An object A of mass 20 kg and travelling at | vedclass

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(A) According to the law of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collisio

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$13$

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(A) According to the law of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision, as no external force acts on the system.$m_A u_A + m_B u_B = m_A v_A + m_B v_B$Given: $m_A = 20 \ kg$, $u_A = 20 \ m \ s^{-1}$, $m_B = 200 \ kg$, $u_B = 10 \ m \ s^{-1}$.After collision, $v_A = -10 \ m \ s^{-1}$ (since it bounces back in the opposite direction).Substituting the values:$(20 	imes 20) + (200 	imes 10) = (20 	imes -10) + (200 	imes v_B)$$400 + 2000 = -200 + 200 v_B$$2400 = -200 + 200 v_B$$2600 = 200 v_B$$v_B = 13 \ m \ s^{-1}$

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