Question

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.)

Short answer

(a) ${\mathrm{M}}_{\mathrm{AD}}={\mathrm{M}}_{\mathrm{AB}}\times {\mathrm{M}}_{\mathrm{BC}}\times {\mathrm{M}}_{\mathrm{CD}}$is proved.

(b) The velocity of the particle A is$1.4\times {10}^8\text{\hspace{0.17em}m/s}$ .

Step-by-step solution

Identification of given data

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${\mathrm{\beta }}_{\mathrm{CD}}=+0.60$ .

Significance of the reference frame

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.

Determination of the relation MAD=MABMBCMCD

(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.

Determination of the velocity of the particle A

(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 $1.4\times {10}^8\text{\hspace{0.17em}m/s}$.