Two bodies A and B of masses 5 kg and 8 kg are moving such that the momentum of body B is twice that of the body A. The ratio of their kinetic energies will be :
Solution
<p>Given,</p>
<p>Mass of body A = 5 kg</p>
<p>Mass of body B = 8 kg</p>
<p>Momentum of body B is twice that of body A,</p>
<p>$\therefore$ ${P_B} = 2{P_A}$</p>
<p>We know,</p>
<p>Kinetic Energy $(K) = {{{P^2}} \over {2m}}$</p>
<p>$\therefore$ $${{{K_A}} \over {{K_B}}} = {\left( {{{{P_A}} \over {{P_B}}}} \right)^2} \times {{{m_B}} \over {{m_A}}}$$</p>
<p>$= {\left( {{1 \over 2}} \right)^2} \times {8 \over 5}$</p>
<p>$= {1 \over 4} \times {8 \over 5}$</p>
<p>$= {2 \over 5}$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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