Derivation of relativistic mass Consider the two frame

Derivation of relativistic mass

Consider the two frame of reference S and S’. As S’ is moves with constant Speed v along x direction. Consider the hard ball A and hard ball B in frame of reference S and S’. Observer O in frame of reference S throw the ball A in upward direction with speed u at the same time observer O’ in frame of reference S’ throw the ball B in downward direction with same speed u. The elastic collision is take place in between ball A and ball B at the midway between two X-axis. Let ua be the velocity of ball A measure by the observer in frame S. And uax & uay be the two component velocity of ua. As velocity of ball A along Y direction therefore , uax = 0 , uay = ua =u

Similarly u’b be the velocity of ball B measure by the observer in frame of reference S’. Therefore u’b = -u u’bx & u’by be the two components of u’b in X-Y plane u’bx = 0 & u’by = u’b = -u. Let ub be the velocity of ball B measuere by the observer in frame S. Let its components ubx & uby. The values of ubx & uby in terms of u’bx & u’by is ux = and uy = = - =v

Let ma and mb be the masses of ball A and ball B measure by the observer o in frame S. The total momentum of two balls before collision is ma uay + mb uby = mau – mb u After collision velocities are reverse uay = -u & uby = u

The momentum after collision is ma uay + mb uby = -mau + mb u According to the principal of conservation of momentum Momentum before collision = momentum after collision m au – m b u = -mau + mb u 2 m au = 2 m b u As ma mass of ball A in frame S Denote it as ma = m 0 and mb = m