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Unformatted text preview: 74. We orient our +x axis along the initial direction of motion, and specify angles in the “standard” way — so θ = –90° for the particle B which is assumed to scatter “downward” and φ > 0 for particle A which presumably goes into the first quadrant. We apply the conservation of linear momentum to the x and y axes respectively. mB vB = mB v′ cos θ + mAv′ cos φ B A 0 = mB v′ sin θ + mAv′ sin φ B A
′ (a) Setting vB = v and vB = v 2 , the y-momentum equation yields mA v ′ sin φ = mB A v 2 and the x-momentum equation yields mA v ′ cos φ = mB v. A Dividing these two equations, we find tan φ =
1 2 which yields φ = 27°. (b) We can formally solve for v′ (using the y-momentum equation and the fact that A φ =1 5)
v′ = A 5 mB v 2 mA but lacking numerical values for v and the mass ratio, we cannot fully determine the final speed of A. Note: substituting cos φ = 2 5 , into the x-momentum equation leads to exactly this same relation (that is, no new information is obtained which might help us determine an answer). ...
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- Spring '08