Aproton in a particle accelerator is traveling at a speed of 0.99c.(a) if you use the approximate nonrelativistic equation for the magnitude of momentum of the proton, what answer do you get? (b) what is the magnitude of the correct relativistic momentum of the proton? (c) the approximate value (the answer to part a) is significantly too low. what is the ratio of magnitudes you calculated (correct/approximate)?

a)  p = 4.96 10⁻¹⁹ kg m / s , b)  p = 35 .18 10⁻¹⁹  kg m / s ,

c)  p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

                 p = m v

Let's calculate its value

               p = 1.67 10⁻²⁷ 0.99 3. 10⁸

               p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

               p = m v / √(1 –v² / c²)

Let's calculate

                p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

                p = 4.96 10⁻¹⁹ / 0.141

                p = 35 .18 10⁻¹⁹  kg m / s

c) the relationship between the two values is

            p_correst / p_approximate = 35.18 / 4.96

            p_correst / p_approximate = 7.09

i would say the answer is food.

hope this !

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length