<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J
max spring compression where both velocities are the same: conserve momentum:
1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s
which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J
The remaining energy went into the spring:
U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x²
x = 0.0076 m ↠(a)</span>
The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.
The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.
The change in momentum of the system of the carts is 0.
<h3>
Initial momentum of the carts before collision</h3>
The momentum of the carts before the collision is calculated as follows;
P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s
P(blue) = 1.5 x 0 = 0
<h3>Momentum of the carts after collision</h3>
The momentum of the carts after the collision is calculated as follows;
P(red) = 0.5 x 0.1 = 0.05 kgm/s
P(blue) = 1.5 0.1 = 0.15 kgm/s
<h3>Change in momentum of the carts</h3>

ΔP = (0.05 + 0.15) - (0.2)
ΔP = 0
Learn more about momentum here: brainly.com/question/7538238