Question 89: 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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