JEE Advanced2023PhysicsMotion in Two DimensionsActual
A particle of mass m is moving in the x y -plane such that its velocity at a point ( x , y ) is given as v → = α y x ∧ + 2 x y ∧ where α is a non-zero constant. What is the force F → acting on the particle?
Options
- AF → = 2 m α 2 x x ∧ + y y ∧
- BF → = m α 2 y x ∧ + 2 x y ∧
- CF → = 2 m α 2 y x ∧ + x y ∧
- DF → = m α 2 x x ∧ + 2 y y ∧
Correct answer
A. F → = 2 m α 2 x x ∧ + y y ∧
Step-by-step solution
As force is given by, F → = m d v → d t . Given: v → = α y x ∧ + 2 x y ∧ , where velocity along x-axis is v x = α y and velocity along y-axis is v y = 2 x α . Therefore, d v → d t = α d y d t x ∧ + 2 d x d t y ∧ = α v y x ∧ + 2 v x y ∧ = α 2 x α x ∧ + 2 α y y ∧ = 2 α 2 x x ∧ + y y ∧ Therefore, required value of F → = 2 m α 2 x x ∧ + y y ∧ .