A body of mass $10 \mathrm{~kg}$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$. If $\mathrm{T}$ is the time of flight, then its momentum vector, at time $\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}}$, is _____________.
[Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]
Solution
<p>m = 10 kg</p>
<p>$\theta$ = 45$^\circ$</p>
<p>$y = x\tan \theta \left( {1 - {x \over R}} \right)$</p>
<p>$\Rightarrow 10 = 20\left( {1 - {{20} \over R}} \right)$</p>
<p>$\Rightarrow R = 40$</p>
<p>$40 = {{{u^2}} \over {10}} \Rightarrow u = 20$</p>
<p>$$ \Rightarrow T = {{20 \times 20 \times {1 \over {\sqrt 2 }}} \over {10}} = {4 \over {\sqrt 2 }}s \Rightarrow t = 2\,s$$</p>
<p>at $$t = 2,\,\overrightarrow v = \left( {10\sqrt 2 \widehat i} \right) + \left( {10\sqrt 2 - 2 \times 10} \right)\widehat j$$</p>
<p>$$ \Rightarrow \overrightarrow p = 10\left[ {10\sqrt 2 \widehat i + \left( {10\sqrt 2 - 20} \right)\widehat j} \right]$$</p>
<p>$= 100\sqrt 2 \widehat i + \left( {100\sqrt 2 - 200} \right)\widehat j$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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