A small mass m is launched from the top of a cliff with speed Vat… - Vidyadip

MCQ

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

✓
$\frac{1}{2}\left( {\sqrt 3 + 1} \right)\ mV$
B
$\sqrt {\frac{3}{2}} \left( {\frac{{\sqrt 2 + 1}}{2}} \right)\ mV$
C
$\frac{1}{2}\left( {\sqrt 3 - 1} \right)\ mV$
D
$\frac{1}{2}\left( {\sqrt {\frac{3}{2}} + 1} \right)\ mV$

✓

Answer

Correct option: A.

$\frac{1}{2}\left( {\sqrt 3 + 1} \right)\ mV$

a
$\mathrm{V} \cos 30^{\circ}=\mathrm{V}_{1} \cos 45^{\circ}$

$V\frac{{\sqrt 3 }}{2} = \frac{{{V_1}}}{{\sqrt 2 }}$

$\mathrm{J}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{1}$

$=\mathrm{m}\left[-\mathrm{V}_{1} \sin 45^{\circ}-\mathrm{V} \sin 30^{\circ}\right]$

$=\mathrm{m}\left[-\mathrm{v} \sqrt{\frac{3}{2}} \times \frac{1}{\sqrt{2}}-\frac{\mathrm{V}}{2}\right]=\mathrm{m} \mathrm{V}\left[\frac{\sqrt{3}+1}{2}\right]$

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