JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 2)
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 $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$$ ]
Explanation
m = 10 kg
$$\theta$$ = 45$$^\circ$$
$$y = x\tan \theta \left( {1 - {x \over R}} \right)$$
$$ \Rightarrow 10 = 20\left( {1 - {{20} \over R}} \right)$$
$$ \Rightarrow R = 40$$
$$40 = {{{u^2}} \over {10}} \Rightarrow u = 20$$
$$ \Rightarrow T = {{20 \times 20 \times {1 \over {\sqrt 2 }}} \over {10}} = {4 \over {\sqrt 2 }}s \Rightarrow t = 2\,s$$
at $$t = 2,\,\overrightarrow v = \left( {10\sqrt 2 \widehat i} \right) + \left( {10\sqrt 2 - 2 \times 10} \right)\widehat j$$
$$ \Rightarrow \overrightarrow p = 10\left[ {10\sqrt 2 \widehat i + \left( {10\sqrt 2 - 20} \right)\widehat j} \right]$$
$$ = 100\sqrt 2 \widehat i + \left( {100\sqrt 2 - 200} \right)\widehat j$$
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