A particle of mass 1 kg is moving in x-y plane with a velocity of 6 m/s at angle of 53° with x-axis. A force of 15 N is applied along x-axis at t = 0 on it. The radius of curvature of the path of the particle just after application of force is __________ m.
Physics · Mixed Chapter
JEE Mains

Question

A particle of mass 1 kg is moving in x-y plane with a velocity of 6 m/s at angle of 53° with x-axis. A force of 15 N is applied along x-axis at t = 0 on it. The radius of curvature of the path of the particle just after application of force is __________ m.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: At the instant after application (t=0⁺), the particle's speed is still v=6 m/s. Step 2: Acceleration from the applied force is a=F/m=15 m s⁻² along +x. The velocity initially makes 53° with +x. Step 3: Perpendicular acceleration to the velocity is a⊥=a sin53°=15×(4/5)=12 m s⁻². Step 4: Radius of curvature satisfies a⊥=v²/R. Hence R=36/12=3 m. Answer: 3.