Two identical spheres, each of mass m are suspended by vertical...

Let us assume that after the collision, the velocity of the incoming ball changes from& $u$~to& $u^{'}$~and the other two balls move in the opposite directions with the same speed& $v$, then

$- 2\left(N ∆ t\right)\cos 30 ° = mu^{'} - mu$

$⇒ - \left(N ∆ t\right)\sqrt{3} = mu^{'} - mu$

For the ball attached to the string,

\left(N ∆ t\right)\sin 30 ° = mv$</p><p>$⇒ N ∆ t = 2mv$</p><p>Eliminating~$N ∆ t$, we obtain</p><p>$- 2\sqrt{3}mv = mu^{'} - mu$</p><p>Using the coefficient of restitution equation we get</p><p>$1 = \dfrac{vcos60 ° - u^{'}\cos 30 °}{ucos30 °}$</p><p>$⇒ \sqrt{3}u = v - \sqrt{3}u^{'}$</p><p>$⇒ 2\sqrt{3}v = 6u + 6u^{'}$</p><p>$6u + 6u^{'} = u - u^{'}$</p><p>$⇒ u^{'} = - \dfrac{5}{7}u$</p><p>The total impulse of tension is</p><p>$- 2T ∆ t = mu^{'} - mu$</p><p>$2T ∆ t = m\dfrac{5u}{7} + mu = \dfrac{12}{7}mu