"If momentum of a body is increased by 20%, then its kinetic energy increases by"

Question

Accepted Answer

"

Let, initial momentum of body $({p_i}) = p$

$\therefore$ Final momentum $({p_f}) = {p_i} + 20\%$ of ${p_i}$

$= p + 0.2~p$

$= 1.2~p$

We know,

Kinetic energy $(E) = {{{p^2}} \over {2m}}$

$\therefore$ ${E_i} = {{{p^2}} \over {2m}}$

and ${E_f} = {{{{(1.2p)}^2}} \over {2m}} = {{1.44\,{p^2}} \over {2m}}$

$\therefore$ % Change in kinetic energy

$= {{{E_f} - {E_i}} \over {{E_i}}} \times 100$

$$ = {{{{1.44\,{p^2}} \over {2m}} - {{{p^2}} \over {2m}}} \over {{{{p^2}} \over {2m}}}} \times 100$$

$$ = {{{{{p^2}} \over {2m}}(1.44 - 1)} \over {{{{p^2}} \over {2m}}}} \times 100 = 0.44 \times 100 = 44$$

"

If momentum of a body is increased by 20%, then its kinetic energy increases by