Question

As an illustration of the principle of relativity in classical mechanics, consider the following generic collision: In inertial frame S, particle A (mass${\mathit{m}}_{\mathbf{A}}$, velocity${\mathit{u}}_{\mathbf{B}}$ ) hits particle B (mass${\mathit{m}}_{\mathbf{B}}$, velocity ${\mathbf{u}}_{\mathbf{B}}$). In the course of the collision some mass rubs off A and onto B, and we are left with particles C (mass${\mathit{m}}_{\mathbf{c}}$, velocity${\mathbf{u}}_{\mathbf{c}}$ ) and D (mass ${\mathbf{m}}_{\mathbf{D}}$, velocity${\mathbf{u}}_{\mathbf{D}}$ ). Assume that momentum $\left(p=\mathrm{mu}\right)$is conserved in S.

(a) Prove that momentum is also conserved in inertial frame$\overline{\mathbf{s}}$, which moves with velocity relative to S. [Use Galileo’s velocity addition rule—this is an entirely classical calculation. What must you assume about mass?]

(b) Suppose the collision is elastic in S; show that it is also elastic in $\overline{\mathbf{S}}$.

Short answer

(a) The momentum is conserved in an inertial frame $\overline{S}$.

(b) The collision is also elastic in the frame $\overline{S}$.

Step-by-step solution

Expression for the conservation of momentum:

Write the expression for the total conserved momentum in any collision in frame S.

${\mathit{m}}_{\mathbf{A}}{\mathit{u}}_{\mathbf{A}}\mathbf{+}{\mathit{m}}_{\mathbf{B}}{\mathit{u}}_{\mathbf{B}}\mathbf{=}{\mathit{m}}_{\mathbf{C}}{\mathit{u}}_{\mathbf{C}}\mathbf{+}{\mathit{m}}_{\mathbf{D}}{\mathit{u}}_{\mathbf{D}}$ …… (1)

Here, is the mass of particle A, is the mass of particle B, is the mass of particle C, is the mass of a particle D,${\mathit{u}}_{\mathbf{A}}$ is the velocity of particle A,${\mathit{u}}_{\mathbf{B}}$ is the velocity of particle B, ${\mathit{u}}_{\mathbf{c}}$is the velocity of particle C and${\mathit{u}}_{\mathbf{D}}$ is the velocity of particle D.

Prove that momentum is also conserved in an inertial frame S :

(a)

Consider a frame $\overline{S}$moving with the constant velocity v with respect to S.

Based on Galileo’s addition rule,

$u=\overline{u}+v$

Hence, the velocity for all the particles will be,

$$\begin{gathered} {\mathit{u}}_A=\overline{u_A}+v \\ {\mathit{u}}_B=\overline{u_B}+v \\ {\mathit{u}}_C=\overline{u_C}+v \\ {\mathit{u}}_D=\overline{u_D}+v \end{gathered}$$

Substitute the value of ${\mathbf{u}}_{\mathbf{A}}\mathbf{,}{\mathbf{u}}_{\mathbf{B}}\mathbf{,}{\mathbf{u}}_{\mathbf{C}}\mathrm{and}{\mathbf{u}}_{\mathbf{D}}$in equation (1).

$$\begin{gathered} m_A\left(\overline{u_A}+v\right)+m_B\left({\overline{u}}_B+v\right)=m_C\left({\overline{u}}_C+v\right)+m_D\left({\overline{u}}_D+v\right) \\ {m}_A{\overline{u}}_A+m_B{\overline{u}}_B+\left(m_A+m_B\right)v=m_c{\overline{u}}_c+m_D{\overline{u}}_D+\left(m_C+m_D\right)v \end{gathered}$$

Assuming mass is conserved, i.e.,

$\left(m_A+m_B\right)=\left(m_C+m_D\right)$

It follows that $m_A{\overline{u}}_A+m_B{\overline{u}}_B+\left(m_A+m_B\right)v=m_C{\overline{u}}_C+m_D{\overline{u}}_D+\left(m_C+m_D\right)v$so, the momentum is conserved in an inertial frame $\overline{S}$.

Therefore, the momentum is conserved in an inertial frame $\overline{S}$.

Show that the collision is elastic in S:

(b)

As the collision is elastic, the kinetic energy is conserved in frame S. Hence,

$\frac{1}{2}{m}_A{\mathit{u}}_{\mathbf{A}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{m}_{\mathbf{B}}{\mathit{u}}_B^2=\frac{1}{2}{m}_C{\mathit{u}}_C^2+\frac{1}{2}{m}_D{\mathit{u}}_D^2$

Substitute the value of ${\mathbf{u}}_{\mathbf{A}}\mathbf{,}\mathbf{}{\mathbf{u}}_{\mathbf{B}}\mathbf{,}\mathbf{}{\mathbf{u}}_{\mathbf{C}}\mathrm{and}{\mathbf{u}}_{\mathbf{D}}$in the above expression.

$$\begin{gathered} \frac{1}{2}{m}_A{\left({\overline{u}}_A+v\right)}^2+\frac{1}{2}{m}_B{\left({\overline{u}}_B+v\right)}^2+\frac{1}{2}{m}_C{\left({\overline{u}}_C+v\right)}^2+\frac{1}{2}{m}_D{\left({\overline{u}}_D+v\right)}^2 \\ \left[\begin{array}{c}\frac{1}{2}{m}_A\left(\overline{u_A^2}+v^2+2{\overline{u}}_{A}v\right)+\\ \frac{1}{2}{m}_B\left(\overline{u_A^2}+v^2+2{\overline{u}}_{B}A\right)\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}{m}_C\left(\overline{u_C^2}+v^2+2{\overline{u}}_{C}v\right)+\\ \frac{1}{2}{m}_D\left(\overline{u_D^2}+v^2+2{\overline{u}}_{D}A\right)\end{array}\right] \\ \left[\begin{array}{c}\frac{1}{2}{m}_A{\overline{u}}_A^2=\frac{1}{2}{m}_B{\overline{u}}_B^2+\frac{1}{2}\left(m_A+m_B\right)v^2+\\ \left(m_A{\overline{u}}_A+m_B{\overline{u}}_B\right)v\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}{m}_C{\overline{u}}_C^2=\frac{1}{2}{m}_D{\overline{u}}_D^2+\frac{1}{2}\left(m_C+m_D\right)v^2+\\ \left(m_C{\overline{u}}_C+m_D{\overline{u}}_D\right)v\end{array}\right] \end{gathered}$$

As , the above equation becomes,

$\frac{1}{2}{m}_A{\overline{u}}_A^2=\frac{1}{2}{m}_B{\overline{u}}_B^2+\frac{1}{2}\left(m_C+m_C\right)v^2+\frac{1}{2}\left(m_D+m_D\right)$

Hence, $\left(m_A+m_B\right)=\left(m_C+m_D\right)$the kinetic energy is conserved in the frame $\overline{S}$.

Therefore, the collision is also elastic in the frame$\overline{S}$.