Three balls A, B and C whose masses are m, km and 4m respectively are kept at rest on the horizontal smooth surface as shown in the figure. The ball A is given a velocity $v_{o}$ rightward and it collides with the ball B elastically. Then ball B collides elastically third ball C. For what value of k, does the third ball C gain the maximum speed?

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Solution

Let the velocities of balls B and A after collision be $v_{1} a n d v_{2} r e s p e c t i v e l y$
Since, collision is elastic.Therefore e=1
$\Rightarrow 1 = - \frac{v_{1} - v_{2}}{0 - v_{0}} \Rightarrow v_{1} - v_{2} = v_{0} ⟶ (e q . 1)$
From conservation of momentum eq. we have,
$k m v_{1} + m v_{2} = m v_{0} \Rightarrow k v_{1} + v_{2} = v_{0} ⟶ (e q . 2)$
So, from the two eq.s we have,
$v_{1} = \frac{2 v_{0}}{k + 1}$
Now suppose velocities of ball B and C after collision are $v_{2} a n d v_{3}$
Again e=1 so,
$1 = - \frac{v_{3} - v_{2}}{0 - v_{1}} = \frac{v_{3} - v_{2}}{0 - 2 v_{0} / (k + 1)} \Rightarrow v_{3} - v_{2} = 2 v_{0} / (k + 1) ⟶ (e q . 3) F r o m c o n s e r v a t i o n o f m o m e n t u m p r i n c i p l e, 4 m v_{3} + k m v_{2} = k m \times 2 v_{0} / (k + 1) \Rightarrow 4 v_{3} / k + v_{2} = 2 v_{0} / (k + 1) ⟶ (e q . 4) S o, f r o m e q . 3 a n d e q . 4 w e h a v e, v_{3} = 4 k v_{0} / {(k + 1) (k + 4)} N o w, \frac{d v_{3}}{d k} = 0 (f o r m a x i m u m v e l o c i t y) \Rightarrow 16 - 8 k^{2} = 0 \Rightarrow k = \sqrt{2}$

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