A body of mass m moving with velocity v makes | vedclass

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(C) In an elastic collision,the total kinetic energy is conserved. However,the question asks for the loss in kinetic energy of the colliding body. Let

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$4/9 \ mv^2$

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(C) In an elastic collision,the total kinetic energy is conserved. However,the question asks for the loss in kinetic energy of the colliding body. Let the mass of the first body be $m_1 = m$ and its initial velocity be $u_1 = v$. Let the mass of the second body be $m_2 = 2m$ and its initial velocity be $u_2 = 0$.After the collision,the final velocity $v_1$ of the first body is given by the formula: $v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2$.Substituting the values: $v_1 = \frac{m - 2m}{m + 2m} v + 0 = \frac{-m}{3m} v = -v/3$.The initial kinetic energy of the first body is $K_i = 1/2 \ mv^2$.The final kinetic energy of the first body is $K_f = 1/2 \ m(v_1)^2 = 1/2 \ m(-v/3)^2 = 1/2 \ m(v^2/9) = 1/18 \ mv^2$.The loss in kinetic energy of the colliding body is $\Delta K = K_i - K_f = 1/2 \ mv^2 - 1/18 \ mv^2$.$\Delta K = (9/18 - 1/18) \ mv^2 = 8/18 \ mv^2 = 4/9 \ mv^2$.

$A$ body of mass $m$ moving with velocity $v$ makes a head-on elastic collision with another body of mass $2m$ which is initially at rest. What is the loss in kinetic energy of the colliding body?

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