JEE Advanced 2022 Paper 2

A particle of mass 1‌kg is subjected to a force which depends on the position as

→

F

= −k(x

∧

ı

+y

∧



)‌kg‌ms−2 with k=1‌kg‌s−2. At time t=0, the particle's position

→

r

= (‌

1

√2

∧

ı

+√2

∧



)m and its velocity

→

v

=(−√2

∧

ı

+√2

∧



+‌

2

π

∧

k

)ms−1. Let vx and vy denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z=0.5m, the value of (xvy−yvx) is m2s−1