Question 92: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Let, initial momentum of body $({p_i}) = p$
$	herefore$ Final momentum $({p_f}) = {p_i} + 20\% $ of ${p_i}$
$ = p + 0.2~p$
$ = 1.2~p$
We know,
Kinetic energy $(E) = {{{p^2}} \over {2m}}$
$	herefore$ ${E_i} = {{{p^2}} \over {2m}}$
and ${E_f} = {{{{(1.2p)}^2}} \over {2m}} = {{1.44\,{p^2}} \over {2m}}$
$	herefore$ % Change in kinetic energy
$ = {{{E_f} - {E_i}} \over {{E_i}}} 	imes 100$
$ = {{{{1.44\,{p^2}} \over {2m}} - {{{p^2}} \over {2m}}} \over {{{{p^2}} \over {2m}}}} 	imes 100$
$ = {{{{{p^2}} \over {2m}}(1.44 - 1)} \over {{{{p^2}} \over {2m}}}} 	imes 100 = 0.44 	imes 100 = 44$

Answer

36%

Answer

40%

Answer

44%

Answer

48%

Centre of Mass and Rotational Motion

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

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