In classical physics, if the momentum of a particle doubles, the kinetic energy increases by a factor of 4. What happens to the kinetic energy of the speedy proton in this example if its momentum doubles?

Question

Accepted Answer

Based on what we have seen so far in relativity, it is likely that you would predict that its kinetic energy does not increase by a factor of 4. If the momentum doubles, the new momentum is

[latex] p_{ ext {new }}=2\left(\sqrt{8} \frac{m_{p} c^{2}}{c} ight)=4 \sqrt{2} \frac{m_{p} c^{2}}{c} [/latex].

Using Equation 39.27, we find the square of the new total energy:

[latex] E^{2}=p^{2} c^{2}+\left(m c^{2} ight)^{2} [/latex] (39.27).

[latex] E_{ ext {new }}^{2}=p_{ ext {new }}^{2} c^{2}+\left(m_{p} c^{2} ight)^{2} [/latex].

[latex] E_{ ext {new }}^{2}=\left(4 \sqrt{2} \frac{m_{p} c^{2}}{c} ight)^{2} c^{2}+\left(m_{p} c^{2} ight)^{2}=33\left(m_{p} c^{2} ight)^{2} [/latex].

[latex] E_{ ext {new }}=\sqrt{33}\left(m_{p} c^{2} ight)=5.7 m_{p} c^{2} [/latex].

Now, using Equation 39.25, we find the new kinetic energy:

[latex] E=K+m c^{2} [/latex] (39.25).

[latex] K_{ ext {new }}=E_{ ext {new }}-m_{p} c^{2}=5.7 m_{p} c^{2}-m_{p} c^{2}=4.7 m_{p} c^{2} [/latex].

Notice that this is only 2.35 times as large as the kinetic energy we found in part (C), not four times as large. In general, the factor by which the kinetic energy increases if the momentum doubles will depend on the initial momentum, but will approach 4 as the momentum approaches zero. In this latter situation, classical physics correctly describes the situation.

Question 39.WI.12: In classical physics, if the momentum of a particle doubles,...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions