Continuation of Problem 65. Use the result of part (b) in Problem 65 for the motion along a single axis in the following situation. Frame A in Fig. 37-31 is attached to a particle that moves with velocity $\mathbf{+}\mathbf{0}\mathbf{.500}\mathbf{c}$ past frame B, which moves past frame C with a velocity of $\mathbf{+}\mathbf{0}\mathbf{.500}\mathbf{c}$. What are (a) ${\mathbf{M}}_{\mathbf{AC}}$, (b) ${\mathbf{\beta }}_{\mathbf{AC}}$, and (c) the velocity of the particle relative to frame C?

The given data can be listed below as:

• The velocity of the particle while passing frame A is ${\mathrm{v}}_1=+0.500\mathrm{c}$.
• The velocity of the particle while passing frame B is${\mathrm{v}}_2=+0.500\mathrm{c}$ .

The motion of a particle is mainly the superposition of the components of a particle because of fluid drag. The motion of the particle is also called as the Brownian motion.

(a)

Due to pretty symmetry and also because of the part computation, the value of ${\mathrm{M}}_{\mathrm{AB}}$ is expressed as:

${\mathrm{M}}_{\mathrm{AB}}=\frac{1-{\mathrm{\beta }}_{\mathrm{AB}}}{1+{\mathrm{\beta }}_{\mathrm{AB}}}$

The value of role="math" localid="1663057884603" ${\mathrm{\beta }}_{\mathrm{AB}}$is given $0.5$in the problem statement.

Substitute$0.5$for${\mathrm{\beta }}_{\mathrm{AB}}$in the above equation.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{AB}}=\frac{1-0.5}{1+0.5}\\ =\frac{0.5}{1.5}\\ =\frac{1}{3}\end{array}$

Similarly, the value of ${\mathrm{M}}_{\mathrm{BC}}$will be$\frac{1}{3}$as the values are same for it.

The equation of ${\mathrm{M}}_{\mathrm{AC}}$is expressed as:

${\mathrm{M}}_{\mathrm{AC}}={\mathrm{M}}_{\mathrm{AB}}\times {\mathrm{M}}_{\mathrm{BC}}$

Substitute$\frac{1}{3}$for${\mathrm{M}}_{\mathrm{BC}}$and${\mathrm{M}}_{\mathrm{AB}}$in the above equation.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{AC}}=\frac{1}{3}\times \frac{1}{3}\\ =\frac{1}{9}\end{array}$

Thus, the value of ${\mathrm{M}}_{\mathrm{AC}}$is $\frac{1}{9}$.

(b)

The equation of the value of ${\mathrm{\beta }}_{\mathrm{AC}}$is expressed as:

${\mathrm{\beta }}_{\mathrm{AC}}=\frac{1-{\mathrm{M}}_{\mathrm{AC}}}{1+{\mathrm{M}}_{\mathrm{AC}}}$

Substitute$\frac{1}{9}$ for${\mathrm{M}}_{\mathrm{AC}}$ in the above equation.

$\begin{array}{c}{\mathrm{\beta }}_{\mathrm{AC}}=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\\ =\frac{0.88}{1.11}\\ =0.79\end{array}$

Thus, the value of ${\mathrm{\beta }}_{\mathrm{AC}}$is $0.79$.

(c)

The equation of the velocity of the particle is expressed as:

$\mathrm{v}={\mathrm{\beta }}_{\mathrm{AC}}\times \mathrm{c}$

Here, $\mathrm{v}$is the velocity of the particle and $c$is the speed of light.

Substitute $0.79$for ${\mathrm{\beta }}_{\mathrm{AC}}$and $3\times {10}^8\text{\hspace{0.17em}m/s}$for $\mathrm{c}$in the above equation.

$\begin{array}{c}\mathrm{v}=0.79\times 3\times {10}^8\text{\hspace{0.17em}m/s}\\ =2.3\times {10}^8\text{\hspace{0.17em}m/s}\end{array}$

Thus, the velocity of the particle relative to frame C is $2.3\times {10}^8\text{\hspace{0.17em}m/s}$.