JEE Advanced 2021 Paper 1 · Physics · Rotational Motion · hard

A particle of mass 1 kg is projected from the origin O of a vertical x-y plane with speed 20 m s^-1 at 60^° above the horizontal. Magnitudes of…

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A particle of mass

1\ \mathrm{kg}

is projected from the origin

O

of a vertical

x

y

plane with speed

20\ \mathrm{m\,s^{-1}}

60^\circ

above the horizontal. Take

g=10\ \mathrm{m\,s^{-2}}

and neglect air resistance. Magnitudes of angular momenta are in

\mathrm{kg\,m^2\,s^{-1}}

. Choose the correct statement(s).

A. The angular momentum about $O$ at $t=1\ \mathrm{s}$ is $50$ .
B. The torque of gravity about $O$ is constant in time.
C. The angular momentum about $O$ just before the particle returns to the ground level is $300$ .
D. About the highest point of the trajectory, the angular momentum at $t=1\ \mathrm{s}$ is about $26.8$ .

Show answer and solution

Answer: A, D

With

x=u\cos\alpha\,t

and

y=u\sin\alpha\,t-\tfrac12gt^2

L_z=m(xv_y-yv_x)=-\tfrac12mg\,u\cos\alpha\,t^2

, so

|L|=\tfrac12\cdot10\cdot10\,t^2=50t^2

. (A) At

t=1

|L|=50

: correct. (B) The torque is

mg\,x=mg\,u\cos\alpha\,t=100t

N m, which grows linearly with time: incorrect. (C) The flight time is

2u\sin\alpha/g=2\sqrt3

s, so

t^2=12

and

|L|=600

, not

300

: incorrect. (D) The apex is at

(10\sqrt3,15)

m. At

t=1

s the particle is at

(10,\,12.32)

with velocity

(10,\,7.32)

, so relative to the apex

r=(-7.32,-2.68)

and

L_z=-7.32\times7.32+2.68\times10=-26.8

, magnitude

26.8

: correct. Hence (A) and (D).

Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake