Question

A particle of mass m is moving in a horizontal circle of radius R with a uniform speed v. What happens when it moves from one point to a diametrically opposite point?
A. Kinetic energy changes by \[\dfrac{{m{v^2}}}{4}\]
B. Momentum doesn’t change
C. Momentum changes by 2mv
D. Kinetic energy changes by \[m{v^2}\]

Answer 1

Hint:Before we start addressing the problem, we need to know about momentum. Momentum is defined as the product of the mass of a particle and the velocity.

Formula Used:
To find the momentum the formula is,
\[P = mv\]
Where, m is mass and v is velocity.

Complete step by step solution:
Consider a particle of mass M that is moving in a horizontal circle of radius R with a uniform speed V. We need to find what happens when it moves from one point to a diametrically opposite point. In the case of circular motion, the velocities of the body at a point and the point just diametrically opposite to it are equal but opposite in direction.

Suppose, at point A, the velocity of the body is v, then on the opposite point, the velocity is – v. Here, the momentum at the initial point is,
\[P = m \times v\]
Then, the momentum at a diametrically opposite point is,
\[P = m \times \left( { - v} \right)\]
So, the change in momentum will be,
\[P = mv - \left( { - mv} \right)\]
\[\Rightarrow P = mv + mv\]
\[\therefore P = 2mv\]
Therefore, when a particle moves from one point to a diametrically opposite point the Momentum changes by 2mv.

Note:In this problem it is important to remember that when a particle moves from one point to a diametrically opposite point, then the direction of velocity changes.