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A particle having a mass $0.5 \mathrm{~kg}$ is projected under gravity with a speed of $98 \mathrm{~m} / \mathrm{sec}$ at an angle of $60^{\circ} .$ The magnitude of the change in momentum (in N-sec) of the particle after 10 seconds is $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{sec}^{2}\right]$
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The correct answer is:
49
$$
\begin{array}{l}
\text { Change in momentum }=\mathrm{F} \times \mathrm{t} \\
\quad=\mathrm{mg} \times 10 \\
\quad \Rightarrow 0.5 \times 9.8 \times 10 \\
\quad \Rightarrow 49 \mathrm{~N}-\mathrm{sec}
\end{array}
$$
\begin{array}{l}
\text { Change in momentum }=\mathrm{F} \times \mathrm{t} \\
\quad=\mathrm{mg} \times 10 \\
\quad \Rightarrow 0.5 \times 9.8 \times 10 \\
\quad \Rightarrow 49 \mathrm{~N}-\mathrm{sec}
\end{array}
$$
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