Mass at Planet51
A space traveler about to leave for Planet 51 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 Planet51 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 and observes that B falls with an acceleration of 1.2 m s–2 .
What is the mass of stone B?
Arriving at the Planet51 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 and observes that B falls with an acceleration of 1.2 m s–2 .
What is the mass of stone B?
Labels: funphysics
posted by Rajesh Lal at
1:31 PM
9 Comments:
more then stoneA? just a guess... :D lol
Using successive approximations, I find that B has a mass of 5.745 kg and g on Planet51 is 1/5.745 of g on Earth. (Yeah, I've given too many decimal points.)
b has a mass of 14 kg
B := mass of our stone
A := our reference mass 1kg
g := is gravitational accel. on the planet
a := is the acceleration of our construction (1,2 m/s^2)
k := 9,8N (just for simplicity)
What we have here is an Atwood-machine. Because of the forces and the acceleration we get
g(B-A)=a(B+A) --> B=A((a+g)/(g-a))
We don't know B nor g, but we do know B*g=k (9,8N on our spring-balance). So we multiply our eq. by g an bring all that stuff to one side we get:
Ag^2+g(aA-k)+ak=0
Solving the quadratic eq. we get two answers: g1=6,9m/s^2, g2=1,7m/s^2
because the gravity is about 1/6 we pick g2 and k/g2=B=5,76 kg
so g is 1,7 m/s^2 and B has a mass of 5,76kg.
thy biatchh - nice one, but if you had worked to more significant figures, you would have got
g = 1.70578 and B = 5.745 kg.
well, that might be, but i see our both results as the same. If you look closely the inaccuracy of the force, the mass and the acceleration is 0.1
I think therefore the precision is completely irrelevant =)
Hi tha biatch. My comment was intended to point out an issue that I was aware of - dropping guard digits too early during calculations can lead to not quite trivial discrepancies.
I had implicitly acknowledged that our results were the same and that your method was far more informative than mine. I was waiting to let someone else have a go before giving a full answer. You had saved me the trouble.
I was actually expecting a whole lot of "it can't be done" type of responses.
I am sorry for not letting others try, I was just excited as I finally got it. I was thinking about it all day, especially to remember that damn atwood-machine ;)
But you're right I rounded up when I calculated the root in the quadratic eq. One should only round up at the end when you get the final results.
I was a lazy b****r. I didn't do the quadratic, but made a pair of equations that I iterated through a few times. It only took me 5 minutes though. I didn't know the pulley construction was called an Atwood machine.
Anyway, I think this problem has run its course. Back to Ragknot's "Algebra Trig" problem.
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