Question
If kinetic energy of a body is increased by $$300\% ,$$ then percentage change in momentum will be
A.
$$100\% $$
B.
$$150\% $$
C.
$$265\% $$
D.
$$73.2\% $$
Answer :
$$100\% $$
Solution :
Kinetic energy
$$\eqalign{
& K = \frac{1}{{2m}}\left( {{p^2}} \right) \cr
& {\text{or}}\,\,p = \sqrt {2mK} \cr} $$
If kinetic energy of a body is increased by $$300\% ,$$ let its momentum becomes $$p'.$$
New kinetic energy \[K' = K + \frac{{300}}{{100}}K = 4K\,\,\left( {\begin{array}{*{20}{l}}
{{\rm{initial }}\,KE = K}\\
{{\rm{final }}\,KE = K'}
\end{array}} \right)\]
Therefore, momentum is given by \[p' = \sqrt {2m \times 4K} \,\,\left( {\begin{array}{*{20}{l}}
{{\rm{initial }}\,{\rm{momentum}} = p}\\
{{\rm{Final }}\,{\rm{momentum}} = p'}
\end{array}} \right)\]
$$ = 2\sqrt {2mK} = 2p$$
Hence, percentage change (increase) in momentum
$$\eqalign{
& \frac{{\Delta p}}{p} \times 100 = \frac{{p' - p}}{p} \times 100 \cr
& = \left( {\frac{{p'}}{p} - 1} \right) \times 100 \cr
& = \left( {\frac{{2p}}{p} - 1} \right) \times 100 \cr
& = 100\% \cr} $$