AP Physics 2 Exam Question: Particle Physics

A particle with an initial mass of 2.5 kg is accelerated to a speed of 0.9c (where c is the speed of light). Assume the particle behaves relativistically.

a) Calculate the relativistic mass of the particle. b) Determine the total energy of the particle. c) Find the kinetic energy of the particle.

Answer:

a) Relativistic mass (m_rel) can be calculated using the following formula:

where

• $m_{rel}$ is the relativistic mass,
• $\gamma$ is the Lorentz factor, and
• $m_0$ is the rest mass of the particle.

The Lorentz factor ($\gamma$) can be calculated as:

Substituting given values: $m_0 = 2.5$ kg $v = 0.9c$

Using $c = 3 \times 10^8$ m/s: $v = 0.9 \times 3 \times 10^8$ m/s

Calculating $\gamma$: [ \gamma = \frac{1}{\sqrt{1 - \left(\frac{0.9 \times 3 \times 10^8}{3 \times 10^8}\right)^2}} ]

Simplifying $\gamma$ gives: [ \gamma = \frac{1}{\sqrt{1 - 0.9^2}} ]

Solving for $\gamma$: [ \gamma = \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}} \approx 1.052/ ]

Now substituting $\gamma$ and $m_0$ in the equation for relativistic mass: [ m_{rel} = \gamma \cdot m_0 = 1.052 \times 2.5 = 2.63 , \text{kg} ]

Therefore, the relativistic mass of the particle is 2.63 kg.

b) The total energy (E) of the particle can be calculated using the formula:

Substituting the values of $\gamma$, $m_0$, and $c$: [ E = 1.052 \times 2.5 \times (3 \times 10^8)^2 ]

Simplifying gives: [ E = 1.052 \times 2.5 \times 9 \times 10^{16} ]

Calculating this expression: [ E = 2.368 \times 10^{17} , \text{J} ]

Hence, the total energy of the particle is $2.368 \times 10^{17}$ J.

c) The kinetic energy (K.E.) of a relativistic particle can be determined using the equation:

Substituting the values of $E$, $m_0$, and $c$: [ K.E. = 2.368 \times 10^{17} - 2.5 \cdot (3 \times 10^8)^2 ]

Simplifying: [ K.E. = 2.368 \times 10^{17} - 2.5 \times 9 \times 10^{16} ]

Calculating this expression: [ K.E. = 1.618 \times 10^{17} , \text{J} ]

Therefore, the kinetic energy of the particle is $1.618 \times 10^{17}$ J.

Note: The above calculations assume the particle's speed is constant throughout the given situation.