A body of mass 10 kg is projected at an angle of 45° with the… | JEE Main 2022 (27 Jul, Shift 2)

Q14JEE Main 2022 · 27 JUL · SHIFT 2Difficult

Velocity, Momentum & Energy in Flight

A body of mass $10\text{ kg}$ 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 $T$ is the time of flight, then its momentum vector, at time $t = \frac{T}{\sqrt{2}}$ , is _____________.

[Take $g = 10\text{ m/s}^{2}$ ]

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Answer: (d)

Trajectory at $45^\circ$ : $y = x - \frac{g x^2}{u^2}$ . Through $(20, 10)$ : $10 = 20 - \frac{10 \times 400}{u^2} \Rightarrow u = 20\text{ m/s}$
$u_x = u_y = 10\sqrt{2}\text{ m/s}$ , $T = \frac{2u_y}{g} = 2\sqrt{2}\text{ s}$ , so $t = \frac{T}{\sqrt{2}} = 2\text{ s}$
$v_y = 10\sqrt{2} - 20$ , $v_x = 10\sqrt{2}$
$\vec p = m\vec v = 100\sqrt{2}\,\hat i + (100\sqrt{2} - 200)\hat j$
So the answer is (d).

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