A particle moves in a circular path. At a certain instant its acceleration is $\vec{a}$ =(4î+3ĵ) m/s² and its momentum is $\vec{p}$ =(8î-6ĵ) kg·m/s. The motion of the particle is:
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Question

A particle moves in a circular path. At a certain instant its acceleration is $\vec{a}$ =(4î+3ĵ) m/s² and its momentum is $\vec{p}$ =(8î-6ĵ) kg·m/s. The motion of the particle is:

Options

G1 Physics · Accepted Answer

Speed increases or decreases according to p⃗·a⃗. Here p⃗·a⃗=(8)(4)+(-6)(3)=14>0, so the tangential acceleration is along the velocity and the particle is speeding up. Hence it is accelerated circular motion, B.

A particle moves in a circular path. At a certain instant its acceleration is a ⃗ \vec{a} a =(4î+3ĵ) m/s² and its momentum is p ⃗ \vec{p} p ​ =(8î-6ĵ) kg·m/s. The motion of the particle is:

A particle moves in a circular path. At a certain instant its acceleration is a ⃗ \vec{a} a =(4î+3ĵ) m/s² and its momentum is p ⃗ \vec{p} p =(8î-6ĵ) kg·m/s. The motion of the particle is: