IIT JEE , AIEEE Forum - Physics , Mechanics

Ank999

Two small bodies of masses 10 kg and 20 kg are kept at a distance 1 m apart and released, Assuming that only mutual gravitation force is acting , find the speedsd of the particles when the separation decreases to .5 m

rhd92781

increase in pot energy=Gm1m2(1/0.5-1/1)=1.334*10^-8
since two particles r concerned, we'll take out the reduced mass
u=m1m2/(m1+m2) =6.67 kg
so 1/2 uv^2=U
1/2 * 6.67 * v^2=1.334*10^-8
v=6.33*10^-5m/s

priyesh

mutual force on each is

G10 * 20

= 200G

acc of 10kg is 20G & 20 kg mass is 10G

acc of 10kg w.r.t to 20kg is 30G

so time taken for the seperation to reduce to .5m will be root(1/30G)

in this time 20kg block will attain speed of

10G * root(1/30G)

and 10kg block will attain speed of 20G * root(1/30G)

lokeshsardana

Consider both masses as single system,so, as no F_ext is acting.........therfore change in gravitational potential energy is equal to change in kinetic energy...........

now, change in gravitational potential energy is 200G. where G is universal gravitational constant

Let v₁ and v₂ be their velocity when their sepration is 0.5m

so, 10v₁²/2 + 20v₂²/2 = 200G---------------------------(1)

also, as f_ext =0, so, we can conserve momentum here..........

10v₁=20v₂ so, v₁=2v_{2------------------------------------------------------------}(2)

solving (1) and (2)

answer comes out to be v₁= 4.22 X 10^-5m/s

v₂= 2.11 X 10^-5m/s

note:- here we have considered both masses as as single system,so, we can conserve momentum

Rate!!!!!!!!!!!! if sastisfied..................if anybody finds any mistake in it......pls nudge me I will glad to know that...........

edison

I think your query is answered

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