Continuation of Problem 65. Let reference frame C in Fig. 37-31 move past reference frame D (not shown). (a) Show that${\mathbf{M}}_{\mathbf{AD}}\mathbf{=}{\mathbf{M}}_{\mathbf{AB}}{\mathbf{M}}_{\mathbf{BC}}{\mathbf{M}}_{\mathbf{CD}}$

(b) Now put this general result to work: Three particles move parallel to a single axis on which an observer is stationed. Let plus and minus signs indicate the directions of motion along that axis. Particle A moves past particle B at${\mathbf{\beta }}_{\mathbf{AB}}\mathbf{=}\mathbf{+}\mathbf{0}\mathbf{.20}$ . Particle B moves past particle C at ${\mathbf{\beta }}_{\mathbf{BC}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.40}$. Particle C moves past observer D at${\mathbf{\beta }}_{\mathbf{CD}}\mathbf{=}\mathbf{+}\mathbf{0}\mathbf{.60}$ . What is the velocity of particle A relative to observer D? (The solution technique here is much faster than using Eq. 37-29.)

The given data can be listed below as:

• The particle A moves past the particle B is ${\mathrm{\beta }}_{\mathrm{AB}}=+0.20$.
• The particle B moves past the particle C is${\mathrm{\beta }}_{\mathrm{BC}}=-0.40$ .
• The particle C moves past the particle D is${\mathrm{\beta }}_{\mathrm{CD}}=+0.60$ .

The reference frame is referred to as the geometric points in which the position of an object is physically or mathematically identified. The reference frame is used to identify a body’s position relative to an object.

(a)

The value of${\mathrm{M}}_{\mathrm{AD}}$is$0.34$according to the question.

The equation of the value of ${\mathrm{M}}_{\mathrm{AB}}$is expressed as:

role="math" localid="1663060081607" ${\mathrm{M}}_{\mathrm{AB}}=\frac{1-{\mathrm{\beta }}_{\mathrm{AB}}}{1+{\mathrm{\beta }}_{\mathrm{AB}}}$

Substitute$+0.20$for${\mathrm{\beta }}_{\mathrm{AB}}$in the above equation.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{AB}}=\frac{1-0.20}{1+0.20}\\ =\frac{0.8}{1.2}\\ =0.6\end{array}$

The equation of the value of ${\mathrm{M}}_{\mathrm{BC}}$is expressed as:

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

Substitute$-0.40$for${\mathrm{\beta }}_{\mathrm{AB}}$in the above equation.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{BC}}=\frac{1+0.40}{1-0.40}\\ =\frac{1.4}{0.6}\\ =2.3\end{array}$

The equation of the value of ${\mathrm{M}}_{\mathrm{CD}}$is expressed as:

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

Substitute$+0.60$for${\mathrm{\beta }}_{\mathrm{CD}}$in the above equation.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{CD}}=\frac{1-0.60}{1+0.60}\\ =\frac{0.4}{1.6}\\ =0.25\end{array}$

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

${\mathrm{M}}_{\mathrm{AD}}={\mathrm{M}}_{\mathrm{AB}}\times {\mathrm{M}}_{\mathrm{BC}}\times {\mathrm{M}}_{\mathrm{CD}}$

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{AD}}=0.6\times 2.3\times 0.25\\ =0.345\end{array}$

Thus,${\mathrm{M}}_{\mathrm{AD}}={\mathrm{M}}_{\mathrm{AB}}\times {\mathrm{M}}_{\mathrm{BC}}\times {\mathrm{M}}_{\mathrm{CD}}$ is proved.

(b)

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

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

Substitute $0.345$for ${\mathrm{M}}_{\mathrm{AD}}$in the above equation.

$\begin{array}{c}{\mathrm{\beta }}_{\mathrm{AD}}=\frac{1-0.345}{1+0.345}\\ =\frac{0.655}{1.345}\\ =0.48\end{array}$

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

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

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

Substitute$0.48$ for${\mathrm{\beta }}_{\mathrm{AC}}$ and$3\times {10}^8\text{\hspace{0.17em}m/s}$ for$\mathrm{c}$ in the above equation.

$\begin{array}{c}\mathrm{v}=0.48\times 3\times {10}^8\text{\hspace{0.17em}m/s}\\ =1.4\times {10}^8\text{\hspace{0.17em}m/s}\end{array}$

Thus, the velocity of the particle A is $1.4\times {10}^8\text{\hspace{0.17em}m/s}$.