Answer:
the final speed of the 0.75 kg cart is 2.86 m/s
Explanation:
Given;
mass of the air glider, mâ = 1.0 kg
initial speed of the air glider, uâ = 2.5 m/s
mass of the stationary cart, mâ = 0.75 kg
initial speed of the stationary cart, uâ = 0
Let the final speed of the stationary cart = vâ
Also, let the final speed of the air glider = vâ
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.
mâuâ Â + Â mâuâ Â = mâvâ Â + Â mâvâ
(1 x 2.5) Â + 0.75(0) Â = Â vâ Â + Â 0.75vâ
2.5 = vâ Â + Â 0.75vâ
vâ = 2.5 - 0.75vâ ----------- (1)
Apply one-dimensional velocity concept since the collision occured in one direction;
uâ + vâ Â = uâ Â + Â vâ
2.5 Â + Â vâ Â = 0 Â + Â vâ
vâ = vâ Â - 2.5 ----- (2)
solve (1) and (2) together;
vâ Â - 2.5 = 2.5 - 0.75vâ
vâ + 0.75vâ = 2.5 Â + Â 2.5
1.75vâ = 5
vâ = 5 / 1.75
vâ = 2.86 m/s
Therefore, the final speed of the 0.75 kg cart is 2.86 m/s