Kinematics

JEE Main level, numerical answer, medium

231 of 308

Question 231 of 308

A particle of mass 100g100g is projected at time t=0t=0 with a speed 20ms–120m{s}^{–1} at an angle 45∘45^{\circ} to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2st=2s is found to be Kkgm2s−1\sqrt{K}kg{m}^{2}{s}^{-1}. The value of KK is ______.
(Take g=10ms−2g=10m{s}^{-2})

Show answer and solution

Answer: 800

As torque due to gravitational force will be variable, therefore we can use
ΔL=∫0tτdt\Delta L={\int }_{0}^{t}\tau dt.
Now, torque of mgmg at time tt =(vxt)mg=\left({v}_{x}t\right)mg and horizontal velocity of the projectile will remain constant(vx=vcos⁡45∘=20×12=102)\left({v}_{x}=v\cos 45^{\circ}=20\times \dfrac{1}{\sqrt{2}}=10\sqrt{2}\right).
L0=∫02mg(vxt)dt{L}_{0}={\int }_{0}^{2}mg\left({v}_{x}t\right)dt
=mgvxt22=(0.1)(10)(102)222=mg{v}_{x}\dfrac{{t}^{2}}{2} =(0.1)(10)(10\sqrt{2})\dfrac{{2}^{2}}{2}
=202=20\sqrt{2}
=800kgm2s−1=\sqrt{800}kg{m}^{2}{s}^{-1}
Therefore, K=800K=800.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Projectile motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.