Question

In Fig. 37-28a, particle P is to move parallel to the$\mathbf{x}$ and$\mathbf{x}\mathbf{\text{'}}$ axes of reference frames$\mathbf{S}$and $\mathbf{S}\mathbf{\text{'}}$, at a certain velocity relative to frame $\mathbf{S}$. Frame$\mathbf{S}\mathbf{\text{'}}$ is to move parallel to the$\mathbf{x}$ axis of frame $\mathbf{S}$at velocity $\mathbf{v}$. Figure 37-28b gives the velocity$\mathbf{u}\mathbf{\text{'}}$ of the particle relative to frame$\mathbf{S}\mathbf{\text{'}}$ for a range of values for$\mathbf{v}$ . The vertical axis scale is set by${{\mathbf{u}}^{\mathbf{\text{'}}}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{0.800}\mathbf{c}$ . What value will $\mathbf{u}\mathbf{\text{'}}$have if (a) $\mathbf{v}\mathbf{=}\mathbf{0.80}\mathbf{c}$and (b)$\mathbf{v}\mathbf{\to }\mathbf{c}$ ?

Short answer

(a) The value of $\mathrm{u}\text{'}$is $-2.5\times {10}^8\text{\hspace{0.17em}m/s}$.

(b) The value of $\mathrm{u}\text{'}$is$-3\times {10}^8\text{\hspace{0.17em}m/s}$ .

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The velocity of the frame$\mathrm{S}$ is$\mathrm{v}$ .
• The velocity of the frame ${\mathrm{S}}^{\text{'}}$is ${\mathrm{u}}^{\text{'}}$.
• The scale in the vertical axis is ${{\mathrm{u}}^{\text{'}}}_{\mathrm{a}}=-0.800\mathrm{c}$.

Significance of the reference frame

The reference frame is described as the position where the geometric points are being identified as physically and mathematically. This frame is set by some set of the reference points.

Determination of the value of u' if v=0.80c

(a)

According to the graph, when$\mathrm{v}=0$, then $\mathrm{u}=-0.20\mathrm{c}$.

The equation of the velocity of the frame${\mathrm{S}}^{\text{'}}$is expressed as:

${\mathrm{u}}^{\text{'}}=\frac{\mathrm{u}-\mathrm{v}}{1-\mathrm{uv}/{\mathrm{c}}^2}$ …(i)

Here, $\mathrm{u}\text{'}$is the velocity of the frame ${\mathrm{S}}^{\text{'}}$, $\mathrm{u}$is the initial and $\mathrm{v}$is the final velocity of the frame. $\mathrm{c}$is the speed of the light.

Substitute $-0.20\mathrm{c}$for role="math" localid="1663049487197" $\mathrm{u}$, $3\times {10}^8\text{\hspace{0.17em}m/s}$for $\mathrm{c}$and$0.80\mathrm{c}$ for $\mathrm{v}$in the above equation.

$\begin{array}{c}{\mathrm{u}}^{\text{'}}=\frac{(3\times {10}^8\text{\hspace{0.17em}m/s})(-0.20-0.80)}{1+0.20\times 0.80}\\ =\frac{-3\times {10}^8\text{\hspace{0.17em}m/s}}{1+0.16}\\ =\frac{-3\times {10}^8\text{\hspace{0.17em}m/s}}{1.16}\\ =-2.5\times {10}^8\text{\hspace{0.17em}m/s}\end{array}$

Thus, the value of${\mathrm{u}}^{\text{'}}$ is $-2.5\times {10}^8\text{\hspace{0.17em}m/s}$.

Determination of the value of u' if  v→c

(b)

The equation (i) has been recalled below:

${\mathrm{u}}^{\text{'}}=\frac{\mathrm{u}-\mathrm{v}}{1-\mathrm{uv}/{\mathrm{c}}^2}$

Substitute $-0.20\mathrm{c}$for $\mathrm{u}$, $3\times {10}^8\text{\hspace{0.17em}m/s}$for $\mathrm{c}$and $3\times {10}^8\text{\hspace{0.17em}m/s}$for $\mathrm{v}$in the above equation.

$\begin{array}{c}{\mathrm{u}}^{\text{'}}=\frac{(3\times {10}^8\text{\hspace{0.17em}m/s})(-0.20-1)}{1+0.20}\\ =\frac{-3.6\times {10}^8\text{\hspace{0.17em}m/s}}{1.20}\\ =\frac{-3.6\times {10}^8\text{\hspace{0.17em}m/s}}{1.20}\\ =-3\times {10}^8\text{\hspace{0.17em}m/s}\end{array}$

Thus, the value of ${\mathrm{u}}^{\text{'}}$is $-3\times {10}^8\text{\hspace{0.17em}m/s}$.