Question

Consider the collision of two particles, each of mass m_o. In experiment A, a particle moving at 0.9cstrikes a stationary particle.

1. What is the total kinetic energy before the collision?
2. In experiment B, both particles are moving at a speed u(relative to the lab), directly towards one another. If the total kinetic energy before the collision in experiment B is the same as that in experiment A, what is u?
3. In both particles, the particles stick together. Find the mass of the resulting single particle in each experiment. In which is more of the initial kinetic energy converted to mass?

Short answer

a) In experiment A, the total kinetic energy before the collision is.$1.29m_o{c}^2$

b) The final mass of the particle after the collision is$2.55m_o$.

c) In experiment B, the speed of both the particles is$0.794c$. The final mass of the particle after the collision is.$3.26m_o$

Step-by-step solution

Determine total kinetic energy before collision in experiment A

In this problem, two experiments of inelastic collision are performed. In experiment A, a particle moving at $0.9c$strikes a stationary particle and two particles stick together. Let us calculate the initial kinetic energy of the system.

$K{E}_{initial}=({\gamma }_{0.9c}-1)m_o{c}^2+({\gamma }_{0c}-1)m_o{c}^2$

Here, for particle 1, the Lorentz factor will be 2.29, and for particle 2, the Lorentz factor will be one as the velocity is zero. Putting these values in the above equation, we get the initial kinetic energy as follows:

.$K{E}_{initial}=1.29m_o{c}^2$

Determine total kinetic energy before collision in experiment B

In experiment B, two particles are coming at each other at velocity u.The total energy before the collision in this experiment is the same as the initial kinetic energy in experiment A. Let us calculate the velocity u from the above information.

$\begin{array}{c}K{E}_B=K{E}_A\\ ({\gamma }_u-1)m_o{c}^2+({\gamma }_u-1)m_o{c}^2=1.29m_o{c}^2\\ 2({\gamma }_u-1)=1.29\\ {\gamma }_u=1.645\\ \end{array}$

Using Lorentz formula to calculate speed,

$\begin{array}{rcl}\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$u^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}& =& 1.645\\ u& =& c\left(\sqrt{1-\frac{1}{{\left(1.645\right)}^2}}\right)\\ u& =& 0.794c\\ & & \\ & & \end{array}$

Apply energy and momentum conservation for both experiments

in both experiments, the total kinetic energy before the collision is equal.The total kinetic energy after the collision can be calculated using the final mass at the end of both the experiments. So,for experiment A,

momentum conserved:

.$\begin{array}{c}{\gamma }_{0.9c}{m}_o\left(0.9c\right)+0={\gamma }_f{m}_f{u}_f\\ {\gamma }_f{m}_f{u}_f=2.06m_{o}c\end{array}$

energy conserved: .

$\begin{array}{c}{\gamma }_{0.9c}{m}_o{c}^2+m_o{c}^2={\gamma }_f{m}_f{c}^2\\ {\gamma }_f{m}_f=3.29m_o\end{array}$

Dividing the both resulting conserved equations,

.$u_f=0.6353c$

And using this value for velocity, one can find the corresponding Lorentz factor as${\gamma }_f=1.29$. Then putting this value in energy conserved equation,

$m_f=2.55m_o$

Now, in experiment B, applying energy conservation,

.$\begin{array}{l}{\gamma }_{0.79c}{m}_o{c}^2+{\gamma }_{0.79c}{m}_o{c}^2={\gamma }_f{m}_f{c}^2\\ \end{array}$

Here, ${\gamma }_f=1$because the final velocity will be zero.

localid="1659091830810" $\begin{array}{c}{m}_f=2{\gamma }_{0.79c}{m}_o\\ =2\left(\frac{1}{\sqrt{1-{\left(0.79\right)}^2}}\right)m_o\\ =3.26m_o\end{array}$

As you can see in both experiments, mass increases after the collision. This implies that the total kinetic energy of the system in both experiments decreases, increasing mass. In experiment B, the change in mass is more than that in experiment A. Therefore, more initial kinetic energy will be converted to mass in experiment B.