A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (where m₂m₁). If the acceleration of the system is g/(√2), then the ratio of the masses (m₁)/(m₂) is

Question

A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (where m₂>m₁). If the acceleration of the system is g/(√2), then the ratio of the masses (m₁)/(m₂) is

Accepted Answer

(1) (√2-1)/(√2+1). Idea: for an Atwood machine a=((m₂-m₁)g)/(m₁+m₂), so the acceleration fixes the ratio directly. (m₂-m₁)/(m₂+m₁)=1/(√2). Write r=(m₁)/(m₂) and divide through by m₂: (1-r)/(1+r)=1/(√2). √2(1-r)=1+r, so √2-1=r(√2+1). r=(√2-1)/(√2+1). Sanity check: this is about 0.17, comfortably less than 1, as it must be when m₂>m₁.

A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (where m₂>m₁). If the acceleration of the system is g/(√2), then the ratio of the masses (m₁)/(m₂) is

Step-by-step solution

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