Answer:
The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
Explanation:
Given;
mass of the first object, m₁ = 0.3 kg
initial velocity of the first ball, u₁ = 2.8 m/s
mass of the second ball, m₂ = 0.4 kg
initial velocity of the second ball, u₂ = 0
let the final velocity of the first ball, = v₁
let the final velocity of the second ball, = v₂
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂
0.84 = 0.3v₁ + 0.4v₂
2.8 = v₁ + 1.333v₂ -------equation (1)
Apply one-direction velocity;
u₁ + v₁ = u₂ + v₂
2.8 + v₁ = 0 + v₂
v₂ = 2.8 + v₁
substitute the value of v₂ into equation (1)
2.8 = v₁ + 1.333v₂
2.8 = v₁ + 1.333(2.8 + v₁)
2.8 = v₁ + 3.732 + 1.333v₁
2.8 - 3.732 = v₁ + 1.333v₁
-0.932 = 2.333v₁
v₁ = -0.932 / 2.333
v₁ = -0.4 m/s
Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
