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Standard XI

Physics

Gravitational Potential of a Ring

Particle 'A' ...

Question

Particle 'A' moves with speed $10 m/s$ in a frictionless circular fixed horiozontal pipe of radius $5$ m and strikes with $^{'} B^{'}$ of double mass that of A. Coefficient of resitution is $\frac{1}{2}$ and particle 'A' starts its journey at $t = 0$ . The time at which second collision occurs is:

\frac{π}{2} s

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\frac{2 π}{3} s

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\frac{5 π}{2} s

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4 π s

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Solution

The correct option is C $\frac{5 π}{2} s$
For first collision $v = 10 m/s . t_{1} = \frac{π (5)}{10} = \frac{π}{2} sec .$
velocity of seperation $= e$ . velocity of approach.
$v_{2} - v_{1} = \frac{1}{2} (10)$
$v_{2} - v_{1} = 5 m / s$
for second collision
$∴ t_{2} = \frac{2 π (5)}{5} = 2 π$
$∴$ total time $t = t + t_{2} = \frac{π}{2} + 2 π$
$t = \frac{5}{2} π$

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Particle 'A' moves with speed 10 m/s in a frictionless circular fixed horiozontal pipe of radius 5 m and strikes with ′B′ of double mass that of A. Coefficient of resitution is 12 and particle 'A' starts its journey at t=0. The time at which second collision occurs is:

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