JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2020 · 9 January · Evening Shift · Physics · Center of Mass, Momentum and Collision

A particle of mass m is projected with a speed u from the ground at...

A particle of mass m is projected with a speed u from the ground at an angle

\theta = \dfrac{\pi}{3}

w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity

u\overset{^}{i} .

The horizontal distance covered by the combined mass before reaching the ground is: A particle of mass m is projected with a speed u from the ground at an angle θ = (π)/(3) w.r.t. horizontal (x-axis).

A particle of mass m is projected with a speed u from the ground at an angle θ = (π)/(3) w.r.t. horizontal (x-axis).

Options

A $\dfrac{3\sqrt{3}}{8}\dfrac{{u}^{2}}{g}$
B $\dfrac{3\sqrt{2}}{4}\dfrac{{u}^{2}}{g}$
C $\dfrac{5}{8}\dfrac{{u}^{2}}{g}$
D $2\sqrt{2}\dfrac{{u}^{2}}{g}$

Correct answer

A. $\dfrac{3\sqrt{3}}{8}\dfrac{{u}^{2}}{g}$

Step-by-step solution

{p}_{i} = {p}_{f}

mu + mucos\theta = 2mv

⇒ v = \dfrac{u\left(1 + cos {60}^{0}\right)}{2} = \dfrac{3}{4}u

So horizontal range after collision

= vt

= v\sqrt{\dfrac{{2H}_\max}{g}}

= \dfrac{3}{4}u\sqrt{\dfrac{2{u}^{2}\sin^{2} \left({60}^{0}\right)}{2{g}^{2}}}

= \dfrac{3}{4}{u}^{2}\dfrac{\sqrt{\dfrac{3}{4}}}{g} = \dfrac{3\sqrt{3}{u}^{2}}{8g}

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