Answer:
the final speed of the 0.75 kg cart is 2.86 m/s
Explanation:
Given;
mass of the air glider, m₁ = 1.0 kg
initial speed of the air glider, u₁ = 2.5 m/s
mass of the stationary cart, m₂ = 0.75 kg
initial speed of the stationary cart, u₂ = 0
Let the final speed of the stationary cart = v₂
Also, let the final speed of the air glider = v₁
Since the air glider has a spring bumper, the collision will be elastic.
Apply the following principle of conservation of linear momentum for elastic collision.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(1 x 2.5) + 0.75(0) = v₁ + 0.75v₂
2.5 = v₁ + 0.75v₂
v₁ = 2.5 - 0.75v₂ ----------- (1)
Apply one-dimensional velocity concept since the collision occured in one direction;
u₁ + v₁ = u₂ + v₂
2.5 + v₁ = 0 + v₂
v₁ = v₂ - 2.5 ----- (2)
solve (1) and (2) together;
v₂ - 2.5 = 2.5 - 0.75v₂
v₂ + 0.75v₂ = 2.5 + 2.5
1.75v₂ = 5
v₂ = 5 / 1.75
v₂ = 2.86 m/s
Therefore, the final speed of the 0.75 kg cart is 2.86 m/s
