A particle of mass 1 kg is acted on by F ⃗ \vec{F} F = -k(x î + y ĵ), where k = 1 kg s⁻². At t = 0, r ⃗ \vec{r} r = (î/√2 + √2 ĵ) m and v ⃗ \vec{v} v = (-√2 î + √2 ĵ + (2/π)k̂) m s⁻¹. Ignore gravity. When z = 0.5 m, find x vᵧ - y vₓ in m² s⁻¹.
Physics · Rotational Motion · torque and rotational equilibrium
JEE Advanced 2022

Question

A particle of mass 1 kg is acted on by F ⃗ \vec{F} F = -k(x î + y ĵ), where k = 1 kg s⁻². At t = 0, r ⃗ \vec{r} r = (î/√2 + √2 ĵ) m and v ⃗ \vec{v} v = (-√2 î + √2 ĵ + (2/π)k̂) m s⁻¹. Ignore gravity. When z = 0.5 m, find x vᵧ - y vₓ in m² s⁻¹.

Physics · Rotational Motion · torque and rotational equilibrium

JEE Advanced 2022

G1 Physics · Accepted Answer

Start from Newton’s laws for translation and rotation: ΣF=Ma and Στ=Iα. Use the geometrical constraint/energy relation stated in the question and solve for the required quantity. The result is 3.