A metal ball of mass 2 kg moves with a speed | vedclass

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(B) Given: $m_1 = 2 \ kg$,$u_1 = 36 \ km/hr = 36 	imes \frac{5}{18} = 10 \ m/s$,$m_2 = 3 \ kg$,$u_2 = 0 \ m/s$. According to the law of conservation

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$60$

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(B) Given: $m_1 = 2 \ kg$,$u_1 = 36 \ km/hr = 36 	imes \frac{5}{18} = 10 \ m/s$,$m_2 = 3 \ kg$,$u_2 = 0 \ m/s$. According to the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$. $(2)(10) + (3)(0) = (2 + 3)v \implies 20 = 5v \implies v = 4 \ m/s$. Initial kinetic energy $K_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} 	imes 2 	imes (10)^2 = 100 \ J$. Final kinetic energy $K_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} 	imes (2 + 3) 	imes (4)^2 = \frac{1}{2} 	imes 5 	imes 16 = 40 \ J$. Loss in kinetic energy $\Delta K = K_i - K_f = 100 \ J - 40 \ J = 60 \ J$.

$A$ metal ball of mass $2 \ kg$ moves with a speed of $36 \ km/hr$. It collides with a stationary ball of mass $3 \ kg$. If both balls move together after the collision,the loss in kinetic energy due to the collision is ........... $J$.

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