A particle moving in the xy plane experiences a velocity dependent force
F⃗ = k( v_yî + vₓĵ ) , where vₓ and v y are the
x and y components of its velocity v⃗ . If a⃗ is the
acceleration of the particle, then
which of the following statements is true for the particle?

Solution

Physics sir · 2 steps

Write aₓ and a_y from the force, then check which quantity has zero time derivative.

1. Write the acceleration. From a⃗ = F⃗/m, we get aₓ = dvₓ/dt = k/mv_y and a_y = dv_y/dt = k/mvₓ. The motion stays in the xy plane, so v⃗ × a⃗ points along k̂.

   v⃗ × a⃗ = (vₓ a_y - v_y aₓ) k̂ = k/m(vₓ² - v_y²)k̂

   Option	Quantity	Its rate of change	Result
   (a)	K ∝ vₓ² + v_y²	2vₓk/mv_y + 2v_yk/mvₓ = 4k/mvₓv_y	changes
   (b)	v⃗ × a⃗ ∝ vₓ² - v_y²	2vₓk/mv_y - 2v_yk/mvₓ = 0	constant
   (c)	v⃗ · a⃗ = 2k/mvₓv_y	2k²/m²(vₓ² + v_y²), which is positive for a moving particle	changes
   (d)	F⃗ · v⃗ = 2k vₓv_y	nonzero in general, so F⃗ does work	not magnetic

2. So. Only vₓ² - v_y² stays fixed, so v⃗ × a⃗ is constant in time. The answer is (b).

The idea to keep. To test whether a quantity is conserved, differentiate it with respect to time and substitute the equations of motion. If the terms cancel to zero, it is constant.

Answer from the NTA final answer key. Solution by PrepWiser.