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two identical balls of charges Q1 and Q2 initally have equal velocity of d same magnitude and direction.after a uniform electric field is applied for sometym ,the direction of d velocity of d first ball changes by 60. and d magnitude is reduced by half.the direction of d velocity of d second ball changes there by 90. .in what proportion will the velocity of second ball changes???? the curect ans is v/ ......where ...means under-root $...!! give me a curect solution wid a curect explaination!! THanKu... n plz do open my profile.....see the posted ques....n reply 'em.....i dint get reply f d ques i'hv sent

two identical balls of charges Q1 and Q2 initally have equal velocity of d same magnitude and direction.after a uniform electric field is applied for sometym ,the direction of d velocity of d first ball changes by 60. and d magnitude is reduced by half.the direction of d velocity of d second ball changes there by 90. .in what proportion will the velocity of second ball changes???? the curect ans is v/<3>......where<$>...means under-root $...!! give me a curect solution wid a curect explaination!! THanKu... n plz do open my profile.....see the posted ques....n reply 'em.....i dint get reply f d ques i'hv sent

Grade:12

2 Answers

Rahul askIITiansExpert.IITR
110 Points
13 years ago
the solution is fairly simple

the point to be noticed here is the fact that their accelerations will be in the ratio to their charges since F=ma

and F=QE

since the field if uniform

QE=ma

so a=QE/m

a=kQ

now,

write the separate equation of of bothe the charges in x direction

q1=>

Vxi=v, Vxf=V/2cos60=V/4;

so, Vxf=Vxi-a1xT

V/4=V-a1T.................................1

similarly for the charge q2

0=V-a2T.......................................2

using 1 and 2, we get

a1/a2=3/4.......................................3

now write the equation for y direction in the same way

equate a1/a2(in x)=a1/a2(in y)

and get the answer.

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Rahul- IIT Roorkee
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Let v1 and v2 be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the angle between the velocity v1 and the initial velocity v is 60°. Therefore, the change in the momentum of the first ball is
Δp1 = q1 E = Δt = m1v sin60
Here we use the condition that v1 = v/2, which implies that the change in the momentum Δp1 of the first ball occurs in a direction perpendicular to the direction of its velocity v1
Since E || Δp1 and the direction of variation of the second ball momentum is parallel to the direction of Δp1, we obtain for the velocity of the second ball (it can easily be seen that the charges on the balls have the same sign)
V2 = vt an30 = v/√3
The corresponding change in the momentum of the second ball is
Δp2 = q2 E Δt = m2V/cos 30
Hence we obtain
q1/q2 = m1 sin60/m2/cos 30
q2/m2 = 4/3 * q1/m1 = 4/4 k1

Thanks and Regards

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