(a) The velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
(b) The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
Velocity of the balls after collision
Apply the principle of conservation of linear momentum to determine the velocity of the balls;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.2(150) + (0.3)(-0.4) = 0.2v₁ + 0.3v₂
29.88 = 0.2v₁ + 0.3v₂
Apply one directional linear velocity
u₁ + v₁ = u₂ + v₂
v₁ = u₂ + v₂ - u₁
v₁ = -0.4 + v₂ - 150
v₁ = v₂ - 150.4
Substitute the value of v₁ into the first equation;
29.88 = 0.2(v₂ - 150.4) + 0.3v₂
29.88 = 0.2v₂ - 30.08 + 0.3v₂
59.96 = 0.5v₂
v₂ = 59.96/0.5
v₂ = 119.92 m/s
v₁ = 119.92 - 150.4
v₁ = -30.48 m/s
Velocity of their center mass before collision
V(cm) = (0.2 x 150 + 0.3 x 0.4) / (0.2 + 0.3)
V(cm) = 60.24 m/s
Velocity of their center mass after collision
V(cm) = (0.2 x 30.48 + 0.3 x 119.92) / (0.2 + 0.3)
V(cm) = 84.14 m/s
Thus, the velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
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