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(B) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ ms^{-1}$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ ms^{-1}$.Since the collision is perfectly inelastic,the bodies move t

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$0.67$

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(B) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ ms^{-1}$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ ms^{-1}$.Since the collision is perfectly inelastic,the bodies move together after the collision.The loss in kinetic energy during a perfectly inelastic collision is given by the formula:$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$Substituting the values:$\Delta K = \frac{0.50 	imes 1.00}{2(0.50 + 1.00)} (2.00 - 0)^2$$\Delta K = \frac{0.50}{2(1.50)} (4)$$\Delta K = \frac{0.50}{3.00} 	imes 4 = \frac{2}{3} \approx 0.67 \ J$.

$A$ block of mass $0.50 \ kg$ is moving with a speed of $2.00 \ ms^{-1}$ on a smooth surface. It strikes another mass of $1.00 \ kg$ (initially at rest) and then they move together as a single body. The energy loss during the collision is .............. $J$

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