Question

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.

Short answer

1. The mass and speed of the other object is $m_4=3.873u$and $u_4=0.25c$respectively .
2. The change in kinetic energy in collision is $K{E}_f-K{E}_i=\mathbf{-}\mathbf{4}.29\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{J}$.

Step-by-step solution

Apply Momentum Conservation

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)

Apply Energy Conservation

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

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

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 $\Delta KE=-\Delta m{c}^2$.