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susus · Dec 10, 2010, 11:35 AM

---- > Two isolated masses, M1 = 1.6 kg and M2 = 649.0 kg are initially rest, a distance d= 131.0 cm apart. Their gravitational attraction is the only force acting. Calculate the time it takes for M1 to move from that distance to 129.0 cm from M2. Assume that M2 does not move and that the force is constant over that small distance, and equal to that at 130.0 cm.

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---- > Find the speed of M1 when d = 129.0 cm.

ebaines · Dec 10, 2010, 12:30 PM

Good old F=ma:

$<br /> F = ma = \frac {Gm_1m_2} {d^2}<br />$

This lets you find the value for a. Then the time needed to travel distance x given acceleration a can be determined from:

$<br /> \Delta x = 0.02 m\ = \ \frac 1 2 a t^2<br />$

susus · Dec 12, 2010, 12:25 PM

well what I did is this :

a = 4.03*10^312
0.01 (u put 0.02 it's 0.01 , )..
0.01= 0.5at^2
t = 70403.3
and I got that it's wrong :(

susus · Dec 12, 2010, 12:56 PM

well I did it again
I got t = 995.5?
I'm I right ?

susus · Dec 12, 2010, 01:09 PM

No it's wrong,
Well , I do not know how to get the right answeer yet !
And sorry u were not wrong it's 0.02

Unknown008 · Dec 13, 2010, 02:43 AM

I don't know how you are getting these...

$F = \frac{Gm_1m_2}{r^2} = \frac{6.67\times10^{-11} \times 1.6 \times 649}{1.31^2} = 4.04 \times 10^{-8}\ N$

From this, F = ma

$a = \frac{F}{m} = \frac{4.04\times10^{-8}}{1.6} = 2.52 \times 10^{-8}\ m/s^2$

Can you complete it now?