A block (mass=300g) resting on a frictionless table is connected by a string to another block (mass=500g). The second block is dropped off the edge of the table. The two-block system will have what initial acceleration?

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BlackTom AI · Accepted Answer

Let me restate the scenario and the options to set the stage for analysis:
- A block of mass 0.300 kg sits on a frictionless table and is connected by a string to another block of mass 0.500 kg that hangs off the table.
- The system will accelerate as the hanging mass pulls the on-table mass via the string.

Option A: 6 m/s^2
- This value aligns with the standard Atwood-like calculation for a block on a frictionless table connected to a hanging mass. The acceleration a is given by a = (m_hanging * g) / (m_on_table + m_hanging) = (0.500 kg * 9.8 m/s^2) / (0.300 kg + 0.500 kg) = 4.9 / 0.800 ≈ 6.125 m/s^2. Therefore, approximately 6 m/s^2 is what you'd expect physically when g ≈ 9.8 m/s^2. The reasoning uses the net external pull from the hanging mass and divides by the total mass being accelerated, which is exactly how a rope-and-pulley system on a frictionless surface behaves.

Option B: 1.5 m/s^2
- If someone proposed 1.5 m/s^2, they would likely be underestimating the driving force or overestimating the inertia. For this system, the driving force is roughly m_hanging * g, and with the given masses, the resulting acceleration is far larger than 1.5 m/s^2. Specifically, even with rough g ≈ 10 m/s^2, a ≈ (0.5 * 10) / (0.8) ≈ 6.25 m/s^2, which is nowhere near 1.5 m/s^2. This option ignores the correct force-to-mass ratio for the two-block system on a frictionless table.

Option C: 3 m/s^2
- A value of 3 m/s^2 would require a misapplication of the formula or an incorrect mass pairing. If one tried to compute a = (m_hanging * g) / (m_on_table + m_hanging) but used a significantly smaller g (or miscounted the total mass), you might land near 3 m/s^2 only by using, say, g ≈ 4.8 m/s^2 or by halving the effective total mass in error. However, with the given masses and standard gravity, the expected result is closer to 6 m/s^2, not 3 m/s^2. So this option reflects a common arithmetic mistake rather than the correct physics for the described setup.

Option D: 12 m/s^2
- A value this large would imply either using a much larger effective acceleration than the physics dictates or miscalculating the force-to-mass ratio. If you took the entire weight of the hanging mass as the net driving force and divided by the smaller total mass or overcounted the acceleration contribution, you could arrive at an overestimated number like 12 m/s^2. But for the system described, the acceleration is limited by the total mass being accelerated, and with m_hanging = 0.5 kg and m_on_table = 0.3 kg, the acceleration is well under 12 m/s^2. Thus 12 m/s^2 is not consistent with the standard derivation for this setup.

Overall, the physically correct approach is to treat the hanging mass as providing the driving force m_hanging * g and to divide by the total mass (m_on_table + m_hanging) to obtain the acceleration. This yields approximately 6 m/s^2 for g ≈ 9.8 m/s^2, and rules out the much smaller value of 1.5 m/s^2, as well as the 3 m/s^2 and 12 m/s^2 options that arise from incorrect applications or arithmetic.

PHYS 2010-001 Fall 2025 Block 1 Practice Exam Physics 2010 Fall 2025

A block (mass=300g) resting on a frictionless table is connected by a string to another block (mass=500g). The second block is dropped off the edge of the table. The two-block system will have what initial acceleration?

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