Question

A steel ball of mass 5 g is thrown downward with a velocity of 10 \[m{s^{ - 1}}\] from a height of 19.5 m and it penetrates sand by 50 cm. The change in mechanical energy will be (take, g =10 \[m{s^{ - 2}}\] ):
A. 1 J
B. 1.25 J
C. 1.5 J
D. 1.75 J

Answer 1

Hint:Mechanical energy is of two types: potential and kinetic. Hence, the total mechanical energy of a system at a point is the sum of potential and kinetic energies at the given point. When kinetic energy is maximum, the potential energy is minimum and vice-versa.

Formula(e) Used:
Kinetic Energy, \[K.E. = \dfrac{1}{2}m{v^2}\]
where m = mass of the object and v = velocity of the object.
Potential Energy, \[P.E. = mgh\]
where m = mass of the object, g= acceleration due to gravity = 10 \[m{s^{ - 2}}\] and h = height.

Complete step by step solution:
Given: mass of the steel ball, m = 5 g = 0.005 kg
Velocity of the ball, v = 10 \[m{s^{ - 1}}\]
Height, h = 19.5 m
Depth penetration, d = 50 cm = 0.5 m
In this situation, the potential energy of the system will be maximum before being thrown and zero when the ball falls on the ground into the sand. At that point, the kinetic energy of the system will be maximum.

We know that the kinetic energy of a system is given by,
\[\begin{array}{l}K.E. = \dfrac{1}{2}m{v^2}\\\end{array}\]---------(1)
So, the kinetic energy of the steel ball will be
\[K.E. = \dfrac{1}{2}X0.005X10X10\]
\[\Rightarrow K.E. = 0.25\] J

We know that the potential energy of a system is given by
\[P.E. = mgh\]---------------(2)
Here, the total height will be (h+d) = (19.5 + 0.5) = 20 m
So, \[P.E. = 0.005 \times 10 \times 20\]
\[P.E. = 1\] J
Total mechanical energy = Potential energy + Kinetic energy ------(3)
\[U = 1 + 0.25\]
\[\therefore U = 1.25\] J
Hence mechanical energy of the system is 1.25 J.

Hence option (B) is the correct option.

Note: The total height acquired by the steel ball for calculation of the potential energy will be the height from which it is thrown and the depression it makes in the sand and not just the height from which the ball is thrown. When a ball penetrates the sand, the potential energy at that point is taken to be zero.