Question

Two balls P and Q are at opposite ends of the diameter of a frictionless horizontal circular groove. P is projected along the groove and at the end of T second, it strikes ball Q. Let difference in their final velocities be proportional to the initial velocity of ball P and coefficient of proportionally is e then second strike occurs at

• A. 2T/e
• B. e/2T
• C. 2Et
• D. T/2e

Answer 1

Answer: (A)

2T/e

Answer 2

Let u be the initial velocity of ball P, then u = $\frac{2\pi r}{2T}$ and difference in final velocities, v' – v = ev(given)

Time for second strike = $\frac{2\pi r}{v' - v} = \frac{2\pi r}{eu}$

= $\frac{2\pi r}{e \times \frac{2\pi r}{2T}} = \frac{2T}{e}$