help pls! spring balance(moon and earth) question.

As part of the Apollo space missions to the Moon, astronauts were required to measure the mass of collected lunar rock samples using a spring balance. The spring balance was calibrated to measure mass in the non-SI unit of the pound. The gravitational field strength on the Moon is 1/6th of that on Earth
. (a) A sample of lunar rock was observed to have a mass of 35 pounds. Calculate the weight of the sample on the Moon. 1 pound = 0.45kg(i already know the answer to part A)
(b) The sample measured in part (a) was brought to Earth. Suggest how the measurements could be scaled so that the same spring balance could be used to measure the mass of the sample on Earth.
PLEASE HELP ME WITH PART B. THE MARK SCHEME SAYS "divide by 6" but i have no idea why that is so,in fact i thought we should multiply by 6... i would really appreciate an explanation,thanks.

As part of the Apollo space missions to the Moon, astronauts were required to measure the mass of collected lunar rock samples using a spring balance. The spring balance was calibrated to measure mass in the non-SI unit of the pound. The gravitational field strength on the Moon is 1/6th of that on Earth
. (a) A sample of lunar rock was observed to have a mass of 35 pounds. Calculate the weight of the sample on the Moon. 1 pound = 0.45kg(i already know the answer to part A)
(b) The sample measured in part (a) was brought to Earth. Suggest how the measurements could be scaled so that the same spring balance could be used to measure the mass of the sample on Earth.
PLEASE HELP ME WITH PART B. THE MARK SCHEME SAYS "divide by 6" but i have no idea why that is so,in fact i thought we should multiply by 6... i would really appreciate an explanation,thanks.

Well if the spring balance was specially made and calibrated to read pounds in moon gravity, it'd read approximately 6 times higher in earth gravity for the same mass... so if you brought the same balance back to earth and weighed the same rock sample, the conversion would be divide by 6. though I'd say it's not 100% clear from the text what's going on.

afaik The Americans use the unit 'Pound' to measure both mass and force, which is confusing.

As part of the Apollo space missions to the Moon, astronauts were required to measure the mass of collected lunar rock samples using a spring balance. The spring balance was calibrated to measure mass in the non-SI unit of the pound. The gravitational field strength on the Moon is 1/6th of that on Earth
. (a) A sample of lunar rock was observed to have a mass of 35 pounds. Calculate the weight of the sample on the Moon. 1 pound = 0.45kg(i already know the answer to part A)
(b) The sample measured in part (a) was brought to Earth. Suggest how the measurements could be scaled so that the same spring balance could be used to measure the mass of the sample on Earth.
PLEASE HELP ME WITH PART B. THE MARK SCHEME SAYS "divide by 6" but i have no idea why that is so,in fact i thought we should multiply by 6... i would really appreciate an explanation,thanks.

Assume that the mass of the rock is $m$ and the gravitational field strength on Earth is $g$.

The weight of the rock registered by the spring balance on Moon would be
$W_{RM} = m\frac{g}{6}$

The weight of the rock registered by the same spring balance on Earth would be
$W_{RE} = mg$

You want to make the value of $W_{RE}$ to be "equal" to $W_{RM}$, so you have to divide $W_{RE}$ by 6 to ensure that the spring balance is measuring mass.

Hope it helps.

wow really good explanation,thank you for your help!
One more question please,how would i know,from the question,that we are supposed to change the value to be equal to the one on the moon? i thought weight was meant to be different

The question has this quote
...the same spring balance could be used to measure the mass of the sample on Earth.

Spring balance measure weight the of an object due to the extension of spring.
Since we are calibrating the spring balance to measure mass, then we need to ensure the weight is the same, is like saying
"weight is equal to mass"
in this situation.
But it is not true... Hope you understand what I am trying to say.