Question

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?

Short answer

(a) The value of ${\mathrm{M}}_{\mathrm{AC}}$is$\frac{1}{9}$ .

(b) The value of ${\mathrm{\beta }}_{\mathrm{AC}}$is $0.79$.

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

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The 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${\mathrm{v}}_2=+0.500\mathrm{c}$ .

Significance of the motion of a particle

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.

Determination of the value of MAC

(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 $\frac{1}{9}$.

Determination of the value of  βAC

(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 $0.79$.

Determination of the velocity of the particle

(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 $2.3\times {10}^8\text{\hspace{0.17em}m/s}$.