Question

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

Answer 1

http://www.youtube.com/watch?v=5C5_dOEyAfk
you can skip the first 30[s] of the video

Answer 2

can u explain it theoritically @UnkleRhaukus

Answer 3

the objects accelerate at the same rate, there is little atmospheric resistance

Answer 4

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).

Answer 5

its different on earth because of the air resistance

Answer 6

oh yes, in sums we generally assume zero air resistance. so i sort of carried it here. thanks for the correction!

Answer 7

ok soo : the answer is a

Answer 8

yup.

Answer 9

But can u explain more.. by diagram or anything ?

Answer 10

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.

Answer 11

How can u say that t is same .?

Answer 12

v=s/t
t=s/v is that the reason ?

Answer 13

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.