A. Calculate the kinetic energy of a proton (1.6710^-27kg) traveling 7.2710^7

ID: 2299874 • Letter: A

Question

A. Calculate the kinetic energy of a proton (1.67*10^-27kg) traveling 7.27*10^7 m/s. Answer should be in J.

B. Calculate the momentum of a proton (1.67*10^27kg ) traveling 7.27*10^7 m/s. Answe should be in kg * m/s

C. By what percentages would your calculations have been in error if you had used classical formulas? KE?KErelKErel,prelprel

Another set of quick questions.

1. What is the speed of an electron whose kinetic energy is 1.85MeV? 26.31

2.What is the speed of a proton accelerated by a potential difference of 114MV ? Thanks!

A)m = mass = 1.67 x 10^-27kg

v = velocity = 7.27 x 10^7 m/s

c = speed of light = 3 x 10^8 m/s

Since the proton is moving so fast we must use relativity to solve for the kinetic energy (KE):

KE = mc^2[1 - 1/sqrt(1 - v^2/c^2)]

= (1.67x10^-27kg)*(3x10^8m/s)^2*[1 - 1/sqrt(1 - (7.27x10^7)^2/(3x10^8)^2)]

= 2.4x 10^-9 Joules

B)P=MV

=1.67 x 10^-27kg * 7.27 x 10^7 m/s

=1.21 x10^35 kgm/s

C) 1/square root(1-(7.27x10^7/3x10^8)^2= 1.03 (use this value for relativistic formula, shown below.)

*classic: (1.67x10^-27)(7.27x10^7)= 1.21x10^-19 (this is the classic value which you need to use to compare to the relativistic value)

*relativistic: (1.03)(1.21x10^-19)<==classic value= 1.24x10^-19 kg*m/s is your answer for MOMENTUM of a proton.

*Percentage error for momentum: 1.21x10^-19/1.24x10^-19= 2.41%==> -2.41% DONT FORGET TO ADD THE NEGATIVE SIGN.

1)KE=0.5mv^2

1.85 x10^6=0.5 x 9.1*10^(-31) x V^2

v=2.01 x10^18 m/s

1/2mv^2 = qV
=>v = sqrt(2qV/m)
=>v = sqrt[(2 x 1.6 x 10^-19 x 114 x 10^6)/(1.67 x 10^-27)]
=>v = 1.47 x 10^8 m/s

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