Question

In the collision shown, energy is conserved because both objects have the same speed and mass after as before the collision. Since the collision merely reserves the velocities, the final (total) momentum is opposite the initial. Thus. momentum can be conserved only if it is zero.

(a) Using the relativistically correct expression for momentum. Show that the total momentum is zero-that momentum is conserved. (Masses are in arbitrary units).

(b) Using the relativistic velocity transformation. find the four velocities in a frame moving to the right at 0.6c.

(c) Verify that momentum is conserved in the new frame.

Short answer

(a) The total momentum is zero and conservation of momentum is proved.

(b) The velocities in a moving frame of first particle before and after collision are 0 and $-0.88c$respectively and the velocities of second particle before and after collision are $-0.95c$and $0.38c$respectively.

(c) The momentum conservation for the moving frame is verified.

Step-by-step solution

Identification of given data

The speed of the first particle before the collision is $u_1=0.6c$

The speed of the second particle before the collision is $u_2=-0.8c$

The speed of the first particle after the collision is $V_1=-0.6c$

The speed of the second particle after the collision is $v_2=0.8c$

The speed of the frame is $v=0.6c$

Relativistic Momentum

The momentum of a particle moving at speeds relative to the speed of light is called relativistic momentum. The relativistic momentum of particles cannot be calculated by using the classical formula of momentum.

Proof for conservation of total momentum

The total relativistic momentum before collision is given as:

$p_b=\frac{m_{16}{u}_1}{\sqrt{1-{\left({\displaystyle \frac{u_1}{c}}\right)}^2}}+\frac{m_9{u}_2}{\sqrt{1-{\left({\displaystyle \frac{u_2}{c}}\right)}^2}}$

Here, c is the speed of light in vaccuum $m_{16}$and localid="1659335882447" $m_9$are the rest mass of the particles.

For $u_1=0.6c$and $u_2=-0.8c$, equation becomes-

$$\begin{gathered} p_b=\frac{m_{16}\left(0.6c\right)}{\sqrt{1-{\left({\displaystyle \frac{0.6c}{c}}\right)}^2}}+\frac{m_9\left(-0.8c\right)}{\sqrt{1-{\left({\displaystyle \frac{-0.8c}{c}}\right)}^2}} \\ {p}_b=\frac{m_{16}\left(0.6c\right)}{0.8}+\frac{m_9\left(-0.8c\right)}{0.6} \\ {p}_b=0.75m_{16}c-1.33m_{9}c\hspace{0.33em}\hspace{0.33em}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right) \end{gathered}$$The total relativistic momentum after collision is given as:

$p_a=\frac{m_{16}{v}_1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}+\frac{m_9{v}_2}{\sqrt{1-\left(\frac{V_2}{c}\right)}}$

For $v_1=-0.6c$and $v_2=0.8c$, equation becomes-

$$\begin{gathered} p_a=\frac{m_{16}\left(-0.6c\right)}{\sqrt{1-{\left({\displaystyle \frac{-0.6c}{c}}\right)}^2}}+\frac{m_9\left(0.8c\right)}{\sqrt{1-{\left({\displaystyle \frac{0.8}{c}}\right)}^2}} \\ {p}_a=\frac{m_{16}\left(-0.6c\right)}{0.8}+\frac{m_9\left(0.8c\right)}{0.6} \\ {p}_a=-0.75m_{16}c+1.33m_{9}c\hspace{0.33em}\hspace{0.33em}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right) \end{gathered}$$

The total momentum of the particles is given as:

$P=p_a+p_b$

Using equation (1) and (2), we get-

$$\begin{gathered} P=(-0.75m_{16}c+1.33m_{9}c)+(0.75m_{16}c-1.33m_{9}c) \\ P=0 \end{gathered}$$

the total momentum of the system is zero. Hence, it is conserved.

Determination of velocities by using relativistic velocity transformations

(b)

The velocity of first ball before collision with a moving frame is given as:

$u_{1b}=\frac{u_1-v}{1-\frac{u_1.V}{c^2}}$

For $u_1=0.6c$and $v=0.6c$, equation becomes-

$$\begin{gathered} u_{1b}=\frac{0.6c-0.6c}{1-\frac{(0.6)\left(0.6c\right)}{c^2}} \\ =0 \end{gathered}$$

Similarly, the velocity of second ball before collision with a moving frame is given as:

$u_{2b}=\frac{u_2-v}{1-\frac{u_2.V}{c^2}}$

For $v=0.6c$and $u_2=-0.8c$, equation becomes

$$\begin{gathered} u_{2b}=\frac{-0.8c-0.6c}{1-\frac{\left(-0.8c\right)\left(06c\right)}{c^2}} \\ =-95c \end{gathered}$$

The velocity of first ball after collision with a moving frame is given as:

$v_{1b}=\frac{v_1-v}{1-\frac{V_1.V}{c^2}}$

For $V_1=-0.6c$and $v=0.6c$, equation becomes-

$\begin{array}{l}{v}_{1b}=\frac{-0.6c-0.6c}{1-\frac{\left(-0.6c\right))\left(0.6c\right)}{c^2}}\\ =0.88c\end{array}$

The velocity of second ball after collision with a moving frame is given as:

$v_{2b}=\frac{v_2-v}{1-\frac{V_2.V}{c^2}}$

For $v=0.6c$and $V_2=0.8c$, equation becomes-

$\begin{array}{l}{v}_{2b}=\frac{0.8c-06c}{1-\frac{\left(0.8\right)\left(0.6c\right)}{c^2}}\\ =038c\end{array}$

Therefore, the four velocities in a moving frame are $u_{1b}=0$, $u_{2b}=-0.95c$, $V_{1b}=-0.88c$and $V_{2b}=0.38c$.

Verification for conservation of momentum in moving frame

(c)

The speed of the moving frame is same for all the particles so the momentum of all the particles before and after collision remains conserved.