A body of mass 0.50 kg is moving on a smooth | vedclass

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(C) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ m/s$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ m/s$.By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 =

Vedclass · Accepted Answer

$0.67$

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(C) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ m/s$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ m/s$.By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$.$(0.50)(2.00) + (1.00)(0) = (0.50 + 1.00)v$.$1.00 = 1.50v \implies v = \frac{1.00}{1.50} = \frac{2}{3} \ m/s$.Initial kinetic energy $K_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (0.50) (2.00)^2 = 1.00 \ J$.Final kinetic energy $K_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (1.50) (\frac{2}{3})^2 = 0.75 	imes \frac{4}{9} = 0.333... \ J$.Energy loss $\Delta K = K_i - K_f = 1.00 - 0.333 = 0.667 \ J \approx 0.67 \ J$.

$A$ body of mass $0.50 \ kg$ is moving on a smooth surface with a speed of $2.00 \ m/s$. It collides with another body of mass $1.00 \ kg$ at rest,and they move together as a single body. The energy loss during the collision is ....... $J$.

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