Answer:
the speed of the ball just after the collision is 1.5 m/s.
Explanation:
Given;
mass of the ball, m₁ = 1.6 kg
initial velocity of the ball, u₁ = 0
mass of the block, m₂ = 0.8 kg
initial velocity of the block, u₂ = 0
final velocity of the block, v₂ = 3 m/s
let the final velocity of the ball after collision = v₁
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
1.6 x 0 + 0.8 x 0 = 1.6 x v₁ + 0.8 x 3
0 = 1.6v₁ + 2.4
-1.6v₁ = 2.4
v₁ = -2.4 / 1.6
v₁ = - 1.5 m/s
v₁ = 1.5 m/s (in opposite direction of the block)
Therefore, the speed of the ball just after the collision is 1.5 m/s.
