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Q.

A small $1 k g$ mass is launched from the top of a cliff with speed $V$ at an angle of $30 °$ above the horizontal. When the mass reaches the ground, its velocity is directed at $45 °$ 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.

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a

$\frac{1}{2} (\sqrt{3} - 1) V$

b

$\sqrt{\frac{3}{2}} (\frac{\sqrt{2} + 1}{2}) V$

c

$\frac{1}{2} (\sqrt{\frac{3}{2}} + 1) V$

d

$\frac{1}{2} (\sqrt{3} + 1) V$

answer is A.

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Detailed Solution

$V \cos 30 ° = V_{1} \cos 45 ° V \frac{\sqrt{3}}{2} = \frac{V_{1}}{\sqrt{2}} V_{1} = V \sqrt{\frac{3}{2}} J = p_{f} - p_{i} = m [- V_{1} \sin 45 ° - V \sin 30 °] = 1 [- v \sqrt{\frac{3}{2}} \times \frac{1}{\sqrt{2}} - \frac{V}{2}] = V [\frac{\sqrt{3} + 1}{2}]$

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