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Two particles, each having a mass m are placed at a separation d in a uniform magnetic field B as shown in they have opposite charges of equal magnitude q. at time t =0, the particles are projectd towards each other,each with a speed v. suppose the coulomb force between the charges is switched off. (a)Find the maximum value v_m ofthe projection speed so that the two particles do not collide. (b)What would be the minimum and maximum separation between the particles if v =v_m /2? (c) At whatinstant will a collision occur betweenthe particles if v = 2v_m ? (d) Suppose v =2v_m and the collision between the particles is completely inelastic. Describe the motion after the collision. |
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Answer» Solution :The particles will not collide if `d = r_(1) + r_(2)` `rArr d = ((mV)/(qB) + (mV)/(qB) + (2mV)/(qB))` `v= (qBd)/(2m)` `d_(1) = r_(1)+r_(2) = 2r` `rArr = (d)/(2)`(minimum distance) Max distance , `d_(2) = + d + 2r = d + (d)/(2)= (3d)/(2)` `( c ) V= 2V _(2) m` `r_(1) = r_(2) = d `The are is `(1)/(6)` `(d )V = 2V_(m)` The particle will collide at point P . At point P , both the particle will have motion `m` in upward direction . Since the particle collide inelastically they stick together distance`i` between centers `= d`, `sinsintheta = (i)/(2r)`velocity upward `= v cos (90 - sin theta ) = (vi)/(2r)` `(mv^(2) )/(r)= QVB` ` rArr r = (mV)/(qB)` Using `(mu^(2) )/(r)= (qvB)/(2m)= V_(m)` HENCE the COMBINE mass will movewith the velocity `V_(m)` |
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