A projectile of mass $100 g$ is fired with a velocity of $20 m / s$ making an angle of $30^{0}$ with the horizontal.As it rises to the highest point of its path,its momentum changes by

\frac{1}{2} k g - m / s

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1 k g - m / s

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2 k g - m / s

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None of these

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Solution

The correct option is B $1 k g - m / s$
Initial momentum $p_{1} = O A = m u$
$\frac{1}{10} \times 20 = 2$
Final momentum at highest point
$p_{2} = O B = m u c o s θ$
$= \frac{1}{10} \times 20 c o s 30^{0}$
$= 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$
Change in momentum=AB
$= O B - O A$
$= \sqrt{O A^{2} - O B^{2}}$
$\sqrt{4 - 3} = 1 k g - m / s$

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A projectile of mass 100g is fired with a velocity of 20m/s making an angle of 300 with the horizontal.As it rises to the highest point of its path,its momentum changes by

The correct option is B 1kg−m/sInitial momentump1=OA=mu110×20=2Final momentum at highest pointp2=OB=mucosθ=110×20cos300=2×√32=√3Change in momentum=AB=OB−OA=√OA2−OB2√4−3=1kg−m/s

A projectile of mass $100 g$ is fired with a velocity of $20 m / s$ making an angle of $30^{0}$ with the horizontal.As it rises to the highest point of its path,its momentum changes by