Two masses M₁ and M₂ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M₂ is twice that of M₁, the acceleration of the system is a₁. When the mass M₂ is thrice that of M₁, the acceleration of the system is a₂. The ratio a₁/a₂ will be :

Solution

Physics sir · 4 steps

In both cases the net pull and the total mass change, so find each acceleration fully before dividing.

1. Atwood machine. The heavier mass M₂ moves down and M₁ moves up, both with the same acceleration. The net driving force is the difference of the weights, (M₂ - M₁)g. It moves the total mass M₁ + M₂.

   a = ((M₂ - M₁) g)/(M₁ + M₂)

2. First case. Put M₁ = m and M₂ = 2m. Then a₁ = (2m - m)g/3m = g/3.
3. Second case. Put M₁ = m and M₂ = 3m. Then a₂ = (3m - m)g/4m = g/2.
4. So. a₁/a₂ = (g/3)/(g/2) = 2/3, which is option (b).

The idea to keep. For two masses over a light, frictionless pulley, a = (M₂ - M₁)g/(M₁ + M₂). When you change a mass, it changes both the net pull on top and the total mass below.

Answer from the NTA final answer key. Solution by PrepWiser.