JEE MAIN - Physics (2005 - No. 66)
A spherical ball of mass $$20$$ $$kg$$ is stationary at the top of a hill of height $$100$$ $$m$$. It rolls down a smooth surface to the ground, then climbs up another hill of height $$30$$ $$m$$ and finally rolls down to a horizontal base at a height of $$20$$ $$m$$ above the ground. The velocity attained by the ball is
$$20$$ $$m/s$$
$$40$$ $$m/s$$
$$10\sqrt {30} \,\,\,m/s$$
$$10\,\,m/s$$
Explanation
Loss in potential energy $$=$$ gain in kinetic energy
$$m \times g \times 80 = {1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $$10 \times 80 = {1 \over 2}{v^2}$$
$$ \Rightarrow $$$${v^2} = 1600$$ or $$v = 40\,m/s$$
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