Two frictionless toy train carts, not of equal mass, are held together by a weightless compressed spring between them as shown below. At this point the spring expands and falls to the floor. After the spring release, the speed of cart B is measured to be 1.32 times of cart A. If cart A has a mass of 858.83 g, calculate the mass of cart B in grams.Give your answer rounded off to 2 decimal places without units.

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The problem involves two frictionless train carts connected by a spring that releases from rest. The internal forces during release do not change the total momentum of the system, so the total momentum before and after release is zero. After the spring releases, the carts move in opposite directions with speeds vA and vB and masses mA and mB, respectively. By conservation of momentum: mA * vA = mB * vB.

You are given that vB is 1.32 times vA, i.e., vB = 1.32 * vA. Substituting into the momentum relation gives:
mA * vA = mB * (1.32 * vA).
Assuming vA ≠ 0, you can divide both sides by vA to obtain:
mA = 1.32 * mB,
which rearranges to:
mB = mA / 1.32.

With mA = 858.83 g, the mass of cart B is:
mB = 858.83 / 1.32 ≈ 650.629..., which rounds to 650.63 g to two decimal places.

Explanation of potential missteps (why other incorrect lines of thinking would fail):
- If one tried to sum kinetic energies or use energy conservation, that would be inappropriate here because the spring’s potential energy converts to kinetic energy, but momentum conservation is the correct governing principle for a closed system with internal forces only. 
- Assuming vB/vA equals mA/mB in the wrong direction or mixing up which cart is heavier would lead to incorrect ratios. The correct relation from momentum conservation is mA * vA = mB * vB, which directly yields vB/vA = mA/mB and then mB = mA / (vB/vA) = mA / 1.32.
- Any rounding before final calculation could introduce small errors; it is best to perform the division with full precision and then round to two decimals at the end, as shown.

MUF0121 Physics Unit 1 - Semester 2, 2025

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