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(D) Given:Mass of ball $A$,$m_A = 0.25\, kg$Mass of ball $B$,$m_B = 0.5\, kg$Height $h = 0.45\, m$$1$. Velocity of ball $A$ just before collision $(v_

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$1\, m/s$ to the left

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(D) Given:Mass of ball $A$,$m_A = 0.25\, kg$Mass of ball $B$,$m_B = 0.5\, kg$Height $h = 0.45\, m$$1$. Velocity of ball $A$ just before collision $(v_{Ai})$:Using conservation of mechanical energy for ball $A$:$m_A g h = \frac{1}{2} m_A v_{Ai}^2$$v_{Ai} = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 0.45} = \sqrt{9} = 3\, m/s$$2$. Velocity of ball $B$ just after collision $(v_B)$:Given kinetic energy $E_B = 1\, J$$E_B = \frac{1}{2} m_B v_B^2$$1 = \frac{1}{2} 	imes 0.5 	imes v_B^2$$v_B^2 = 4 \implies v_B = 2\, m/s$$3$. Velocity of ball $A$ just after collision $(v_A)$:Using conservation of linear momentum:$m_A v_{Ai} = m_A v_A + m_B v_B$$0.25 	imes 3 = 0.25 	imes v_A + 0.5 	imes 2$$0.75 = 0.25 v_A + 1$$0.25 v_A = 0.75 - 1 = -0.25$$v_A = -1\, m/s$The negative sign indicates that the velocity of ball $A$ is $1\, m/s$ to the left.

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