AP Physics B&C – Answer to Free Response Practice Question on Newton’s Laws

In the post dated 15^th June 2009, the following free-response question for practice was given to you:

In the adjoining figure P1 and P2 are light frictionless pulleys. P1 is fixed and P2 is movable. The strings used are inextensible and of negligible mass. The fixed pulley P₁ carries a mass M₁ and the movable pulley P₂ carries a mass M₂ which has one and a half times the mass of M₁. Now, answer the following questions:

(a) How is the tension T₁ in the string attached to the mass M₁ related to the tension T₂ in the string attached to the mass M₂?

Justify your answer.

(b) How is the magnitude of acceleration a₁ of the mass M₁ related to the magnitude of acceleration a₂ of the mass M₂?

Justify your answer

(c) From the following options regarding the motion of the mass M₁, select the correct one by putting a tick (√) mark:

Mass M₁ will be accelerated upwards ______

Mass M₁ will be accelerated downwards ______

Mass M₁ will move with uniform velocity ______

Justify your answer.

(d) Determine the accelerations a₁ and a₂ in of the masses M₁ and M₂ respectively.

(e) Determine the tensions T₁ and T₂ in the strings attached to the masses M₁ and M₂ respectively.

As promised, I give below the answer:

(a) The forces acting on the system of masses are shown in the figure. The string passing over the pulleys P₁ and P₂ has the same tension T₁ everywhere. Since the tension T₂ in the string connected to the mass M₂ is produced because of the tensions T₁ and T₁ it follows that

T₂ = T₁ + T₁ = 2 T₁

(b) When the mass M₁ moves through a distance x, the mass M₂ moves through a distance x/2 only. Therefore the magnitude of acceleration a₁ of the mass M₁ is twice the magnitude of acceleration a₂ of the mass M₂:

a₁ = 2a₂

(c) If the masses were such that the system were stationary we would have

M₁g = T₁ and M₂g = T₂ = 2 T₁

From the above equations M₂ = 2M₁.

Since it is given in the question that M₂ = 1.5 M₁, it follows that the system cannot be stationary. Evidently the mass M₁ will be accelerated downwards (and M₂ will be accelerated upwards).

(d) The net force acting on M₁ = M₁g – T₁

Therefore, M₁g – T₁ = M₁a₁ ………………(i)

Similarly, considering mass M₂ we obtain

2T₁ – M₂g = M₂ a₁/2 …………….(ii)

[We have used T₂ = 2T₁ and a₂ = a₁/2].

Multiplying Eq.(i) by 2 and adding to Eq.(ii) we obtain

2M₁g – M₂g = a₁[2M₁ + (M₂/2)]

This gives

a₁ = 2g(2M₁– M₂) /(4M₁ + M₂)

This is directed downwards.

Since a₂ = a₁/2 we have

a₂ = g(2M₁– M₂) /(4M₁ + M₂), which is directed upwards.

(e) Substituting for a₁ in Eq.(i),

T₁ = M₁(g – a₁) = M₁[g – {2g(2M₁– M₂) /(4M₁ + M₂)}]

Or, T₁ = M₁[(4M₁ + M₂)g– 2g(2M₁– M₂)] /(4M₁ + M₂)

This simplifies to T₁ = 3M₁M₂g /(4M₁ + M₂)

Since T₂ = 2T₁ we obtain

T₂ = 6M₁M₂g /(4M₁ + M₂)