Post Utme Afe Babalola University 2017 Physics Question 9

Practice objective / multiple choice question 9 from the 2017 Post Utme Afe Babalola University Physics examination.

Post Utme Afe Babalola University 2017 Physics Objective / Multiple Choice
Question 9 POST_UTME_AFE_BABALOLA_UNIVERSITY • 2017 • PHYSICS • objective

A system consists of two particles of mass $m_1$ and $m_2$ connected by a massless string of length $L$. The particles are moving in a circular orbit around their common center of mass. If the speed of $m_1$ is $v_1$ and the speed of $m_2$ is $v_2$, what is the total energy of the system?

Answer Options

A. E = \frac{1}{2}m_1v_1^2 - \frac{1}{2}m_2v_2^2
Correct Answer B. E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2
C. E = \frac{1}{2}m_1v_1^2 - \frac{1}{2}m_2v_2^2 - \frac{k}{L}
D. E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{k}{L}
Correct Answer
B
Correct Option:
E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

About This Question

This is Post Utme Afe Babalola University 2017 Physics Question 9. It is one of the objective questions from the 2017 Post Utme Afe Babalola University Physics examination.

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