Question

A ball is released along the wall of a hemispherical bowl of diameter 0.2m. Velocity of the ball at the bottom of the bowl is...​

Answer 1

A ball is released along the wall of a hemispherical bowl of diameter 0.2m. Velocity of the ball at the bottom of the bowl is 1.4 m/s

Explanation:

Given

Diameter of the hemispherical bowl = 0.2 m

Therefore, radius of the hemispherical bowl r = 0.1 m

if the ball is initially the top of the edge of the hemispherical bowl then its potential energy

$PE=mgr$, where m is the mass of the ball

Therefore,

$PE==0.1mg$

Now at the bottom of the bowl, its whole potential energy would have converted into Kinetic Energy, if the velocity at the bottom is v

Then

$0.1mg=\frac{1}{2}mv^2$

$\implies v^2=2\times 0.1\times 9.8$

$\implies v^2=\frac{2\times 2\times 49}{100}$

$\implies v=\sqrt{\frac{2\times 2\times 49}{100}}$

$\implies v=\frac{2\times 7}{10}$

$\implies v=1.4$ m/s

Hope this answer is helpful.

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