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Physics-

General

Easy

Question

In the given figure four identical spheres of equal mass $m$ are suspended by wires of equal length $l subscript 0 end subscript$ , so that all spheres are almost touching to each other. If the sphere 1 is released from the horizontal position and all collisions are elastic, the velocity of sphere 4 just after collision is

$\sqrt{2 g l_{0}}$
$\sqrt{3 g l_{0}}$
$\sqrt{g l_{0}}$
$\sqrt{\frac{g l_{0}}{2}}$

The correct answer is: $square root of 2 g l subscript 0 end subscript end root$

When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height $l subscript 0 end subscript equals$ kinetic energy at bottom
Or $m g l subscript 0 end subscript equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent$
Or $v equals square root of 2 g l subscript 0 end subscript end root$
Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to $square root of 2 g l subscript 0 end subscript end root$

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