101. The mass of each ball is m, and the initial speed of one of the balls is 12.2m s.iv=We apply the conservation of linear momentum to the xand yaxes respectively. 111222 2cos0siniffffmvθθ=+=−The mass mcancels out of these equations, and we are left with two unknowns and two equations, which is sufficient to solve. (a) The y-momentum equation can be rewritten as, using 260θ=°and21.1 m/sfv=,sin(1.1 m/s)sin 600.95 m/s.fv=and the x-momentum equation yields cos(2.2 m/s) (1.1 m/s)cos601.65 m/s.fv°=Dividing these two equations, we find tan1= 0.576 which yields 1= 30°. We plug the value into either equation and find 1fv≈1.9 m/s. (b) From the above, we have 1= 30°, measured
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