71. We orient our +xaxis along the initial direction of motion, and specify angles in the“standard” way — so θ= +60° for the proton (1) which is assumed to scatter into the first quadrant and φ= –30° for the target proton (2) which scatters into the fourth quadrant (recall that the problem has told us that this is perpendicular to ). We apply the conservation of linear momentum to the xand yaxes respectively. mv111 12 20=+' cos' cos'sinθφWe are given v1= 500 m/s, which provides us with two unknowns and two equations, which is sufficient for solving. Since
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