Question

Two solids A and B of mass $1\text{k}g$ and $2\text{k}g$ respectively are moving with equal linear momentum. The ratio of their kinetic energies $(K.E.{)}_{\text{A}}$ : $(K.E.{)}_{\text{B}}$ will be $\frac{A}{1}$. So the value of $A$ will be .

Answer 1

Given that, $\frac{M_1}{M_2}=\frac{1}{2}$

We know that $K=\frac{p^2}{2M}$
$\Rightarrow \frac{K_1}{K_2}=\frac{p^2}{2M_1}\times \frac{2M_2}{p^2}=\frac{2}{1}$

$\Rightarrow \frac{K_1}{K_2}=\frac{M_2}{M_1}=\frac{2}{1}$

$\Rightarrow \frac{A}{1}=\frac{2}{1}$

Therefore , $A$ =2.