Question
Two masses $$1\,g$$ and $$9\,g$$ are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is
A.
$$1:9$$
B.
$$9:1$$
C.
$$1:3$$
D.
$$3:1$$
Answer :
$$1:3$$
Solution :
Given, $$K{E_1} = K{E_2}$$
$$\eqalign{
& \frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2 \cr
& {\text{or}}\,\,\frac{{v_2^2}}{{v_1^2}} = \frac{{{m_1}}}{{{m_2}}}\,\,{\text{or}}\,\,\frac{{{v_2}}}{{{v_1}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \cr
& {\text{As,}}\,\,{p_2} = {m_2}{v_2}{\text{ and }}{p_1} = {m_1}{v_1} \cr
& \therefore \frac{{{p_2}}}{{{p_1}}} = \frac{{{m_2}{v_2}}}{{{m_1}{v_1}}} = \frac{{{m_2}}}{{{m_1}}}\sqrt {\frac{{{m_1}}}{{{m_2}}}} = \sqrt {\frac{{m_2^2{m_1}}}{{m_1^2{m_2}}}} \cr
& \frac{{{p_2}}}{{{p_1}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} \cr
& {\text{or}}\,\,\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \cr
& {\text{Here,}}\,\,{m_1} = 1\;g,{m_2} = 9\;g \cr
& \therefore \frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{1}{9}} = \frac{1}{3} \cr} $$
Alternative
The relation between $$KE$$ and $$p$$ is given by
$$\eqalign{
& KE = \frac{{{p^2}}}{{2m}} \cr
& \Rightarrow {p^2} = 2mKE \cr
& \Rightarrow p = \sqrt {2mKE} \cr} $$
If $$KE$$ of two bodies are equal.
$$\eqalign{
& {\text{So,}}\,\,{p_1} \propto \sqrt {{m_1}} \cr
& {\text{and}}\,\,{p_2} \propto \sqrt {{m_2}} \cr
& \Rightarrow \frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \cr
& = \sqrt {\frac{1}{9}} = \frac{1}{3} \cr} $$