A uranium atom traveling at speed$\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$collides elastically with a stationary hydrogen molecule, head-on. What is the approximate final speed of the hydrogen molecule?

Mass of the car:$m_1=1000\text{kg}$

Mass of the truck:$m_2=3000\text{kg}$

Initial velocity of the car:$v_1=30\text{m}/\text{s}$

Initial velocity of the truck:$v_2=-20\text{m}/\text{s}$

If the total system kinetic energy before the collision equals the energy after the collision, then it is an elastic collision and If the total kinetic energy is not conserved then it is referred to as an inelastic collision.

All vectors in this system are 1-dimensional, in the x direction.

The velocity of the centre of momentum system of two particles:

$$\begin{gathered} v_{CM}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2} \\ {v}_{CM}=\frac{(1000\times 30)+(3000\times -20)}{4000} \\ {v}_{CM}=-7.5\text{m}/\text{s} \end{gathered}$$

We find the relative velocities of the car and the truck by subtracting the centre-of-momentum velocity from the original velocities, so the relative velocity of the car:

$v_{1,rel}=v_1-v_{CM}=30-(-7.5)=37.5\text{m}/\text{s}$

And the relative velocity of the truck:

$v_{2,rel}=v_2-v_{CM}=-20-(-7.5)=-12.5\text{m}/\text{s}$

So, the total initial kinetic energy in the CM frame:

$\begin{array}{rcl}K{E}_{rel,i}& =& K{E}_{{\text{rel}}_l,1}+K{E}_{\text{rel}/2}\\ & =& \frac{1}{2}{m}_1{v}_{\text{rel},1}^2+\frac{1}{2}{m}_2{v}_{rel,2}^2\end{array}$

$$\begin{gathered} K{E}_{\text{rel,i}}=\left(\frac{1}{2}\times 1000\times {(37.5)}^2\right)+\left(\frac{1}{2}\times 3000\times {(-12.5)}^2\right) \\ K{E}_{\text{rel,i}}=9.4\times {10}^5\text{J} \end{gathered}$$

The vehicles stick together, so in the CM frame $K{E}_{rel,f}=0$and the change in kinetic energy:

$\begin{array}{rcl}\Delta KE& =& K{E}_{\text{rel},f}-K{E}_{\text{rel},i}\\ & =& -K{E}_{\text{rcl},i}\end{array}$

$\Delta KE=-9.4\times {10}^5\text{J}$