Two 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 0.78 times of cart A. 100 grams of sand was dumped onto cart B before release. If cart A has a mass of 736.04 g, calculate the mass of cart B in grams. Give your answer rounded off to 2 decimal places without units.

Question

BlackTom AI · Accepted Answer

Momentum is conserved because the spring is internal and exerts equal and opposite impulses on the two carts; with no external horizontal impulse the total horizontal momentum remains zero. Let mA be the mass of cart A, mB the original mass of cart B, and mB' the mass of B after 100 g of sand is added: mB' = mB + 100 g. If vA and vB are the speeds after release, conservation of momentum gives:

1) mA vA = mB' vB.

The measured relation vB = 0.78 vA can be substituted into (1):

2) mA vA = (mB + 100) (0.78 vA).

Cancel vA (nonzero) to get an algebraic relation between masses:

3) mA = 0.78 (mB + 100).

Solve for mB:

4) mB + 100 = mA / 0.78
   mB = (mA / 0.78) - 100.

Insert the given mA = 736.04 g and carry out the arithmetic with intermediate steps:

5) mA / 0.78 = 736.04 / 0.78
   = 73604 / 78 (moving decimal)
   = 943.6410256... g.

6) Subtract 100 g:
   mB = 943.6410256... - 100 = 843.6410256... g.

Round to two decimal places as requested:

7) mB ≈ 843.64 (to two decimal places).

Why this is correct: the spring’s internal force cannot change the system’s total momentum, so the impulse balance directly links masses and velocities; using the measured velocity ratio yields the equation above and the numeric solution. Common mistakes include forgetting to include the added 100 g on cart B (which would give the larger value 943.64 g) or arithmetic errors in the division. The final result 843.64 g matches the provided answer.

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