body}\mathrm{A}\mathrm{by}\mathrm{B}\mathrm{is}\mathrm{equal}\mathrm{and}\mathrm{opposite}\mathrm{to}\mathrm{the} \\ \mathrm{force}\mathrm{on}\mathrm{body}\mathrm{B}\mathrm{by}\mathrm{A}.\mathrm{Therefore},\mathrm{force}\mathrm{of}\mathrm{the}\mathrm{nail}\mathrm{on}\mathrm{the}\mathrm{hammer}\mathrm{will}\mathrm{be}2500\mathrm{N}. \end{gathered}$$