- Home
- Standard 11
- Physics
A small mass $m$ is launched from the top of a cliff with speed Vat an angle of $30^o$ above the horizontal. When the mass reaches the ground, its velocity is directed at $45^o$ below the horizontal. Which one of the following choices is the magnitude of the total impulse that was imparted to the mass during its flight ? Ignore air resistance
$\frac{1}{2}\left( {\sqrt 3 + 1} \right)\ mV$
$\sqrt {\frac{3}{2}} \left( {\frac{{\sqrt 2 + 1}}{2}} \right)\ mV$
$\frac{1}{2}\left( {\sqrt 3 - 1} \right)\ mV$
$\frac{1}{2}\left( {\sqrt {\frac{3}{2}} + 1} \right)\ mV$
Solution
$\mathrm{V} \cos 30^{\circ}=\mathrm{V}_{1} \cos 45^{\circ}$
$V\frac{{\sqrt 3 }}{2} = \frac{{{V_1}}}{{\sqrt 2 }}$
$\mathrm{J}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{1}$
$=\mathrm{m}\left[-\mathrm{V}_{1} \sin 45^{\circ}-\mathrm{V} \sin 30^{\circ}\right]$
$=\mathrm{m}\left[-\mathrm{v} \sqrt{\frac{3}{2}} \times \frac{1}{\sqrt{2}}-\frac{\mathrm{V}}{2}\right]=\mathrm{m} \mathrm{V}\left[\frac{\sqrt{3}+1}{2}\right]$
