Question

The equation for kinetic energy is $K.E. = \dfrac{1}{2}m{v^2}$. What is the correct equation when this equation is solved for mass?
A. $m = \dfrac{1}{2}m{\left( {KE} \right)^2}$
B. $m = {\left( {\dfrac{{2\left( {KE} \right)}}{v}} \right)^2}$
C. $m = \sqrt {\dfrac{{2\left( {KE} \right)}}{v}}$
D. $m = \dfrac{{2\left( {KE} \right)}}{{{v^2}}}$
E. $m = \dfrac{{2{{\left( {KE} \right)}^2}}}{v}$

Answer 1

The mass m can be obtained by shifting all the values to the left hand side except m. It can be also obtained by dimensional analysis i.e., by putting the symbols on both the sides in the given options.

Complete step by step answer:
The kinetic energy is defined as the energy possessed by a body by virtue of its state of motion. It is denoted by K. For example: A fast moving stone has the capacity of breaking a window pane on striking it and thus it has kinetic energy. The equation of kinetic energy can be obtained by applying an opposing force,
Kinetic energy (K) = Work done by retarding force (F) in stopping a body moving with initial velocity (v).
$K = F \times S$ [S is displacement]
$F = ma$ where a is acceleration of the body and m is mass
From the kinematic equation,
${v^2} = {u^2} + 2aS$
$0 = {v^2} - 2aS$ [ v = 0 as body comes to rest]
${v^2} = 2aS$
$S = \dfrac{{{v^2}}}{{2a}}$
Now, we substitute the value in K,
$K = F \times \left( {\dfrac{{{v^2}}}{{2a}}} \right)$
$K = ma \times \left( {\dfrac{{{v^2}}}{{2a}}} \right)\left[ {F = ma} \right]$
$K = \dfrac{1}{2}m{v^2}$
Now, $\dfrac{K}{{{v^2}}} = \dfrac{1}{2}\left( m \right)$
$\dfrac{{2K}}{{{v^2}}} = m \Rightarrow \dfrac{{2\left( {KE} \right)}}{{{v^2}}} = m$ where KE is kinetic energy.
Therefore, option D is correct.

Note: Another method to find the value of m is from dimensional analysis, If we check by putting the symbols in both the sides, then we’ll get option D as a correct answer. In option D, on left hand side, it will be M and on right hand side, $\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{{{\left[ {L{T^{ - 1}}} \right]}^2}}}$ which is $\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^2}{T^{ - 2}}} \right]}} = M$.