Question

What happens to the kinetic energy when:
1) The mass of the body is doubled at constant velocity.
2) The velocity is doubled at constant mass.
3) The mass of the body is doubled but velocity is reduced to half.

Answer 1

Hint: For this question, you must know the concept of kinetic energy, formula of kinetic energy and velocity. After that just by doing a simple calculation according to the question we will come to know all the answers to our question.

Formula used:
\[K.E = \dfrac{1}{2} \times m \times {v^2}\]
Where,
$K.E$is the kinetic energy,
$m$is the mass of the object and
$v$is the velocity of the object.

Complete step by step solution:
As we know that kinetic energy is energy which the body gets while the body is in motion. The energy of a body depends on mass and velocity. Kinetic energy is equal to half of the product mass and velocity. SI unit is joules.
1) If mass doubled then kinetic energy also gets doubled.
$\because K.E = \dfrac{1}{2} \times m \times {v^2}$
And according to the question the body is doubled at constant velocity.
$
  \therefore K.E = \dfrac{1}{2} \times (2m) \times {v^2} \\
   \Rightarrow K.E = 2\left( {\dfrac{1}{2} \times m \times {v^2}} \right) \\
 $
i.e.Twice the kinetic energy.
2) If velocity doubled with constant mass, then kinetic energy becomes four times the initial kinetic energy. Let us see how:
$\because K.E = \dfrac{1}{2} \times m \times {v^2}$
And according to the question the velocity is doubled at constant mass.
\[
  \therefore K.E = \dfrac{1}{2} \times m(2{v^2}) \\
   \Rightarrow K.E = \dfrac{1}{2} \times m \times 4{v^2} \\
  K.E = 4\left( {\dfrac{1}{2} \times m \times {v^2}} \right) \\
 \]
i.e.Four times the initial kinetic energy.
3) If the mass of the body is doubled but velocity is reduced to half, then kinetic energy becomes half the initial kinetic energy. Let us see how:
$\because K.E = \dfrac{1}{2} \times m \times {v^2}$
And according to the question the mass of the body is doubled but velocity is reduced to half.
\[
  \therefore K.E = \dfrac{1}{2} \times (2m) \times {\left( {\dfrac{v}{2}} \right)^2} \\
   \Rightarrow K.E = \dfrac{1}{2} \times (m)2 \times \dfrac{{{v^2}}}{4} \\
   \Rightarrow K.E = \dfrac{1}{2} \times \dfrac{1}{2} \times m \times {v^2} \\
   \Rightarrow K.E = \dfrac{1}{2}\left( {\dfrac{1}{2} \times m \times {v^2}} \right) \\
 \]
i.e.Half the initial kinetic energy.

Note:
According to the principle of Conservation of Energy, “energy can neither be created nor can be destroyed but can only be converted from one form to another”. So, kinetic energy is not created or destroyed. A body possesses kinetic energy when the body is in motion and a body possesses potential energy when it is at some height.