balance moon stone A space traveler about to leave for the moon has a spring balance and a 1.0 kg mass A, which when hung on the balance on the Earth gives the reading of 9.8 newtons. Arriving at the moon at a place where the acceleration of gravity is not known exactly but has a value of about 1/6 the acceleration of gravity at the Earth's surface, he picks up a stone B which gives a reading of 9.8 newtons when weighed on the spring balance. He then hangs A and B over a pulley as shown in the figure and observes that B falls with an acceleration of 1.2 m s–2 . What is the mass of stone B?

Solution by Valentin Eberhardt

Let
gE    = acceleration of gravity on the earth,
gM   = acceleration of gravity on the moon ≈ gE / 6,
mA  = mass of stone A = 1.0 kg,
mB  = mass of stone B = 9.8N / gM,
a   = acceleration of stones hung in pulley on moon = 1.2 m s–2.

When both masses hang (on the Moon) over the pulley as in the drawing, the system of two masses mA and mB connected by the thread, which thread we just think of being not elastic as not mentioned other way, moves with an acceleration of a = 1.2 m

F = (mA +mB ) x a.

The resulting force F acting on the system is the gravitational force of the moon which in this case (pulley) is

F = (mB - mA ) x gM.

So we can write

(mB - mA ) x gM = (mA + mB ) x a.

(where we take mB > mA, as B falls and not A). We can rewrite this as

( mB - 1.0 ) x ( 9.8 / mB ) = ( 1.0 + mB ) x 1.2.

Multiplying by mB (mB ≠ 0), expanding and reordering of the terms gives:

1.2 mB2 - 8.6 mB + 9.8 = 0.

This quadratic equation has two solutions,

mB = 1.421kg, with  gM = gE / 1.421,

and

mB = 5.745kg with gM = gE / 5.745.

As it is given that gM ≈ gE / 6, the second solution must be the correct one. Therefore,

mB = 5.745 kg.