Question

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for speed of 6.00×107m/s.

Answer 1

Answer:

$KE = 5.01 \times 10^{-13}Joules$

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion. It is expressed as;

$KE = \frac{1}{2}mv^2$ where:

m is the mass of proton

v is the speed

Substitute:

$KE = \frac{1}{2}(1.67\times 10 ^{-27})(6.0\times10^7)^2\\\\KE = \frac{1}{2}(1.67\times 10 ^{-27})(6.0\times10^{14})\\\\\\KE = \frac{1}{2}(1.67\times 6.0 \times 10 ^{-27}\times10^{14})\\\\KE = \frac{1}{2}(1.67\times 6.0 \times 10 ^{-27+14})\\\\KE = 3 \times 1.67 \times 10^{-13}\\\\KE = 5.01 \times 10^{-13}Joules$

Hence the kinetic energy of a proton is $5.01 \times 10^{-13}Joules$