33[1]
Two point charges P1and P2 are moving along positive x-axis. Charge and mass of P1 and P2 are (2uc,60ug)and (-5uc,30ug). At particular instant , if the distance of separation is 9mm and the respective speeds are 400ms-1and 1300ms-1 the distance of maximum separation, then should be???
(No other forces, except electrostatic are acting mutually)
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23 Answers
33whatever but any prob can be solved without it.. and effictively..
13hmmm... now d statement is correct pirate... [3]
reduced mass is not blind... u r blind at reduced mass... [4]
33reduced mass and COM they r diff naa
COM me ankh jada khula rahta hai... reduced mass is blind... B-(
1pirate thats what subhash's doing..."reduced mass" solving frm the frame of centre of mass
33(x+y) will be max separation..
May be above post is bakwaas because its all biomolecules in my mind for mock tomorrow..
If wrong simply ignore :P
33Let velocity be v1'and v2' wrt COM
and dist frm COM be l1 and l2..
now let at max sep x and y be dist of q1 and q2 frm COM
Now Kq1q2/(l1+l2)+.5m2(v2')2=kq1q1/(l1+x)
Kq1q2/(l1+l2)+.5m1(v1')2=kq1q1/(l1+y)
1if u could obtain the general expression for separation .....(using coulumnbs law)..then use maxima n minima
62subash i think what u are doing is correct..
but i dont like reduced mass concept personally...
If you can understnad it "completely" then it is fine.. other wise it is a recipie for suicide!
11bhaiyya is correct in the eqn relative velocity is zero
and vcm is constant for max separation
62akshay kinetic energy will not necessarily be zero...
It might so turn out but it is not a necessity!
Take an example in normal colission question
First body is moving with velocity of 5 m/s
2nd by 2 m/s
will the kinetic energy ever be zero???
11i had thought that one should use either use centre of mass
or reduced mass to reduce the system of masses
can both be used together
11in the previos post v is relative velocity
and take u to be reduced mass
11it is in the equation of energy conservation that i m getting a doubt
in the solution it is given like
total energy=1/2(m1+m2)vcm^2+1/2uv^2+q1q2/4piE0r
