A cosmic ray muon with mass mμ = 1.88 ✕ 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 2.40 ✕ 106 m/s in a straight line, decreases in speed to 1.56 ✕ 106 m/s over a distance of 1.22 km. (a) What is the magnitude of the force experienced by the muon?
(b) How does this force compare to the weight of the muon?
|F|/Fg =

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

the components of a vector are the parts of the vectors that are to each other.

b perpendicular