Question

A spherical ball of mass $20kg$ is stationary at the top of a hill of height $100m$. It rolls down a smooth surface to the ground, then climbs up another hill of height of $20m$ above the ground. The velocity attained by the ball is

• A. $40m{s}^{-1}$
• B. $20m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. $10\sqrt{30}m{s}^{-1}$

Answer 1

The correct option is A

$40m{s}^{-1}$

Step 1. Given Data:

Mass of the ball $m=20kg$

Height of the first hill $h_1=100m$

Height of the second hill $h_2=20m$

Step 2. Find the velocity of the ball:

If $v$ is the speed of the ball and $g$ is the acceleration due to gravity then,

According to the law of conservation of energy we have

$mg{h}_1=mg{h}_2+\left(\frac{1}{2}m{v}^2\right)$

$\Rightarrow$ $mg\left(h_1-h_2\right)=\frac{1}{2}m{v}^2$

$\Rightarrow$ $v=\sqrt{2g\left(h_1-h_2\right)}$

$=\sqrt{2\times 10\times \left(100-20\right)}$

$=40m{s}^{-1}$

Hence option (A) is correct.