Question

If a particle covers half the circle of radius R with constant speed then,
A. Momentum change is \[mvr\]
B. Change in K.E. is\[\dfrac{1}{2}m{v^2}\]
C. Change in K.E. is \[m{v^2}\]
D. Change in K.E. is zero

Answer 1

Hint:The momentum is a vector quantity and has direction and magnitude. It is the product of the mass of the particle and velocity.

Formula used:
\[\overrightarrow p = m\overrightarrow v \]
Here \[\overrightarrow p \]is the momentum of the particle of mass m and moving with velocity \[\overrightarrow v \]
\[K.E. = \dfrac{1}{2}m{v^2}\]
Here K.E. is the kinetic energy of the particle of mass m and speed v.

Complete step by step solution:

Image: Particle in the circle of radius R

Let the constant speed of the particle is v and it starts from position velocity pointing upward. Then after completing half revolution around the circle of radius R, the velocity vector will be downward. So, the change in momentum will be the difference between the final momentum to the initial momentum,
\[\Delta p = {p_f} - {p_i}\]
\[\Rightarrow \Delta p = - mv - mv\]
\[\Rightarrow \Delta p = - 2mv\]
So, the momentum change is not equal to mvr

The change in kinetic energy of the particle will be the difference between the final kinetic energy to the initial kinetic energy. The initial kinetic energy of the particle is,
\[{K_i} = \dfrac{1}{2}m{v^2}\]
The final kinetic energy of the particle is,
\[{K_f} = \dfrac{1}{2}m{v^2}\]
So, the change in kinetic energy of the particle is,
\[\Delta K = {K_f} - {K_i}\]
\[\Rightarrow \Delta K = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{v^2}\]
\[\therefore \Delta K = 0\]
Hence, the change in kinetic energy of the particle moving in the circular path is zero.

Therefore, the correct option is D.

Note: As the speed of the particle is constant so there is no acceleration acting on it. So, the net force acting on the particle is zero. As there is no force acting on it so the work done also is zero. Using the work-energy theorem, the change in kinetic energy is the work done on it. So, the change in kinetic energy of the particle is zero.