Question

What is the percent change in momentum of a proton that accelerates from 0.40c to 0.83c?

a. Calculated value
b. 25%
c. 48%
d. 108%

Answer 1

The percent change in momentum of a proton that accelerates from
`\(0.40c\)` to
`\(0.83c\)` is approximately
`\(108%\).`%. (Option d)

Step-by-step explanation:

The formula for calculating the relativistic momentum
`(\(p\))` of an object moving at a velocity
`\(v\)` is given by the equation
`\(p = \frac{m_0v}{\sqrt{1 - (v^2)/(c^2)}}\)`, where
`\(m_0\)` is the rest mass of the proton,
`\(v\)` is its velocity, and
`\(c\)` is the speed of light.

First, we find the relativistic momentum
`(\(p_1\))` at the initial velocity
`\(0.40c\)` and the relativistic momentum
`(\(p_2\))` at the final velocity
`\(0.83c\)`. The percent change in momentum is then calculated using the formula
`\(\%\Delta p = (p_2 - p_1)/(p_1) * 100\)`.

In this case, the percent change in momentum is found to be approximately
`\(108%\)`%. This indicates that the momentum of the proton increases significantly as it accelerates from
`\(0.40c\)` to
`\(0.83c\)`. The relativistic effects become more pronounced at higher speeds, and this change in momentum aligns with the principles of special relativity.

Understanding the relativistic change in momentum is crucial in high-speed scenarios, providing insights into the behavior of particles as they approach the speed of light.(Option d)