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 P_{1} carries a mass M_{1} and the movable pulley P_{2} carries a mass M_{2} which has *one and a half times* the mass of M_{1}. Now, answer the following questions:

**(a)** How is the tension T_{1} in the string attached to the mass M_{1} related to the tension T_{2} in the string attached to the mass M_{2}?

Justify your answer.

**(b) **How is the magnitude of acceleration *a*_{1} of the mass M_{1} related to the magnitude of acceleration *a*_{2} of the mass M_{2}?

Justify your answer

**(c)** From the following options regarding the motion of the mass M_{1}, select the correct one by putting a tick (√) mark:

Mass M_{1} will be accelerated upwards ______

Mass M_{1} will be accelerated downwards ______

Mass M_{1} will move with uniform velocity ______

Justify your answer.

**(d)** Determine the accelerations *a*_{1} and *a*_{2} in of the masses M_{1} and M_{2} respectively.

**(e)** Determine the tensions* *T_{1} and T_{2 }in the strings attached to the masses M_{1} and M_{2} 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 P_{1} and P_{2 }has the same tension T_{1} everywhere. Since the tension T_{2} in the string connected to the mass M_{2} is produced because of the tensions T_{1} and T_{1} it follows that

T_{2} = T_{1 }+ T_{1} = 2 T_{1}

(b) When the mass M_{1} moves through a distance *x*, the mass M_{2} moves through a distance *x*/2 only. Therefore the magnitude of acceleration *a*_{1} of the mass M_{1} is *twice* the magnitude of acceleration *a*_{2} of the mass M_{2}:

*a*_{1} = 2*a*_{2}

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

M_{1}g = T_{1} and M_{2}g = T_{2} = 2 T_{1}

From the above equations M_{2} = 2M_{1}.

Since it is given in the question that M_{2} =_{ }1.5 M_{1}, it follows that the system cannot be stationary. Evidently the mass M_{1} will be accelerated *downwards *(and M_{2} will be accelerated upwards).

(d) The net force acting on M_{1} = M_{1}g – T_{1}

Therefore, M_{1}g – T_{1} = M_{1}*a*_{1} ………………(i)

Similarly, considering mass M_{2} we obtain

2T_{1} – M_{2}g = M_{2 }*a*_{1}/2 …………….(ii)

[We have used T_{2} = 2T_{1} and *a*_{2} = *a*_{1}/2].

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

2M_{1}g – M_{2}g = *a*_{1}[2M_{1} + (M_{2}/2)]

This gives

*a*_{1} = 2g(2M_{1}**– ****M _{2}) /(4M_{1} + M_{2}) **

This is directed *downwards*.

Since *a*_{2} = *a*_{1}/2 we have

*a*_{2} = g(2M_{1}**– ****M _{2}) /(4M_{1} + M_{2})**,

**which is directed**

*upwards*.

(e) Substituting for *a*_{1} in Eq.(i),

T_{1} = M_{1}(g – *a*_{1}) = M_{1}[g – {2g(2M_{1}– M_{2}) /(4M_{1} + M_{2})}]

Or, T_{1} = M_{1}[(4M_{1} + M_{2})g– 2g(2M_{1}– M_{2})] /(4M_{1} + M_{2})

This simplifies to **T _{1} = 3M_{1}M_{2}g /(4M_{1} + M_{2})**

Since T_{2} = 2T_{1} we obtain

**T _{2} = 6M_{1}M_{2}g /(4M_{1} + M_{2})**

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