(1997)
Solution:
Initial kinetic energy $K_i = \frac{1}{2}m_1 u_1^2 = 100\text{ J}$ (with $u_1 = 10\text{ m/s}$). Final velocity $v = \frac{m_1 u_1}{m_1+m_2} = 4\text{ m/s}$, and final kinetic energy $K_f = \frac{1}{2}(m_1+m_2)v^2 = 40\text{ J}$. Loss in K.E. = $100 - 40 = 60\text{ J}$.
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