What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass?
(Assume the collision to be head-on elastic collision)
Solution
<p>For a head on elastic collision</p>
<p>${v_2} = {{m{u_1}} \over {m + 5m}} + {{m{u_1}} \over {m + 5m}}$</p>
<p>$= {{2{u_1}} \over 6}$ or ${{{u_1}} \over 3}$</p>
<p>Initial kinetic energy of first mass $= {1 \over 2}mu_1^2$</p>
<p>Final kinetic energy of second mass</p>
<p>$= {1 \over 2} \times 5m{\left( {{{{u_1}} \over 3}} \right)^2}$</p>
<p>$= {5 \over 9}\left( {{1 \over 2}mu_1^2} \right)$</p>
<p>$\Rightarrow$ kinetic energy transferred = 55% of initial kinetic energy of first colliding mass</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Conservation of Angular Momentum
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