Question

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 and then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height 20 m above the ground. The velocity attained by the ball is:
A. $40\,m{s^{ - 1}}$
B. $20\,m{s^{ - 1}}$
C. $10\,m{s^{ - 1}}$
D. $10\sqrt {30} \,m{s^{ - 1}}$

Answer 1

The loss in the potential energy of the spherical ball is equal to the gain in kinetic energy of the ball. We calculate the change in potential energy, the negative of which is the kinetic energy. The kinetic energy is related to the velocity of the spherical ball.

Formula used:
$\Delta P.E = - \Delta K.E$
Here, $\Delta P.E$ is the change in potential energy and $\Delta K.E$ is the change in kinetic energy.

Complete step by step answer:
We know that, according to conservation of total energy, the change in potential energy of the body is equal to change in kinetic energy of the body. Therefore, if we calculate the change in potential energy of the ball, we can calculate the final kinetic energy of the spherical ball. The kinetic energy is proportional to square of velocity of the body.
Suppose the initial height of the spherical ball from the ground is ${h_1}$ and the final height of the spherical ball from the ground is ${h_2}$.
The change in potential energy of the spherical ball is,
$\Delta P.E = mg{h_2} - mg{h_1}$
$\Rightarrow \Delta P.E = mg\left( {{h_2} - {h_1}} \right)$
Here, m is the mass of the spherical ball and g is the acceleration due to gravity.
Substitute 20 kg for m, $10\,m/{s^2}$ for g, 20 m for ${h_2}$ and 100 m for ${h_1}$ in the above equation.
$\Delta P.E = \left( {20\,kg} \right)\left( {10\,m/{s^2}} \right)\left( {20\,m - 100\,m} \right)$
$\therefore \Delta P.E = - 16000\,J$
We know that,
$\Delta P.E = - \Delta K.E$
$\Rightarrow \Delta P.E = - \left( {\dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2} \right)$
$\Rightarrow \Delta P.E = \dfrac{1}{2}mv_1^2 - \dfrac{1}{2}mv_2^2$
Here, ${v_1}$ is the initial velocity of the ball and ${v_2}$ is the final velocity of the ball.
The initial velocity of the ball is zero. Therefore, we can write the above equation as follows,
$\Delta P.E = - \dfrac{1}{2}mv_2^2$
Substitute $- 16000\,J$ for $\Delta P.E$ and 20 kg for m in the above equation.
$- 16000 = - \dfrac{1}{2}\left( {20\,kg} \right)v_2^2$
$\Rightarrow 1600 = v_2^2$
$\Rightarrow {v_2} = \sqrt {1600}$
$\therefore {v_2} = 40\,m/s$

So, the correct answer is “Option A”.

Note:
Use $\Delta P.E = - \Delta K.E$ and not $\Delta P.E = \Delta K.E$.
The negative sign implies that the gain in potential energy is equal to the loss in the kinetic energy.
In this question, the potential energy lost by the ball is equal to the kinetic energy gained by the ball.