help in gravitation : Two objects of different masses falling freely near the surface of moon would (a) have same velocities at any instant (b) have different accelerations (c) experience forces of same magnitude (d) undergo a change in their inertia
http://www.youtube.com/watch?v=5C5_dOEyAfk you can skip the first 30[s] of the video
can u explain it theoritically @UnkleRhaukus
the objects accelerate at the same rate, there is little atmospheric resistance
its the same case as that on earth. irrespective of their masses, the acceleration due to gravity (in this case, that of moon) acting on both the bodies will be the same. so the answer will be option (a).
its different on earth because of the air resistance
oh yes, in sums we generally assume zero air resistance. so i sort of carried it here. thanks for the correction!
ok soo : the answer is a
yup.
But can u explain more.. by diagram or anything ?
i don't know abt a diag...but see if this helps: lets use the eq, v = u + at. here, both bodies are falling freely, so assuming they start from rest, u = 0 also a = g1.........(accel due to gravity of moon) now, since they both are experiencing the same acceleration, and assuming they cover the same distance, the time taken to reach the moon's surface will be the same for both. so t1 = t2 = t for body1, v1 = 0 + g1 t1 = g1 t........(i) for body2, v2 = 0 + g1 t2 = g1 t.........(ii) so from (i) and (ii), v1 = v2.
How can u say that t is same .?
v=s/t t=s/v is that the reason ?
well yes. but u can't go that way now, coz we used t1 = t2 = t to derive v1 = v2. go this way: s = ut + 1/2 at^2 s = 0 + 1/2 at^2 s = 1/2 at^2 now s is the same for both. a = g1 is the same for both. so t has to be the same for both.
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