Question

A particle of mass m, moving at speed $\mathit{v}\mathbf{=}\frac{\mathbf{4}}{\mathbf{5}}\mathit{c}$, collides with an identical particle that is at rest. The two particles react to produce a new particle of mass M and nothing else. (a) What is the speed V of the composite particle? (b) What is its mass M?

Short answer

1. The speed of composite particle is 0.5c.
2. The mass of the composite particle is 2.31m.

Step-by-step solution

Identification of given data

The given data can be listed below,

• Speed of the particle is $v=\frac{4}{5}c$.
• Mass of the particle is m.
• Mass of the new particle after collision is M.

Concept/Significance of the relativistic energy

A mass’s energy is increased by increasing its velocity. Increasing its velocity, in turn, increases its relativistic mass. The relativistic energy contains both kinetic energy and the rest mass-energy of the particle.

(a) Determination of the speed V of the composite particle

The energy of the particle is given by,

$E_1=\frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{u_1^2}{c^2}}}}$

Here, $u_1$is the speed of particle, c is the speed of light and m is the mass of the particle.

Substitute values in the above,

$$\begin{gathered} E_1=\frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{{\left({\displaystyle \frac{4}{5}}c\right)}^2}{c\left(2\right)}}}} \\ =\frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{16}{25}}}} \\ =\frac{5m{c}^2}{3} \end{gathered}$$

The identical particle is at rest so the energy of the particle is given by,

$E_2=m{c}^2$

The relativistic energy of the new particle is given by,

$E_3=\frac{M{c}^2}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

From law of conservation of energy.

$E_1+E_2=E_3$

Substitute all the values in the above,

$$\begin{gathered} \frac{5}{3}m{c}^2+m{c}^2=\frac{m{c}^2}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}} \\ \frac{5}{3}m+m=\frac{M}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}} \\ \frac{8m}{3}=\frac{M}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}} \end{gathered}$$

The initial momentum of the first particle is given by,

$p_1=\frac{m{u}_1}{\sqrt{1-{\displaystyle \frac{u_1^2}{c^2}}}}$

Substitute the values in the above,

$$\begin{gathered} p_1=\frac{m(4/5)c}{\sqrt{1-{\displaystyle \frac{{(4/5)}^2{c}^2}{c^2}}}} \\ =\frac{4}{3}mc \end{gathered}$$

The initial velocity of second particle is zero. So, the initial momentum of the particle is also zero.

The initial momentum of the new particle is given by,

localid="1657868663005" $p_3=\frac{Mv}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

Here, Mis the mass of the new particle, v is the speed of new particle.

From law of conservation of momentum.

$p_1+p_2=p_3$

Substitute all the values in the above,

$\frac{4}{3}mc+0=\frac{Mv}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

Compare both momentum and energy equations.

$$\begin{gathered} \frac{4}{3}mc=\frac{8}{3}mv \\ c=2v \\ v=0.5c \end{gathered}$$

Thus, the speed of composite particle is 0.5c.

Step 4(b) determination of the mass of composite particle.

From relativistic energy equation mass of the particle is given by,

$$\begin{gathered} \frac{8}{3}m=\frac{M}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ M=\frac{8}{3}m\sqrt{1-\frac{v^2}{c^2}} \end{gathered}$$

Here, mis the mass of the first particle and v is the speed of composite particle.

Substitute all the values in the above,

$$\begin{gathered} M=\frac{8}{5}m\sqrt{1-\frac{{\left({\displaystyle \frac{c}{2}}\right)}^2}{c^2}} \\ =\frac{4}{\sqrt{3}}m \\ =2.31m \end{gathered}$$

Thus, the mass of the composite particle is 2.31m.