Chapter 26, Problem 34P

An unstable particle with a mass equal to 3.34 × 10⁻²⁷ kg is initially at rest. The particle decay's into two fragments that fly off with velocities of 0.987c and −0.868c, respectively. Find the masses of the fragments. Hint: Conserve both mass-energy and momentum.

To determine

The masses of the fragments.

Answer to Problem 34P

  The masses of the fragments are $2.51\times {10}^8\text{kg}$ and $8.84\times {10}^{-28}\text{kg}$.

Explanation of Solution

Given info:

The speed of first segment is $0.868c$.

The speed of second segment is $0.987c$.

The formula used to calculate the rest energy of the particle is,

$E_R=m{c}^2$

• $E_R$ is energy of the particle
• $c$ is the speed of light
• $m$ is the mass of the particle

The formula used to calculate the rest energy of the first fragment is,

$E_{R_1}=m_1{c}^2$

• $E_R{}_{{}_1}$ is energy of the first fragment
• $m_1$ is the mass of the first fragment

The formula used to calculate the rest energy of the second fragment is,

$E_{R_2}=m_2{c}^2$

• $E_R{}_{{}_2}$ is energy of the second fragment
• $m_2$ is the mass of the second fragment

The formula used to calculate the gamma factor for first fragment is,

${\gamma }_1=\frac{1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}$

• ${\gamma }_1$ is the gamma factor of the first fragment
• $v_1$ is the speed of the first fragment

The formula used to calculate the gamma factor for second fragment is,

${\gamma }_2=\frac{1}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}$

• ${\gamma }_2$ is the gamma factor of the second fragment
• $v_2$ is the speed of the second fragment

The formula used to calculate the total energy of the first fragment is,

$\begin{array}{c}{E}_1={\gamma }_1{E}_{R_1}\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}\right)\left(m_1{c}^2\right)\end{array}$

The formula used to calculate the total energy of the second fragment is,

$\begin{array}{c}{E}_2={\gamma }_2{E}_{R_1}\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}\right)\left(m_2{c}^2\right)\end{array}$

The formula define the conservation of energy is,

$\begin{array}{c}{E}_R=E_1+E_2\\ m{c}^2=\left(\frac{1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}\right)\left(m_1{c}^2\right)+\left(\frac{1}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}\right)\left(m_2{c}^2\right)\\ m=\frac{m_1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}+\frac{m_2}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}\end{array}$

Substitute $3.34\times {10}^{-27}\text{kg}$ for $m$ , $\text{0}\text{.987}c$ for $v_1$ and $-\text{0}\text{.868}c$ for $v_2$ to find the masses.

$\begin{array}{l}3.34\times {10}^{-27}\text{kg}=\frac{m_1}{\sqrt{1-{\left(\frac{\text{0}\text{.987}c}{c}\right)}^2}}+\frac{m_2}{\sqrt{1-{\left(\frac{-\text{0}\text{.868}c}{c}\right)}^2}}\\ 3.34\times {10}^{-27}\text{kg}=6.22m_1+2.01m_2\end{array}$

Substitute $3.34\times {10}^{-27}\text{kg}$ for $m$ , $\text{0}\text{.987}c$ for $v_1$ and $-\text{0}\text{.868}c$ for $v_2$ to find the masses.

$\begin{array}{l}3.34\times {10}^{-27}\text{kg}=\frac{m_1}{\sqrt{1-{\left(\frac{\text{0}\text{.987}c}{c}\right)}^2}}+\frac{m_2}{\sqrt{1-{\left(\frac{-\text{0}\text{.868}c}{c}\right)}^2}}\\ 3.34\times {10}^{-27}\text{kg}=6.22m_1+2.01m_2\end{array}$ I

The formula used to calculate the momentum of the first fragment is,

$\begin{array}{c}{P}_1={\gamma }_1{m}_1{v}_1\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}\right)m_1{v}_1\end{array}$

The formula used to calculate the momentum of the second fragment is,

$\begin{array}{c}{P}_2={\gamma }_2{m}_2{v}_2\\ =\left(\frac{1}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}\right)m_2{v}_2\end{array}$

The formula define the conservation of momentum is,

$\begin{array}{c}{P}_1=P_2\\ {\gamma }_1{m}_1{v}_1={\gamma }_2{m}_2{v}_2\\ \left(\frac{1}{\sqrt{1-{\left(\frac{0.987c}{c}\right)}^2}}\right)\left(m_1\right)\left(0.987c\right)=\left(\frac{1}{\sqrt{1-{\left(\frac{0.868c}{c}\right)}^2}}\right)\left(m_2\right)\left(0.868c\right)\\ m_2=3.52m_1\end{array}$ II

Substitute $3.52m_1$ for $m_2$ in equation I to find $m_1$.

$\begin{array}{c}3.34\times {10}^{-27}\text{kg}=6.22m_1+2.01\left(3.52m_1\right)\\ m_1=2.51\times {10}^{-28}\text{kg}\end{array}$

Substitute $2.51\times {10}^{-28}\text{kg}$ for $m_1$ in equation II to find $m_2$.

$\begin{array}{c}{m}_2=3.52\left(2.51\times {10}^{-28}\text{kg}\right)\\ =8.84\times {10}^{-28}\text{kg}\end{array}$

Thus, the masses of the fragments are $2.51\times {10}^8\text{kg}$ and $8.84\times {10}^{-28}\text{kg}$.

Conclusion:

The masses of the fragments are $2.51\times {10}^8\text{kg}$ and $8.84\times {10}^{-28}\text{kg}$