A body of mass 10 kg is… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Mechanics

2022mcqeasy

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 $T$ is the time of flight, then its momentum vector, at time $t=\frac{T}{\sqrt{2}}$ , is _____ .

[Take $g=10m{s}^{-2}$ ]

Official previous-year question

Held on 27 Jul 2022 · Verified 6 Jul 2026.

Options

A
$100\hat{i}+(100\sqrt{2}-200)\hat{j}Ns$
B
$100\sqrt{2}\hat{i}+(100-200\sqrt{2})\hat{j}Ns$
C
$100\hat{i}+(100-200\sqrt{2})\hat{j}Ns$
D
$100\sqrt{2}\hat{i}+(100\sqrt{2}-200)\hat{j}Ns$

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Solution

It is given that the projectile passes through $(20,10)$ and angle of projection is $\theta =45^{\circ}$ . Using the trajectory equation,

$y=x\mathrm{tan}\theta -\frac{g{x}^{2}}{2{u}^{2}{\mathrm{cos}}^{2}\theta }$

$\Rightarrow 10=20-\frac{(10)(100)}{{u}^{2}}$

$\Rightarrow u=20m{s}^{-1}$

Now the time of flight of projectile,

$T=\frac{2u\mathrm{sin}\theta }{g}$

$\Rightarrow T=\frac{(2)(20)}{\sqrt{2}(10)}=2\sqrt{2}s \Rightarrow \frac{T}{\sqrt{2}}=2s$

Now velocity of the projectile at any time $t$ can be written as,

$\vec{v}=u\mathrm{cos}\theta \hat{i}+(u\mathrm{sin}\theta -gt)\hat{j}$

$\Rightarrow \vec{v}=10\sqrt{2}\hat{i}+(10\sqrt{2}-10(2)]\hat{j}m{s}^{-1}$

Therefore, momentum $\vec{p}=M\vec{v}=100\sqrt{2}\hat{i}+(100\sqrt{2}-200)\hat{j}Ns$

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Options

$100\hat{i}+(100\sqrt{2}-200)\hat{j}Ns$

$100\sqrt{2}\hat{i}+(100-200\sqrt{2})\hat{j}Ns$

$100\hat{i}+(100-200\sqrt{2})\hat{j}Ns$

$100\sqrt{2}\hat{i}+(100\sqrt{2}-200)\hat{j}Ns$

Solution

It is given that the projectile passes through $(20,10)$ and angle of projection is $\theta =45^{\circ}$ . Using the trajectory equation,

$y=x\mathrm{tan}\theta -\frac{g{x}^{2}}{2{u}^{2}{\mathrm{cos}}^{2}\theta }$

$\Rightarrow 10=20-\frac{(10)(100)}{{u}^{2}}$

$\Rightarrow u=20m{s}^{-1}$

Now the time of flight of projectile,

$T=\frac{2u\mathrm{sin}\theta }{g}$

$\Rightarrow T=\frac{(2)(20)}{\sqrt{2}(10)}=2\sqrt{2}s \Rightarrow \frac{T}{\sqrt{2}}=2s$

Now velocity of the projectile at any time $t$ can be written as,

$\vec{v}=u\mathrm{cos}\theta \hat{i}+(u\mathrm{sin}\theta -gt)\hat{j}$

$\Rightarrow \vec{v}=10\sqrt{2}\hat{i}+(10\sqrt{2}-10(2)]\hat{j}m{s}^{-1}$

Therefore, momentum $\vec{p}=M\vec{v}=100\sqrt{2}\hat{i}+(100\sqrt{2}-200)\hat{j}Ns$

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics