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Hemant Gond Grade: 12
        quest---two particles of mass

1kg n 3kg move towards each
other under their mutual force
of attraction. no other force
acts on dem.tjheir rel...velocity
of approach is 2m/s. find vel.of
centr of mass/.
well mjhe quest galat lag raha
tha......wat u thnk.......since its
ans is nt given.. so m nt
confirm
6 years ago

Answers : (3)

Aakash Dutta
29 Points
										

Well I also think that it is incomplete


The initial conditions are not given 


Are they at rest?


If yes then 


The v of COM remains unchanged = 0

6 years ago
AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

Dear Hemant


Question is correct


suppose particles(m and 3m) are placed at some distance r. take origin at m


position of centre of mass= m*0+3m*r/4m = 3r/4 from m


now if they start moving then from action reaction force on both will be same but due to mass difference acceleration will not be same. if acc of m is a then acc of 3m will be a/3. so there will be relative acceleration they will move towards each other with relative acc of 4a/3 and the value of a will not be constant. as gravitational force is proportioal to 1/r2 hence a is also proportional to 1/r2. centre of mass will also change its position as both move with diff acc and diff velocity.


so please confirm the question that is the force constant or v=2m/s is at what instant.


All the best.                                                           


AKASH GOYAL


AskiitiansExpert-IIT Delhi


 


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6 years ago
SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


Here concept of reduced mass wll be used..


Mr=m1m1/m1+m2



















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Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi


















6 years ago
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