Question

A $\mathbf{10}\mathbf{}\mathbf{kg}$object is moving to the right at $\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{c}$. It explodes into two pieces, one of mass ${\mathit{m}}_{\mathbf{1}}$moving left at $\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathit{c}$and one of mass ${\mathit{m}}_{\mathbf{2}}$moving right at $\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}\mathit{c}$

(a) Find ${\mathit{m}}_{\mathbf{1}}$ and ${\mathit{m}}_{\mathbf{2}}$.

(b) Find the change in kinetic energy in this explosion.

Short answer

1. The masses m₁ and m₂ is 1.43 kg and 6.43 kg respectively.
2. The change in kinetic energy is 1.93 x 10¹⁷J.

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 us consider that the $10\mathrm{kg}$object is traveling along the positive $\mathrm{x}$-axis and after the explosion, an object of mass $m_1$is moving along the negative x-axis at $0.6c$and $m_2$along the positive x-axis at $0.8c$

Use relativistic conservation of momentum as below:

${\gamma }_{0.6c}mu={\gamma }_{0.6c}{m}_1{u}_1+{\gamma }_{0.8c}{m}_2{u}_2$

$\left(\frac{5}{4}\right)(10\mathrm{kg})(0.6c)=\left(\frac{5}{4}\right)(-0.6c)m_1+\left(\frac{5}{3}\right)(0.8c)m_2$

$90=-9m_1+16m_2$ … (1)

Apply Energy Conservation

Relativistic energy conservation,

${\gamma }_{0.6c}m{c}^2={\gamma }_{0.6c}{m}_1{c}^2+{\gamma }_{0.8c}{m}_2{c}^2$

$\left(\frac{5}{4}\right)(10\mathrm{kg})=\left(\frac{5}{4}\right)m_1+\left(\frac{5}{3}\right)m_2$

$30=3m_1+4m_2$… (2)

Multiplying equation (2) by 3 and add it with equation (1) as given by,

localid="1659091030679" $\begin{array}{rcl}90+90& =& 3\left(3m_1+4m_2\right)+\left(-9m_1+16m_2\right)\\ 180& =& 9m_1+12m_2+-9m_1+16m_2\\ 180& =& 28m_2\\ m_2& =& \text{6.43 kg}\end{array}$

Substitute 6.43 kg for m₂ into equation (2).

localid="1659091034690" $\begin{array}{rcl}30& =& 3m_1+4\left(\text{6.43 kg}\right)\\ 3m_1& =& 30-\text{25.72}\\ {\text{m}}_1& =& \frac{4.28}{3}\\ {\text{m}}_1& =& \text{1.426}kg\end{array}$

Hence, the massesm₁ and m₂ is 1.43 kg and 6.43 kg respectively.

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

Notice the decrease in total mass which indicates the conversion of internal energy into kinetic energy. That is

$\begin{array}{rcl}\Delta KE& =& -\Delta m{c}^2\\ & =& -\left[\left(m_1+m_2\right)-m\right]c^2\\ & =& -\left[\left(1.43+6.43\right)kg-10kg\right]{\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\\ & =& -\left(7.86kg-10kg\right)\times 9\times {10}^{16}{\left(\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\end{array}$

$\begin{array}{rcl}\Delta KE& =& 2.14kg\times 9\times {10}^{16}{\left(\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\\ & =& 1.93\times {10}^{17}J\\ & & \\ & & \end{array}$

Hence, 2.14 kg mass was converted to $1.93\times {10}^{17}$kinetic energy, which registers as an increase in velocity of both objects.