Two particles A and B each of mass m are attached by a light inextensible string of length 2 l . The whole…

Two particles $\mathrm{A}$ and $\mathrm{B}$ each of mass $m$ are attached by a light inextensible string of length $2l$. The whole system lies on a smooth horizontal table with $\mathrm{B}$ initially at a distance $l$ from $\mathrm{A}$. The particle at the end $\mathrm{B}$ is projected across the table with speed $u$ perpendicular to $\mathrm{A}\mathrm{B}$. The velocity of the ball $\mathrm{A}$ just after the string is taut, is

1. $\frac{\mathrm{u}\sqrt{3}}{4}$
2. $\mathrm{u}\sqrt{3}$
3. $\frac{\mathrm{u}\sqrt{3}}{2}$
4. $\frac{\mathrm{u}}{2}$

Solution

When the string jerk tight both particles begin to move with velocity components $\mathrm{v}$ in the direction $\mathrm{A}\mathrm{B}$ . Using conservation of momentum in the direction $\mathrm{A}\mathrm{B}$

$\mathrm{m}\mathrm{u}\cos \mathrm{ }30°=\mathrm{m}\mathrm{v}+\mathrm{m}\mathrm{v}$

or $\mathrm{v}=\frac{\mathrm{u}\sqrt{3}}{4}$

Hence, the velocity of ball $\mathrm{A}$ just after the jerk is $\frac{\mathrm{u}\sqrt{3}}{4}$

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