Question

Lead bullet of mass 20 g, travelling with a velocity of 250 ms1, comes to rest after penetrating 40 cm in a target at rest. Find the resistive force offered by the target

Answer 1

$$\begin{gathered} Accordingtoquestion \\ giveninitialvelocityu=250m/s,finalvelocityv=0,displacement=40cm=0.4m,a=acceleration=?andmass\left(m\right)=20\mathrm{gram}=\frac{20}{1000}=0.02\mathrm{kg} \\ Nowu\sin gformula \\ {v}^2-u{}^2=2as \\ {0}^2-(250{)}^2=2\times a\times 0.4 \\ \Rightarrow a=-\frac{62500}{2\times 0.4} \\ a=-\frac{31250}{0.4} \\ a=-78125\mathrm{m}/{\mathrm{s}}^2(herenegetivesignshowsretardationmeansbodyisloo\sin gitsvelocity) \\ Nowforce=mass\times acceleration \\ =0.02\times -781250=-1562.5\mathrm{N} \\ Herenegetivesignshowsforceisrestitiveinnature. \end{gathered}$$