Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P23 040

Transcript

40. We assume there are no forces or force-components along the x direction. We combine Eq. 23-28 with Newton’s second law, then use Eq. 4-21 to determine time t followed by Eq. 4-23 to determine the final velocity (with −g replaced by the ay of this problem); for these purposes, the velocity components given in the problem statement are re-labeled as v0x and v0y respectively. (a) We have a =  qE m =−
e m  which leads to E  a = − 1.60 × 10−19 C 9.11 × 10−31 kg   N ˆ 120 j = −2.1 × 1013 m/s2 ˆj . C (b) Since vx = v0x in this problem (that is, ax = 0), we obtain t vy ∆x 0.020 m = = 1.3 × 10−7 s v0x 1.5 × 105 m/s 

 = v0y + ay t = 3.0 × 103 m/s + −2.1 × 1013 m/s2 1.3 × 10−7 s = which leads to vy = −2.8 × 106 m/s. Therefore, in unit vector notation (with SI units understood) the final velocity is v = 1.5 × 105 ˆi − 2.8 × 106 ˆj .