- #1
Tanya Sharma
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Two small particles have electric charges of equal magnitude and opposite signs. The masses of the particles are m and 2m. Initially, the distance between the particles is d, and the velocities of the particles have equal magnitude v. However, the velocity of particle 2m is directed away from particle m, whereas the velocity of particle m is directed perpendicular to the line connecting the particles. In the subsequent motion of the particles, they are found to be at a distance 3d from each other—twice. Find the possible values of the charge of each particle.
My approach is to find the maximum distance ‘x ‘ between the charges and impose the condition that x>d ,as then the charges will be at a distance d- twice .
I will work in CM frame of reference.
## V_{cm} = \frac{1}{3m}(2mv\hat{i}+mv\hat{j}) ##
## V_{cm} = \frac{2v}{3}\hat{i}+\frac{v}{3}\hat{j} ##
Now , Let ##V_{2i}## be initial velocity of 2m and ##V_{1i}## be initial velocity of m in CM frame.
##V_{2i} = \frac{v}{3}\hat{i}-\frac{v}{3}\hat{j} ##
##V_{i} = \frac{-2v}{3}\hat{i}+\frac{2v}{3}\hat{j} ##
Let ##V_{2f}## be initial velocity of 2m and ##V_{1f}## be velocity of m in CM frame ,when the particles are at maximum distance.
Since no external forces are acting,linear momentum is conserved .
Initial momentum in CM frame = Final momentum in CM frame = 0
##2mV_{2f} = mV_{1f}##
Let ##V_{1f} = v'## ,so ##V_{1f} = \frac{v'}{2}##
Since no external torques are present ,angular momentum is conserved in CM frame .
## 2m\frac{v}{3}\frac{d}{3} + m\frac{2v}{3}\frac{2d}{3} = (2m)\frac{v'}{2}\frac{x}{3} + (m)(v')\frac{2x}{3} ##
This gives ##v' = \frac{2vd}{3x}##
Now applying conservation of energy ,
##\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x}##
##\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}m(\frac{2vd}{3x})^2+\frac{1}{2}2m(\frac{1}{2}\frac{2vd}{3x})^2 - \frac{kq^2}{x}##
##\frac{1}{3}mv^2- \frac{kq^2}{d} = \frac{1}{3}\frac{mv^2d^2}{x^2}- \frac{kq^2}{x}##
##\frac{mv^2d-3kq^2}{3d} = \frac{mv^2d^2-3kq^2x}{3x^2}##
I am not sure whether I am approaching the problem correctly .I would be grateful if somebody could help me with the problem.
Homework Equations
The Attempt at a Solution
My approach is to find the maximum distance ‘x ‘ between the charges and impose the condition that x>d ,as then the charges will be at a distance d- twice .
I will work in CM frame of reference.
## V_{cm} = \frac{1}{3m}(2mv\hat{i}+mv\hat{j}) ##
## V_{cm} = \frac{2v}{3}\hat{i}+\frac{v}{3}\hat{j} ##
Now , Let ##V_{2i}## be initial velocity of 2m and ##V_{1i}## be initial velocity of m in CM frame.
##V_{2i} = \frac{v}{3}\hat{i}-\frac{v}{3}\hat{j} ##
##V_{i} = \frac{-2v}{3}\hat{i}+\frac{2v}{3}\hat{j} ##
Let ##V_{2f}## be initial velocity of 2m and ##V_{1f}## be velocity of m in CM frame ,when the particles are at maximum distance.
Since no external forces are acting,linear momentum is conserved .
Initial momentum in CM frame = Final momentum in CM frame = 0
##2mV_{2f} = mV_{1f}##
Let ##V_{1f} = v'## ,so ##V_{1f} = \frac{v'}{2}##
Since no external torques are present ,angular momentum is conserved in CM frame .
## 2m\frac{v}{3}\frac{d}{3} + m\frac{2v}{3}\frac{2d}{3} = (2m)\frac{v'}{2}\frac{x}{3} + (m)(v')\frac{2x}{3} ##
This gives ##v' = \frac{2vd}{3x}##
Now applying conservation of energy ,
##\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x}##
##\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}m(\frac{2vd}{3x})^2+\frac{1}{2}2m(\frac{1}{2}\frac{2vd}{3x})^2 - \frac{kq^2}{x}##
##\frac{1}{3}mv^2- \frac{kq^2}{d} = \frac{1}{3}\frac{mv^2d^2}{x^2}- \frac{kq^2}{x}##
##\frac{mv^2d-3kq^2}{3d} = \frac{mv^2d^2-3kq^2x}{3x^2}##
I am not sure whether I am approaching the problem correctly .I would be grateful if somebody could help me with the problem.