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The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
The given parameters;
Let the final velocity of the 0.8 kg ball immediately after collision = v₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)
2 = 2.5 + (0.8)v₁
-0.5 = (0.8)v₁
Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.
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