Question

If the linear momentum of a body is increased by $ 50\% $ , its kinetic energy will increase by:
(A) $ 50\% $
(B) $ 100\% $
(C) $ 125\% $
(D) $ 150\% $

Answer 1

Hint
The linear momentum of a body is the product of mass of the body and its velocity. The kinetic energy of the body is the product of mass of the body with its velocity squared divided by 2. We notice here that the term mv is common in both. Try to replace this term in the expression for KE of the body.

Complete step by step answer
The linear momentum of a body is given by:
  $ p\, = \,mv $
And its kinetic energy is given by:
  $ E\, = \,\dfrac{1}{2}m{v^2} $
Dividing/multiplying the kinetic energy by mass of the body we get:
  $ E\, = \,\dfrac{1}{{2m}}{m^2}{v^2} $
As we know that mv is called the linear momentum of the body, so we get:
  $ E\, = \,\dfrac{1}{{2m}}{p^2} $
When the momentum is increased by $ 50\% $ , its new value becomes : $ p + 0.5p = \dfrac{3}{2}p $
Therefore, the new kinetic energy becomes,
  $
  {E_N}\, = \,\dfrac{1}{{2m}}{(\dfrac{3}{2}p)^2} \\
  {E_N}\, = \,\dfrac{{9{p^2}}}{{2m \times 4}} \\
  {E_N}\, = \,\dfrac{{9{p^2}}}{{8m}} \\
  $
Therefore the change in kinetic energy becomes:
  $
  KE\, = \,\dfrac{{9{p^2}}}{{8m}} - \dfrac{{{p^2}}}{{2m}} \\
  KE\, = \,\dfrac{9}{4}E - E \\
  KE\, = \,\dfrac{5}{4}E \\
  $
Therefore this means that there will be $ 125\% $ increase in the KE of the body. So the correct answer is option (C).

Note
This formula relating the KE of the body with its momentum is very useful and widely used. So it is recommended that the student learns this formula. This formula in rotational mechanics is written as: $ KE\, = \,\dfrac{1}{{2I}}{L^2} $
Where $KE$ is the rotational kinetic energy,
$I$ in the moment of inertia of the body and,
$L$ is the angular momentum of the body.