Two bodies A and B of masses 20 kg and 5 kg | vedclass

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(B) Given: Mass of body $A$ $(m_A = 20 \ kg)$,mass of body $B$ $(m_B = 5 \ kg)$,force $(F = 40 \ N)$.Since both bodies start from rest,their kinetic e

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$2: 1$

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(B) Given: Mass of body $A$ $(m_A = 20 \ kg)$,mass of body $B$ $(m_B = 5 \ kg)$,force $(F = 40 \ N)$.Since both bodies start from rest,their kinetic energies are $K_A = K_B = K$.Kinetic energy is given by $K = \frac{p^2}{2m}$,where $p$ is momentum.Since $K_A = K_B$,we have $\frac{p_A^2}{2m_A} = \frac{p_B^2}{2m_B} \implies \frac{p_A}{p_B} = \sqrt{\frac{m_A}{m_B}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$.From the impulse-momentum theorem,$F \cdot t = \Delta p$. Since initial momentum is zero,$p = F \cdot t$.Thus,$\frac{p_A}{p_B} = \frac{F \cdot t_A}{F \cdot t_B} = \frac{t_A}{t_B}$.Therefore,$\frac{t_A}{t_B} = 2$,which means $t_A: t_B = 2: 1$.

Two bodies $A$ and $B$ of masses $20 \ kg$ and $5 \ kg$ respectively are at rest. Due to the action of a force of $40 \ N$ separately,if the two bodies acquire equal kinetic energies in times $t_A$ and $t_B$ respectively,then $t_A: t_B=$

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