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Question

(A) The kinetic energy $(K.E.)$ of a body is related to its momentum $(p)$ by the formula: $K.E. = \frac{p^2}{2m}$.When the momentum increases by $20

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) The kinetic energy $(K.E.)$ of a body is related to its momentum $(p)$ by the formula: $K.E. = \frac{p^2}{2m}$.When the momentum increases by $20 \%$,the new momentum $p'$ becomes:$p' = p + 0.20p = 1.2p$.The new kinetic energy $(K.E.')$ is:$K.E.' = \frac{(p')^2}{2m} = \frac{(1.2p)^2}{2m} = \frac{1.44p^2}{2m}$.Since $K.E. = \frac{p^2}{2m}$,we have $K.E.' = 1.44 	imes K.E$.The percentage increase in kinetic energy is given by:$\Delta K.E. \% = \frac{K.E.' - K.E.}{K.E.} 	imes 100 \%$$= \frac{1.44 K.E. - K.E.}{K.E.} 	imes 100 \%$$= 0.44 	imes 100 \% = 44 \%$.

If the momentum of a body of mass $m$ is increased by $20 \%$,then its kinetic energy increases by: (in $\%$)

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