JEE Advanced Model Paper 8 with solutions for online practice

Question : 77 of 84

Marks: +1, -0

A heavy ball of mass 2m rests on the horizontal surface and a lighter ball of mass m isdropped from a height h > 2l . At the instant the string gets taut, upward velocity of heavy ballwill be

√gl . Find the value of ‘x’.

Your Answer:

Validate

Solution:

Just before getting taut, speed of m is
V1 = √2g×2l = 2 √gl
Just after getting taut let the velocity of m be V2 (downwards) and so velocity of 2m will be V2
(upwards) upwards is taken as + ve.
For m1 l = (−mV2) - (-mV1) = mV1 - mV2
For 2m1 l = (2mV2) - (0) = 2mV2
∴ mV1 - mV2 = 2mV2 ⇒ V2 =

√gl

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