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A solid sphere of mass 1.5 kg rolls on a frictionless horizontal surface, its centre of mass moving at a speed of 5 ms^–1. Then it rolls up an incline of 30⁰ to horizontal. The height attained by the sphere before it stops is (g = 10 ms^–2)

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3.65m

1.75 m

1.5m

5 m

answer is C.

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Detailed Solution

$x = ?, m = 1.5 kg, v = 5 m / s, θ = 30^{°}$

We know that,

$\frac{1}{2} {mv}^{2} + \frac{1}{2} {Iω}^{2} = mgh$

$\frac{1}{2} {mv}^{2} + \frac{1}{2} \frac{2}{5} {mr}^{2} \frac{v^{2}}{r^{2}} = mgh$

$m [\frac{1}{2} v^{2} + \frac{1}{5} v^{2}] = mgh$

$\frac{7}{10} v^{2} = gh$

$h = \frac{7 v^{2}}{10 g} = \frac{7 \times 5^{2}}{10 \times 10} = 1.75 m$

Hence the correct answer is 1.75m.

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