Kinetic energy and momentum change in uniform circular motion – Rankers Physics

Kinetic Energy and Momentum: Practice Problem & Solution

A particle of mass \(M\) is moving in a horizontal circle of radius \(R\) with uniform speed \(v\). When it moves from one point to a diametrically opposite point, its: (1992)
Kinetic energy change by \(Mv^2/4\)
Momentum does not change
Momentum change by \(2Mv\)
Kinetic energy changes by \(Mv^2\)

Solution Explained:

To solve this problem, we apply the core principles of Kinetic Energy and Momentum. Understanding the underlying formula is key to arriving at the correct answer below:

Concept: Momentum and kinetic energy in uniform circular motion. Formula: Momentum \(p = Mv\), Kinetic Energy \(KE = \frac{1}{2}Mv^2\). Since speed \(v\) is uniform, KE remains constant (\(\Delta KE = 0\)). At diametrically opposite points, the direction of velocity reverses. If initial momentum is \(\vec{p_1} = M\vec{v}\), then final momentum is \(\vec{p_2} = -M\vec{v}\). The change in momentum is \(\Delta\vec{p} = \vec{p_2} - \vec{p_1} = -2M\vec{v}\). The magnitude of the change is \(2Mv\).

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