Question

What change should be affected in the velocity of the body to maintain the same kinetic energy if its mass is increased by four times?

Answer 1

Hint: Kinetic energy of the given body is the energy due to the motion of the body. Kinetic energy is directly proportional to the product of mass and square of the velocity of the body. Kinetic energy increases with an increase in mass; to keep it the same to its initial value, we have to reduce the body's velocity.

Complete step by step solution:
Assume:
The initial mass of the body is \[{m_1}\].
The initial velocity of the body is \[{v_1}\].
The final mass of the body is \[{m_2}\].
The final velocity of the body is \[{v_2}\].

The expression for initial kinetic of the body is given as:
\[K.{E_1} = \dfrac{1}{2}{m_1}v_1^2\]……(1)

The expression for the final kinetic energy of the body is:
\[K.{E_2} = \dfrac{1}{2}{m_2}v_2^2\]……(2)

On dividing equation (1) and equation (2), we get:
\[\begin{array}{l}
\dfrac{{K.{E_2}}}{{K.{E_1}}} = \dfrac{{\dfrac{1}{2}{m_2}v_2^2}}{{\dfrac{1}{2}{m_1}v_1^2}}\\
 = \dfrac{{{m_2}v_2^2}}{{{m_1}v_1^2}}
\end{array}\]……(3)

It is given that the mass of the body is increased by four times.
\[{m_2} = 4{m_1}\]

Substitute \[4{m_1}\] for \[{m_2}\] in equation (3).
\[\begin{array}{l}
\dfrac{{K.{E_2}}}{{K.{E_1}}} = \dfrac{{\left( {4{m_1}} \right)v_2^2}}{{{m_1}v_1^2}}\\
 = \dfrac{{4v_2^2}}{{v_1^2}}
\end{array}\]……(4)

It is also given that initial and final kinetic energies of the body are the same, so their ratio can be written as:
\[\dfrac{{K.{E_2}}}{{K.{E_1}}} = 1\]

Substitute \[1\] for \[\dfrac{{K.{E_2}}}{{K.{E_1}}}\] in equation (4).
\[\begin{array}{l}
1 = \dfrac{{4v_2^2}}{{v_1^2}}\\
v_1^2 = 4v_2^2
\end{array}\]
Taking the square root of both sides of the above expression, we will find the relation between the given body's initial and final velocity.
\[\begin{array}{l}
{v_1} = 2{v_2}\\
{v_2} = \dfrac{{{v_1}}}{2}
\end{array}\]

Therefore, to have the same kinetic energy, the body's final velocity must be half of its initial velocity if its mass is increased by four times.

Note: Alternate method: Rather than substituting the ratio of initial and final kinetic energy, we can also equate the right-hand side of equation (2) and equation (2) because it is given initial and final kinetic energies of the body are the same.