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A sphere of radius R is in contact with a wedge. The point of contact is $\frac{R}{5}$ from the ground as shown in the figure. Wedge is moving with velocity 20 ${ms}^{- 1}$ towards left, then the velocity of the sphere at this instant will be

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$20 {ms}^{- 1}$

$15 {ms}^{- 1}$

$16 {ms}^{- 1}$

$12 {ms}^{- 1}$

answer is B.

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Detailed Solution

$\cos θ = \frac{4 R / 5}{R} = 0.8$

$\frac{v_{1}}{v_{2}} = \tan θ = \tan 37 ° = \frac{3}{4}$

$∴ v_{1} = \frac{3}{4} v_{2} = \frac{3}{4} \times 20 = 15 m / s$

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