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

Two particles A and 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 B initially at a distance l from A. The particle at end B is projected across the table with speed u perpendicular to AB. Velocity of ball A just after the string is taut, is –

A

u34\frac{u\sqrt{3}}{4}

B

u3u\sqrt{3}

C

u32\frac{u\sqrt{3}}{2}

D

u2\frac{u}{2}

Answer

u34\frac{u\sqrt{3}}{4}

Explanation

Solution

When the string jerk tight both particles begin to move with velocity components v in the direction AB. Using conservation of momentum in the direction AB

mu cos 300 = mv + mv

or v = u34\frac{u\sqrt{3}}{4}

Hence, the velocity of ball A just after the jerk is u34\frac{u\sqrt{3}}{4}