NDA General Ability Question (2026) — Solution

Step-by-step solution

Given position vector r = 3 t^2 i + 2 t j + 5 k Velocity v = d r dt = 2 3 t i + 2 j Instantaneous momentum p = m v = m(2 3 t i + 2 j ) Acceleration a = d v dt = 2 3 i Force F = m a = 2 3 m i Taking the cross product of force and momentum: F p = (2 3 m i ) m(2 3 t i + 2 j ) = 2 6 m^2 k 0 Since the cross product is non-zero, the force is not parallel to the instantaneous momentum. Also, F p = 12m^2t 0, so they are not perpendicular either. Answer: The force acting on the particle is not parallel to the instantaneous momentum of the particle.