Question

Particle $\mathbf{1}$, of mass role="math" localid="1657551290561" ${\mathit{m}}_{\mathbf{1}}$, moving at role="math" localid="1657551273841" $\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}\mathit{c}$relative to the lab, collides head-on with particle $\mathbf{2}$, of mass ${\mathit{m}}_{\mathbf{2}}$, moving at $\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathit{c}$relative to the lab. Afterward, there is a single stationary object. Find in terms of role="math" localid="1657551379188" ${\mathit{m}}_{\mathbf{1}}$

(a) ${\mathit{m}}_{\mathbf{2}}$

(b) the mass of the final stationary object; and

(c) the change in kinetic energy in this collision.

Short answer

1. The mass of m₂ is 16/3 m₁ .
2. The mass of the final stationary object is m_f = 25/3 m₁ .
3. The change in kinetic energy in this collsion is $\Delta KE-2m_1{c}^2$.

Step-by-step solution

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.

(a) Applying Momentum Conservation:

As the collision is completely inelastic, kinetic energy will be lost, and the internal energy to increase.

Consider the known data as below.

The relativistic factors is ${\gamma }_{0.6c}=\frac{5}{4}$and ${\gamma }_{0.8c}=\frac{5}{3}$.

To determine m₂ in this case of completely inelastic collision let’s apply momentum conservation as below.

$$\begin{gathered} {\gamma }_1{m}_1{u}_1+{\gamma }_2{m}_2{u}_2={\gamma }_f{m}_f{u}_f \\ {\gamma }_{0.8c}{m}_1\left(0.8c\right)+{\gamma }_{0.6c}{m}_2\left(-0.6c\right)=0 \\ \left(\frac{5}{3}\right)\left(0.8\right)m_1=\left(\frac{5}{4}\right)\left(0.6\right)m_2 \\ {m}_2=\frac{16}{3}{m}_1 \end{gathered}$$

(c) Determine the kinetic energy

Now applying relativistic kinetic energy conservation equation,

$\begin{array}{rcl}\Delta KE& =& \left[\left({\gamma }_0-1\right)m_f{c}^2\right]-\left[\left({\gamma }_{0.8c}-1\right)m_1{c}^2+\left({\gamma }_{0.6c}-1\right)m_2{c}^2\right]\\ & =& 0-\left[\left(\frac{5}{3}-1\right)m_1+\left(\frac{5}{4}-1\right)m_2\right]c^2\\ & =& -\left[\frac{2}{3}{m}_1+\frac{1}{4}\left(\frac{16}{3}{m}_1\right)\right]c^2\\ & =& -2m_1{c}^2\end{array}$

The change in relativistic kinetic energy will result in change in Internal energy which implies that mass will change.

(c) Determine the mass of the final stationary object

In this case, the mass will increase because kinetic energy decreases.

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

Put this term in above result, and you get,

$$\begin{gathered} -\Delta m{c}^2=-2m_1{c}^2 \\ \left[m_f-\left(m_1+m_2\right)\right]=2m_1 \\ \left[m_f-m_1-\left(\frac{16}{3}{m}_1\right)\right]=2m_1 \end{gathered}$$

$\begin{array}{rcl}3m_f-3m_1-16m_1& =& 6m_1\\ 3m_f& =& 25m_1\\ m_f& =& \frac{25}{3}{m}_1\end{array}$

Hence, there is an increase in mass by 2m₁of the system which is a result of decrease in the kinetic energy by 2m₁c² of the system.