A $\mathbf{3}\mathbf{.}\mathbf{000}\mathbf{u}$object moving to the right through a laboratory at $\mathbf{}\mathbf{0}\mathbf{.}\mathbf{6}\mathit{c}$collides with a $\mathbf{4}\mathbf{.}\mathbf{000}\mathbf{u}$object moving to the left through the laboratory at $\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{c}$. Afterward, there are two objects, one of which is a $\mathbf{6}\mathbf{.}\mathbf{000}$$\mathbf{}\mathit{u}\mathbf{}$mass at rest.

(a) What are the mass and speed of the other object?

(b) Determine the change in kinetic energy in this collision.

Conservation of momentum and energy:

A property of a moving body that a body has by virtue of its mass and motion, which is equal to the product of the body's mass and velocity.

A fundamental law of physics and chemistry states that the total energy of an isolated system is constant despite internal changes.

Let’s consider the following values:

The object of mass $3.000\mathrm{u}$is moving toward the positive x-axis and the second object of mass $4.000\mathrm{u}$is moving towards the negative x-axis.

After the collision, two objects are formed, one object of mass $6.000\mathrm{u}$is at rest, and the speed and mass of the other object are required.

The relativistic factors arelocalid="1659090202034" ${\gamma }_1=\left(\frac{5}{3}\right)$and localid="1659090205890" ${\gamma }_2=\left(\frac{5}{4}\right)$.

The velocity of mass m₁ and m₂ before collision is u₁ = 0.87c and u₂ = -0.6c respectively.

The velocity of mass m₃ after collision is u₃ = 0.

Momentum conserved:

${\left({\gamma }_1{m}_1{u}_1+{\gamma }_2{m}_2{u}_2\right)}_{\text{before}}={\left({\gamma }_3{m}_3{u}_3+{\gamma }_4{m}_4{u}_4\right)}_{\text{after}}$

$\left[\begin{array}{c}\left(\frac{5}{3}\right)(3.000u)(0.8c)+\\ \left(\frac{5}{4}\right)(4.000u)(-0.6c)\end{array}\right]=0+{\gamma }_4{m}_4{u}_4$

$7_4{m}_4{u}_4=1c$… (1)

Energy conserved:

${\gamma }_1{m}_1{c}^2+{\gamma }_2{m}_2{c}^2=m_3{c}^2+{\gamma }_4{m}_4{c}^2$

$\left(\frac{5}{3}\right)\left(3\right)+\left(\frac{5}{4}\right)\left(4\right)=6+{\gamma }_4{m}_4$

${\gamma }_4{m}_4=4$ … (2)

To Determine the mass and speed of the other object:

Dividing equation (1) by (2), we get,

localid="1659090192330" $$\begin{gathered} \frac{{\gamma }_4{m}_4{u}_4}{{\gamma }_4{m}_4}=\frac{1c}{4} \\ {u}_4=0.25c \end{gathered}$$

Inserting in Lorentz factor,

localid="1659090196355" $\begin{array}{rcl}{\gamma }_4& =& \frac{1}{\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$0.25c$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}}\\ & =& \frac{1}{\sqrt{1-{\left(0.25\right)}^2}}\\ & =& 1.03\end{array}$

then inserting${\gamma }_4$in equation (2), we get the mass of the other object

localid="1659090186343" $$\begin{gathered} 1.03\times m_4=4 \\ {m}_4=\frac{4}{1.03} \\ =3.873u \end{gathered}$$

Now change in kinetic energy,

$\begin{array}{rcl}K{E}_f-K{E}_i& =& \left[\left({\gamma }_4-1\right)m_4{c}^2+\left({\gamma }_3-1\right)m_3{c}^2\right]-\left[\left({\gamma }_1-1\right)m_1{c}^2+\left({\gamma }_2-1\right)m_2{c}^2\right]\\ & =& \left[\left(1.03-1\right)\left(3.873u\right)+0\right]c^2-\left[\left(\frac{5}{3}-1\right)\left(3\right)+\left(\frac{5}{4}-1\right)\left(u\right)\right]c^2\\ & =& \mathbf{-}2676.2MeV\\ & & \end{array}$

$\begin{array}{rcl}K{E}_f-K{E}_i& =& \left(\mathbf{-}2676.2\times 1.6\mathbf{\times }{\mathbf{10}}^{-19}\right)\mathbf{J}\\ & =& \mathbf{-}\mathbf{4}.29\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{J}\\ & & \end{array}$

Therefore, the amount of kinetic energy is converted into a mass is,

$\begin{array}{rcl}\Delta m& =& \left(6+3.873\right)-7\\ & =& \mathbf{2}\mathbf{.}\mathbf{873}\mathbf{u}\\ & & \end{array}$

This can be verified using the equation $\Delta KE=-\Delta m{c}^2$.