Question

A particle of mass 2 kg is moving in a circular path with constant speed 20 m/s. The magnitude of change in velocity when particle travels from A to P will be:
 $ A)20\sqrt 2 m/s $
 $ B)40m/s $
 $ C)40\sqrt 2 m/s $
 $ D)zero $

Answer 1

Hint: According to the question, a particle of mass $ 2kg $ is moving in a circular path with constant speed $ 20m/s $ and we need to calculate the magnitude of change in velocity when the particle travels from A to P. So here we need to apply the concept of centripetal force and centripetal acceleration.

Complete step by step answer:
Centripetal acceleration, the acceleration of a body traversing a circular path. Because velocity is a vector quantity (that is, it has both the magnitude, the speed, and a direction), when a body travels on a circular path, its direction constantly changes and thus its velocity changes, producing an acceleration. The acceleration is directed radially toward the centre of the circle. The centripetal acceleration ac has a magnitude equal to the square of the body’s speed v along the curve divided by the distance r from the centre of the circle to the moving body; that is, $ {a_c} = \dfrac{{{v^2}}}{r} $ . Centripetal acceleration has units of metre per second squared. The force causing this acceleration is directed also toward the centre of the circle and is named centripetal force.
After completing half revolution:
The particle speed remains constant.
So, it can be said that the change in speed will be $ 0m/s $
However, there is a change in velocity vector as both have the same magnitude but opposite directions.
So the magnitude of change in velocity will be $ 20m/s $
So the change in velocity vector= $ 20\widehat i - 20\widehat j $ ...... $ (x\widehat i - y\widehat {j)} $
So, the magnitude of this change in velocity vector= $ \sqrt {({{20}^2}) + ( - {{20}^2})} $
= $ \sqrt {400 + 400} m/{s^2} $
= $ \sqrt {800} m/{s^2} $
= $ 20\sqrt 2 m/{s^2} $
So, the final answer is $ 20\sqrt 2 m/{s^2} $ .

Additional Information:
From his law of centrifugal force and Kepler's third law of planetary motion, Newton deduced that the centrifugal (and hence centripetal) force of the Moon or of any planet must decrease as the inverse square of its distance from the centre of its motion. Newton coined the term centripetal force in his discussions of gravity in his De motu corporum in gyrum, a 1684 manuscript which he sent to Edmond Halley.

Note:
It is very important to understand the concept of speed and velocity to solve such questions. As seen in the question, the particle speed is constant, so, the change in speed is zero but since there is a change in direction, there will be a change in velocity vector.