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 30m and finally rolls down to a horizontal base at a height of 20m above the ground. The velocity attained by the ball is:
A) $40 m/s$
B) $20 m/s$
C) $10 m/s$
D) $10\sqrt {30} m/s$

Answer 1

Hint: Here, we have to apply the law of conservation of energy. In the subject of Physics, there are several quantities that are conserved.
Conserved quantity is a function of the dependent variables in a dynamically changing system whose value remains equal along the entire phase and the path that the system undergoes. Examples of the conserved quantities are mass, charge, energy, and momentum.

Complete step by step answer:
The energy is defined as the ability to perform work.
There are many forms and manifestations of energy, however, the most common and most dispensable form of energy is mechanical energy.
The mechanical energy is the energy possessed by a body under motion which is the combination of the energy possessed by virtue of the position and the energy possessed by the virtue of its speed.
The energy possessed by the virtue of its position is called potential energy and the energy possessed by the virtue of its velocity is called kinetic energy.
Mechanical Energy = Potential Energy + Kinetic energy
Since, we have seen that the quantity energy is conserved here, let us understand the law of conservation of energy.
It states that,
Energy can neither be created nor destroyed but can be converted from one form to another.
This means that the total mechanical energy of the system remains the same and the potential and kinetic energy are interconvertible.
Total energy,
$E = KE + PE$
Since energy is conserved, we have –
$KE = PE$
Given that the ball is at the top of a hill at height 100m.
The potential energy of the ball at this height, –
${P_1} = mgh$
where m = mass, g = acceleration due to gravity and h = height.
When it descends to the height of 20 m in the end, the loss in potential energy is given by –
$PE = mg\left( {100 - 20} \right) = 80mg$
Given,
Mass, m = 20 kg
Acceleration due to gravity, $g = 10m{s^{ - 2}}$
Substituting, we get –
$PE = 80 \times 20 \times 10 = 16000J$
As per the conservation of energy, this loss in potential energy is equal to the gain in the kinetic energy of the ball.
Kinetic energy,
$KE = \dfrac{1}{2}m{v^2}$
where m = mass and v = velocity of the ball.
Thus, $\dfrac{1}{2}m{v^2} = PE$
$ \Rightarrow \dfrac{1}{2} \times 20 \times {v^2} = 16000$
$ \Rightarrow {v^2} = \dfrac{{16000 \times 2}}{{20}} = 1600$
$ \Rightarrow v = \sqrt {1600} = 40m{s^{ - 1}}$

Thus, the velocity is equal to $40m{s^{ - 1}}$.

Note: Even though the law of conservation of energy states that the potential energy gets converted to kinetic energy, it is an ideal case to assume that all of the potential energy is converted to kinetic energy. There is a loss of energy in the form of heat, light or sound in the real scenarios.