Question:
Two solids $\mathrm{A}$ and $\mathrm{B}$ of mass $1 \mathrm{~kg}$ and $2 \mathrm{~kg}$ respectively are moving with equal linear momentum. The ratio of their kinetic energies
$(\text { K.E. })_{\mathrm{A}}:(\text { K.E. })_{\mathrm{B}}$ will be $\frac{\mathrm{A}}{1}$, so the value of $\mathrm{A}$ will be
Solution:
Kinetic energy $\mathrm{K}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}},\left(\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{B}}\right)$
$K \propto \frac{1}{m}$
$\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}$
$=\frac{2}{1}$
Ans. (2)
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