Two solids $\mathrm{A}$ and $\mathrm{B}$ of mass $1 \mathrm{~kg}$ and $2 \mathrm{~kg}$ respectively are moving withequal linear
momentum. The ratio of their kinetic energies $(\mathrm{K} . \mathrm{E} .)_{\mathrm{A}}:(\mathrm{K} . \mathrm{E} .)_{\mathrm{B}}$ will be $\frac{\mathrm{A}}{1} .$ So the
value of $\mathrm{A}$ will be
(2)
Given that, $\frac{M_{1}}{M_{2}}=\frac{1}{2}$
Also, $\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{p}$
$\Rightarrow \mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2}=\mathrm{p}$
Also, we know that
$\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}} \Rightarrow \mathrm{K}_{1}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}_{1}} \& \Rightarrow \mathrm{K}_{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}_{2}}$
$\Rightarrow \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}_{1}} \times \frac{2 \mathrm{M}_{2}}{\mathrm{p}^{2}} \Rightarrow \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}=\frac{2}{1}$
$\Rightarrow \frac{\mathrm{A}}{1}=\frac{2}{1} \Rightarrow \therefore \mathrm{A}=2$
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