Answer:
The final speed of 0.150-kg glider after collision is 3.2 m/s to the left
The final speed of 0.300-kg glider after collision is 0.20 m/s to the left
Explanation:
Given;
mass of glider moving to the right, m₁ = 0.150-kg
mass of glider moving to the left, m₂ = 0.300-kg
initial speed of glider moving to the right, u₁ = 0.80 m/s
initial speed of glider moving to the left, u₂ = 2.20 m/s
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.15 x 0.8) + (0.3 x - 2.2) = 0.15v₁ + 0.3v₂
-0.54 = 0.15v₁ + 0.3v₂
0.15v₁ + 0.3v₂ = - 0.54 -----------equation (i)
Again, their relative velocity after collision is given as;
u₁ - u₂ = v₂ - v₁
0.8 - (-2.2) = v₂ - v₁
3 = v₂ - v₁
v₂ = v₁ + 3 ------------equation (ii)
Substitute v₂ in equation (i)
0.15v₁ + 0.3v₂ = - 0.54
0.15v₁ + 0.3(v₁ + 3 ) = - 0.54
0.15v₁ + 0.3v₁ + 0.9 = - 0.54
0.45v₁ = - 0.54 - 0.9
0.45v₁ = -1.44
v₁ = -1.44 /0.45
v₁ = - 3.2 m/s
Thus, the final speed of 0.150-kg glider after collision is 3.2 m/s to the left
From equation (ii), v₂ = v₁ + 3
v₂ = -3.2 + 3
v₂ = - 0.20 m/s
Therefore, the final speed of 0.300-kg glider after collision is 0.20 m/s to the left
