As shown below, bob A of a pendulum having a | vedclass

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(A) $1$. Velocity of bob $A$ just before hitting bob $B$:Using the principle of conservation of mechanical energy, the change in height $h$ is $R - R

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$\frac{1}{3} \sqrt{Rg}$

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(A) $1$. Velocity of bob $A$ just before hitting bob $B$:Using the principle of conservation of mechanical energy, the change in height $h$ is $R - R \cos(60^{\circ}) = R - R/2 = R/2$.$v = \sqrt{2gh} = \sqrt{2g(R/2)} = \sqrt{gR}$.$2$. Let $u = \sqrt{gR}$ be the velocity of $A$ just before the collision.Let $v_1$ and $v_2$ be the velocities of $A$ and $B$ respectively just after the elastic collision.$3$. By the Law of Conservation of Momentum $(COM)$:$mu = mv_1 + (m/2)v_2$$u = v_1 + v_2/2$$2u = 2v_1 + v_2$ --- $(i)$$4$. For an elastic collision, the coefficient of restitution $e = 1$:$e = (v_2 - v_1) / u = 1$$v_2 - v_1 = u$ --- (ii)$5$. Solving equations $(i)$ and (ii):From (ii), $v_2 = u + v_1$.Substitute into $(i)$: $2u = 2v_1 + (u + v_1)$$2u = 3v_1 + u$$u = 3v_1$$v_1 = u/3 = \frac{1}{3} \sqrt{gR}$.

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