Final answer:
The speed of the 2 kg cart is found using the conservation of momentum, which results in a speed of 1.5 m/s. This is due to the requirement that the total momentum of the system remains zero after the spring is released.
Step-by-step explanation:
The question is about a 2 kg cart and a 1 kg cart that are initially at rest with a compressed spring between them. When the spring is released, conservation of momentum must hold because no external forces are acting on the carts-system. The total momentum of the system before the spring is released is zero since both carts are at rest. After the spring is released, the total momentum of the system must still be zero because momentum is conserved in the absence of external forces.
Let's designate the velocities of the 1 kg and 2 kg carts after the spring is released as v1 and v2, respectively. We know that the 1 kg cart moves off with a velocity of 3 m/s, so v1 = 3 m/s. The 2 kg cart's velocity (v2) can be found using the conservation of momentum:
m1v1 + m2v2 = 0,
where m1 = 1 kg and m2 = 2 kg.
Plugging the known values into this equation, we get:
1 kg × 3 m/s + 2 kg × v2 = 0,
Solving for v2 we find:
v2 = -1.5 m/s.
The speed of the 2 kg cart is the absolute value of the velocity, which is 1.5 m/s. So, the correct answer to the student's question is (d) 1.5 m/s.