Two solids A and B of mass 1, kg and 2, kg re | vedclass

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(B) The kinetic energy $K$ of a body with mass $m$ and linear momentum $P$ is given by the formula $K = \frac{P^2}{2m}$.Since both solids $A$ and $B$

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$2$

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(B) The kinetic energy $K$ of a body with mass $m$ and linear momentum $P$ is given by the formula $K = \frac{P^2}{2m}$.Since both solids $A$ and $B$ have equal linear momentum,we have $P_A = P_B = P$.Therefore,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.Taking the ratio of the kinetic energies of $A$ and $B$,we get $\frac{(K.E.)_A}{(K.E.)_B} = \frac{m_B}{m_A}$.Given $m_A = 1\, kg$ and $m_B = 2\, kg$,we substitute these values: $\frac{(K.E.)_A}{(K.E.)_B} = \frac{2}{1}$.Comparing this with the given ratio $\frac{A}{1}$,we find that $A = 2$.

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

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