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**Unformatted text preview: **6.70: a) This is similar to Problem 6.64, but here > 0 (the force is repulsive), and x2 < x1 , so the work done is again negative; 1 1 W = - = (2.12 10- 26 N m 2 ((0.200 m -1 ) - (1.25 109 m -1 )) x x 2 1 = -2.65 10-17 J. Note that x1 is so large compared to x 2 that the term (6.13)) and solving for v2 ,
2 1 1 x1 is negligible. Then, using Eq. 2W 2(-2.65 10-17 J) 5 2 v2 = v + = (3.00 10 m / s) + = 2.41 105 m / s. - 27 m (1.67 10 kg) b) With K 2 = 0, W = - K1 . Using W = - x 2 , 2 2(2.12 10-26 N m 2 ) = = = 2.82 10 -10 m. 2 - 27 5 2 K1 mv1 (1.67 10 kg )(3.00 10 m / s) c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00 105 m / s . x2 = ...

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