Answer:
The velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.
Explanation:
Given;
mass of first block, m₁ = 6.4 kg
initial speed of first block, u₁ = 5.4 m/s
mass of second block, m₂ = 12.8 kg
initial speed of second block, u₂ = 3.6 m/s
final speed of second block, v₂ = 4.5 m/s
To determine the final speed of 6.4 kg block immediately after the collision, we apply principle of conservation linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
v₁ is the final speed of 6.4 kg block
v₂ is the final speed of 12.8 kg block
Substitute the given values of m₁, u₁, m₂, u₂ , v₂ and calculate v₁
6.4 x 5.4 + 12.8 x 3.6 = 6.4v₁ + 12.8 x 4.5
34.56 + 46.08 = 6.4v₁ + 57.6
80.64 = 6.4v₁ + 57.6
80.64 - 57.6 = 6.4v₁
23.04 = 6.4v₁
v₁ = 23.04 / 6.4
v₁ = 3.6 m/s
Therefore, the velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.
