Find the velocity of separation of the spheres 1 and 2 if the sphere 3 comes down with a velocity of 3m/s. Assume all spheres to be identical and frictionless.
A
2√3m/s
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B
4√3m/s
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C
6√3m/s
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D
8√3m/s
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Solution
The correct option is C6√3m/s The spheres are identical, the centre to centre distance between the spheres will be equal and thus, they will form an equilateral triangle. So, the value of θ will be 30∘ in the above figure shown.
By condition of wedge constraint, acceleration along normal to the common interface should be same.
So, V3cosθ=V2sinθ ⇒V2=V3cotθ=3cot30∘ ∵θis30∘ ⇒V2=3√3
Also, due to similarity V1 will be 3√3
Now, the spheres 1 and 2 will move apart in the opposite direction ∴ velocity of separation will be V12=V1−(−V2)=3√3+3√3=6√3m/s