Dynamics – Effect of Scaling Both Masses in an Atwood-type System In the two-mass pulley system (both masses connected by a light inextensible string over a frictionless pulley), if both masses are doubled, how does the string acceleration change?

Introduction / Context:
The classic Atwood machine illustrates how acceleration depends on the difference and the sum of the two masses. This question checks understanding of proportionality when both masses are scaled by the same factor.

Given Data / Assumptions:

• Two masses m1 and m2 on a light inextensible string over a frictionless pulley.
• Acceleration formula in magnitude: a = (m2 − m1) * g / (m1 + m2).
• Both masses are scaled: m1 → k m1 and m2 → k m2 with k = 2.

Concept / Approach:
Substitute the scaled masses into the expression for acceleration and simplify. Because both numerator and denominator scale by the same factor, the ratio remains unchanged.

Step-by-Step Solution:
Original acceleration: a = (m2 − m1) * g / (m1 + m2). After scaling by k: a′ = (k m2 − k m1) * g / (k m1 + k m2). Factor k: a′ = k (m2 − m1) * g / [k (m1 + m2)] = (m2 − m1) * g / (m1 + m2) = a.

Verification / Alternative check:
Choose numbers, e.g., m1 = 1, m2 = 3 ⇒ a = (2/4) g = 0.5 g. Doubling masses (2 and 6) ⇒ a = (4/8) g = 0.5 g (unchanged).

Why Other Options Are Wrong:
Half, double, or one-fourth would require changing the mass ratio, not uniform scaling. Adding g/2 is physically meaningless in this context.

Common Pitfalls:
Forgetting that both numerator and denominator scale by the same factor.

Final Answer:
same