Question

A cosmic ray muon with mass mu = 1.88 x 10-28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matter it passes through. One such particle, initially traveling at 2.50 x 106 m/s in a straight line, decreases in speed to 1.44 x 106 m/s over a distance of 1.28 km. (a) What is the magnitude of the force experienced by the muon? 5.025 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N (b) How does this force compare to the weight of the muon? |F = FO

Answer 1

Solution : - Geven that, kg Mass 1.88 x 1028 mu Intial velocity, final relocity u= 2.50x Lob reps 1.44 x 100 m/s Distance d = 1.2 8 km 07/ 1.2 8 x 10 m Past @ We kinematics equation, know that from ez u2+ zad re? – uz 2 d a = so Acceleration, a = (1.44X108 32 - (2.50x108² 2x 1.28 x 103 m 6. 25x10² 12 2.07 36 X 10' 2.56 xlo 3 a. 12 4.17 6 4 x 10 2.564 103 m/? a = a = - 1.63 1 x 10 m/s² force experienced by the so the muona m a or -28 F = 1.88 x 10 kg 61.631% 10ºm/2) - 3.066 X 10 - 19 kg m / 32 F

F = – 3.07 x 10 -19 N so, The magnitude of the force experience muon is, by the IFI 7-19 N 3.07 x 10 Part 6 = دا The weight of the muon is, w Fg = w= mg Fg = 1.88 x 162 kg * 9.80 m/s2 Fg = W= 18. 424 X 10 28 N know from above, force is, FI- 3.07 x to cue 19 so compare the weight of muon is, IF 19 3.07 xlo -28 Fg 18. 4 24 x 10 9 O. 1667 X 10' TEL Fg 8. FI 1.67 x 10 ܪܥܐ