**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 Michael A. Gottlieb__

Let g_{E }= acceleration of gravity on the earth = 9.81 m s^{–2}g_{M}= acceleration of gravity on the moon ≈ g_{E }/ 6, m_{A}= mass of stone A = 1.0 kg, m_{B}= mass of stone B, a = acceleration of stones hung in pulley on moon = 1.2 m s^{–2}.

Stones A and B and the pulley form an
Atwood's Machine for which

a = [ (m_{B} - m_{A}) / (m_{B}
+ m_{A}) ] g_{M} . (1)

Stone B on the moon gives the same scale reading as a 1 Kg stone on Earth, so

g_{M}*m_{B} = 9.81.
(2)

Substituting m_{B} =
9.81/g_M into (1) yields a quadratic equation in g_{M},

g_{M}^{2} - 8.61 g_{M} +
11.772 = 0,

which has one solution close to g_{E }/ 6, namely g_{M}
= 1.7048. Substituting this value into (2) gives m_{B} = 5.75434
and rounding to two significant digits (as per the spring balance readings and
the acceleration) gives

m_{B} ≈ 5.8.