Question Number 197155 by otchereabdullai@gmail.com last updated on 09/Sep/23
$$\:{A}\:{bullet}\:{of}\:{mass}\:\mathrm{180}{g}\:{is}\:{fired}\: \\ $$$$\:{horizontally}\:{into}\:{a}\:{fixed}\:{wooden}\: \\ $$$$\:{block}\:{with}\:{a}\:{speed}\:{of}\:\mathrm{24}{m}/{s}.\:{if}\:{the}\: \\ $$$${bullet}\:{is}\:{brought}\:{to}\:{rest}\:{in}\:\mathrm{0}.\mathrm{4}{sec}\:{by}\:{a} \\ $$$${constant}\:{resistance},\:{calculate}\:{the} \\ $$$${distance}\:{moved}\:{by}\:{the}\:{bullet}\:{in}\:{the} \\ $$$${wood} \\ $$
Answered by deleteduser1 last updated on 09/Sep/23
$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{as}\Rightarrow−\mathrm{24}^{\mathrm{2}} =\mathrm{2}{as} \\ $$$${s}={ut}+\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \Rightarrow{s}=\mathrm{9}.\mathrm{6}+\frac{{a}\left(\mathrm{0}.\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{9}.\mathrm{6}+\mathrm{0}.\mathrm{08}{a} \\ $$$$\Rightarrow\frac{−\mathrm{576}}{\mathrm{2}{a}}=\mathrm{9}.\mathrm{6}+\mathrm{0}.\mathrm{08}{a}\Rightarrow\mathrm{0}.\mathrm{16}{a}^{\mathrm{2}} +\mathrm{19}.\mathrm{2}{a}+\mathrm{576}=\mathrm{0} \\ $$$$\Rightarrow{a}=−\mathrm{60}{ms}^{−\mathrm{2}} \Rightarrow{s}=\mathrm{9}.\mathrm{6}−\mathrm{4}.\mathrm{8}=\mathrm{4}.\mathrm{8}{m} \\ $$
Commented by otchereabdullai@gmail.com last updated on 09/Sep/23
$${thanks}\:{sir} \\ $$