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two balls of charges q1 and q2 initially have exactly same velocity, both are subjected to equal electric fields at the same time. as a result the velocity of the first ball is reduced to half and the direction of velocity id deflected by an angle of 60 degrees. whereas the second ball gets deflected by an angle of 90.the electric fields and the initial velocity are inclined at.? |
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Answer» Answer: Explanation: 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 |
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