Chapter 39, Problem 39P

(a)

To determine

Speed of first particle before collision.

Explanation of Solution

Given:

Particle of mass $1.00\text{MeV}/c^2$ and kinetic energy $2.00\text{MeV}$ collides with stationary particle of mass $2.00\text{MeV}/c^2$ .

The particles stick together after collision.

Formula used:

Write the expression of total energy of particle

  $\epsilon =\left(k+{\epsilon }_0\right)$

Here, $k$ is kinetic energy and ${\epsilon }_0$ is rest energy.

Write the expression of total energy in terms of gamma factor

  $\epsilon =\gamma {\epsilon }_0$

Here, $\gamma$ is gamma factor.

Substitute $\left(k+{\epsilon }_0\right)$ for $\epsilon$ in above expression

  $\left(k+{\epsilon }_0\right)=\gamma {\epsilon }_0$

Simplify the above expression for $\gamma$

  $\gamma =1+\frac{k}{{\epsilon }_0}$ …… (1)

Write the expression of gamma factor

  $\gamma =\frac{1}{{\left(1-\frac{v^2}{c^2}\right)}^{\frac{1}{2}}}$

Here, $v$ is speed of first particle before collision.

Simplify above expression for $v$

  $v={\left(1-\frac{1}{{\gamma }^2}\right)}^{\frac{1}{2}}c$ …… (2)

Calculation:

The value of ${\epsilon }_0$ is:

  $\begin{array}{c}{\epsilon }_0=\left(1.00\frac{\text{MeV}}{c^2}\right)c^2\\ =1.00\text{MeV}\end{array}$

Substitute $1.00\text{MeV}$ for ${\epsilon }_0$ and $2.00\text{MeV}$ for $k$ in expression (1)

  $\begin{array}{c}\gamma =1+\frac{\left(2.00\text{MeV}\right)}{\left(1.00\text{MeV}\right)}\\ =3\end{array}$

Substitute 3 for $\gamma$ in expression (2)

  $\begin{array}{c}v={\left(1-\frac{1}{{\left(3\right)}^2}\right)}^{\frac{1}{2}}c\\ =0.943c\end{array}$

Conclusion:

Thus, the speed of first particle before collision is $0.943c$ .

(b)

To determine

Total energy of first particle before collision.

Explanation of Solution

Given:

Particle of mass $1.00\text{MeV}/c^2$ and kinetic energy $2.00\text{MeV}$ collides with stationary particle of mass $2.00\text{MeV}/c^2$ .

The particles stick together after collision.

Formula used:

Write the expression of total energy of particle

  $\epsilon =\gamma {\epsilon }_0$ …… (3)

Here, $\epsilon$ is total energy of first particle before collision.

Calculation:

Substitute 3 for $\gamma$ and $1.00\text{MeV}$ for ${\epsilon }_0$ in expression (3)

  $\begin{array}{c}\epsilon =\left(3\right)\left(1.00\text{MeV}\right)\\ =3.00\text{MeV}\end{array}$

Conclusion:

Thus, the total energy of first particle before collision is $3.00\text{MeV}$ .

(c)

To determine

Initial total momentum of system.

Explanation of Solution

Given:

Particle of mass $1.00\text{MeV}/c^2$ and kinetic energy $2.00\text{MeV}$ collides with stationary particle of mass $2.00\text{MeV}/c^2$ .

The particles stick together after collision.

Formula used:

Write the expression of the total relativistic energy

  ${\epsilon }^2=p^2{c}^2+{\epsilon }_0{}^2$

Here, $p$ is initial momentum.

Rearrange the above expression

  $p=\frac{{\left({\epsilon }^2-{\epsilon }_0{}^2\right)}^{\frac{1}{2}}}{c}$ …… (4)

Calculation:

Substitute $3.00\text{MeV}$ for $\epsilon$ and $1.00\text{MeV}$ for ${\epsilon }_0$ in expression (4)

  $\begin{array}{c}p=\frac{{\left({\left(3.00\text{MeV}\right)}^2-{\left(1.00\text{MeV}\right)}^2\right)}^{\frac{1}{2}}}{c}\\ =2.83\text{MeV}/c\end{array}$

Conclusion:

Thus, the initial total momentum of system is $2.83\text{MeV}/c$ .

(d)

To determine

Total kinetic energy after collision.

Explanation of Solution

Given:

Particle of mass $1.00\text{MeV}/c^2$ and kinetic energy $2.00\text{MeV}$ collides with stationary particle of mass $2.00\text{MeV}/c^2$ .

The particles stick together after collision.

Formula used:

Since the total energy is conserved during collision therefore

  ${\epsilon }_1=5.0\text{MeV}$

Here, ${\epsilon }_1$ is total energy after collision.

Since the momentum is conserved during the collision therefore

  $p_1=2.83\text{MeV}/c$

Here, $p_1$ is total final momentum of system after collision.

Write the expression for rest mass energy ${\epsilon }_0{}^{\prime }$ after collision

  ${\epsilon }_0{}^{\prime }={\left({\epsilon }_1{}^2-p_1{}^2{c}^2\right)}^{\frac{1}{2}}$ …… (5)

Write the expression for total kinetic energy $k_1$ after collision

  $k_1={\epsilon }_1-{\epsilon }_0{}^{\prime }$ …… (6)

Calculation:

Substitute $5.0\text{MeV}$ for ${\epsilon }_1$ and $2.83\text{MeV}/c$ for $p_1$ in expression (5)

  $\begin{array}{c}{\epsilon }_0{}^{\prime }={\left({\left(5.0\text{MeV}\right)}^2-{\left(2.83\frac{\text{MeV}}{c}\right)}^2{c}^2\right)}^{\frac{1}{2}}\\ =4.12\text{MeV}\end{array}$

Substitute $5.0\text{MeV}$ for ${\epsilon }_1$ and $4.12\text{MeV}$ for ${\epsilon }_0{}^{\prime }$ in expression (6)

  $\begin{array}{c}{k}_1=\left(5.0\text{MeV}\right)-\left(4.12\text{MeV}\right)\\ =0.9\text{MeV}\end{array}$

Conclusion:

Thus, the total kinetic energy after collision is $0.9\text{MeV}$ .

(e)

To determine

Mass of system after collision.

Explanation of Solution

Given:

Particle of mass $1.00\text{MeV}/c^2$ and kinetic energy $2.00\text{MeV}$ collides with stationary particle of mass $2.00\text{MeV}/c^2$ .

The particles stick together after collision.

Formula used:

Write the expression of mass of system

  $m=\frac{{\epsilon }_0{}^{\prime }}{c^2}$ …… (7)

Here, is mass of the system after collision

Calculation:

Substitute $4.12\text{MeV}$ for ${\epsilon }_0{}^{\prime }$ in expression (7)

  $\begin{array}{c}m=\frac{\left(4.12\text{MeV}\right)}{c^2}\\ =4.1\text{MeV}/c^2\end{array}$

Conclusion:

Thus, the mass of system after collision is $4.1\text{MeV}/c^2$ .