Question Number 29212 by NECx last updated on 05/Feb/18
$${A}\:{block}\:{of}\:{wood}\:{of}\:{mass}\:\mathrm{0}.\mathrm{6}{kg}\:{is} \\ $$$${balanced}\:{on}\:{top}\:{of}\:{vertical}\:{port} \\ $$$$\mathrm{2}{m}\:{high}.{A}\:\mathrm{10}{g}\:{bullet}\:{is}\:{fired} \\ $$$${horizontally}\:{into}\:{the}\:{block}\:{and} \\ $$$${the}\:{embedded}\:{bullet}\:{land}\:{at}\:{a}\:\mathrm{4}{m} \\ $$$${from}\:{the}\:{base}\:{of}\:{the}\:{port}.{Find} \\ $$$${the}\:{initial}\:{velocity}\:{of}\:{the}\:{bullet}. \\ $$
Commented by NECx last updated on 06/Feb/18
$${please}\:{help} \\ $$
Answered by mrW2 last updated on 07/Feb/18
$${v}_{\mathrm{0}} ={initial}\:{velocity}\:{of}\:{bullet} \\ $$$${v}_{\mathrm{1}} ={velocity}\:{of}\:{wood}\:{with}\:{bullet} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$${L}={v}_{\mathrm{1}} {t}={v}_{\mathrm{1}} \sqrt{\frac{\mathrm{2}{h}}{{g}}} \\ $$$$\Rightarrow{v}_{\mathrm{1}} ={L}\sqrt{\frac{{g}}{\mathrm{2}{h}}} \\ $$$${m}_{{B}} {v}_{\mathrm{0}} =\left({m}_{{B}} +{m}_{{W}} \right){v}_{\mathrm{1}} \\ $$$$\Rightarrow{v}_{\mathrm{0}} =\frac{{m}_{{B}} +{m}_{{W}} }{{m}_{{B}} }×{v}_{\mathrm{1}} \\ $$$$=\frac{{m}_{{B}} +{m}_{{W}} }{{m}_{{B}} }×{L}\sqrt{\frac{{g}}{\mathrm{2}{h}}} \\ $$$$=\frac{\mathrm{10}+\mathrm{600}}{\mathrm{10}}×\mathrm{4}\sqrt{\frac{\mathrm{10}}{\mathrm{2}×\mathrm{2}}} \\ $$$$=\mathrm{122}\sqrt{\mathrm{10}} \\ $$$$\approx\mathrm{386}\:{m}/{s} \\ $$
Commented by NECx last updated on 08/Feb/18
$${now}\:{its}\:{understood}.\boldsymbol{{Gracia}}{s}! \\ $$
Commented by mrW2 last updated on 08/Feb/18
$${de}\:{nada},\:{se}\overset{\sim} {{n}or}! \\ $$