1please check again whether relative velocity or relative acceleration is asked for ???
At the 2 ends of a massless string of length 'l' 2 particles of masses 'm' each are attatched.The system is kept on a frictionless surface.
Now the sring is pulled by a horizontal force'f' at the middle
such that the particles start moving towards each other.Find the relative velocity of the particles when their separation is 'x'.
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8 Answers
11I am getting upto this stage :
vapp y. = Fm∫x√(l2-x2) dt
What is the answer ???
21i m not able to post the whole soln since i m not able to upload the image in which i have done......
in this question firstly find acceleration at any arbitary separation r.......
that acceleration of mass m comes out to be.....a(r)=(F/2m)r/2((l/2)^2 -(r/2)^2))^ .5
relative accel will be 2a(r)......
v(r) be rel velo at any general separation r
now since a=vdv/dx
so...2a(r)=v(r)dv(r)/dr
2a(r)dr=v(r)dv(r)
integrating lhs wrt r and rhs wrt v...with limits from 0 to x and 0 to v....wher v is rel velo when sep is x
we get the answer
11@aveek: In the problem, we have to find the relative velocity of approach of the particles.
Well after trying the solution I have got upto this point:
If we find the work done by the force on any of the particles for a small displacement dx and then integrate it from l to x,we would find out the total work done on any one of the particles.
Now,simply applying work-energy theorem:Change in k=work done by the apllied force(f here)
Thus we can find the individual velocy of each particle and hence find their relative velocity.
66Hint:
Take a reference frame which is fixed to the mid point of the thread. It is a non-inertial one (as the entire system is accelerating in the direction of F with an acceleration a0=F2m.)
Now, in this reference frame, we have a "effective gravity" of a0 in the opposite direction of a0.
1@Kaymant Sir and saggy....can't we proceed like this ....
vdvdx = F2mx√(l2-x2)
hence relative velocity = 2v