Question

$\mathbf{200}\mathbf{\text{\hspace{0.17em}km}}$On July 10, 1996, a granite block broke away from a wall in Yosemite Valley and, as it began to slide down the wall, was launched into projectile motion. Seismic waves produced by its impact with the ground triggered seismographs as far away as$\mathbf{200}\mathbf{\text{\hspace{0.17em}km}}$. Later measurements indicated that the block had a mass between$\mathbf{7}\mathbf{.3}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\text{\hspace{0.17em}kg}}$and$\mathbf{1}\mathbf{.7}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{8}}\mathbf{\text{\hspace{0.17em}kg}}$and that it landedvertically below the launch point and$\mathbf{30}\mathbf{\text{\hspace{0.17em}m}}$horizontally from it.

(The launch angle is not known.)

(a) Estimate the block’s kinetic energy just before it landed.

Consider two types of seismic waves that spread from the impact point—a hemispherical body wave traveled through the ground in an expanding hemisphere and a cylindrical surface wave traveled along the ground in an expanding shallow vertical cylinder (Fig. 17-49). Assume that the impact lasted, the vertical cylinder had a depth d of$\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$, and each wave type received 20% of the energy the block had just before impact. Neglecting any mechanical energy loss the waves experienced as they traveled, determine the intensities of (b) the body wave and

(c) the surface wave when they reach the seismograph$\mathbf{200}\mathbf{\text{\hspace{0.17em}km}}$away.

(d) On the basis of these results, Which wave is more easily detected on a distant seismograph?

Short answer

a. The block’s kinetic energy just before it comes to land will be$3.6\times {10}^{11}\text{\hspace{0.17em}J}<KE<8.3\times {10}^{11}\text{\hspace{0.17em}J}$.

b. Intensity of the body wave will be between $0.58\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$and $1.33\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

c. Intensity of the surface wave when they reach the seismograph 200 km away will be Between$23\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}to53\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$

d. The surface wave will be detected on the distant seismograph$r=200\text{\hspace{0.17em}km}$

Step-by-step solution

The given data

1. Mass of the block varies between $7.9\times {10}^7\text{\hspace{0.17em}kg}<m<1.7\times {10}^8\text{\hspace{0.17em}kg}$.
2. Height from which the block is dropped,$h=500\text{\hspace{0.17em}m}$.
3. Impact lasted for time,$\Delta t=0.5\text{\hspace{0.17em}s}$.
4. Depth of vertical cylinder,$d=5.0\text{\hspace{0.17em}m}$.
5. Distance of grounded triggered seismographs,$r=200\text{\hspace{0.17em}km}$ .

Understanding the concept of energy and intensity

We know the height from which the block is falling, and its mass is also known. Now as the block is under free fall, the potential energy will be converted into kinetic energy when the block falls to the ground. From this, we can find the kinetic energy of the block just before it landed.

We are given enough information to find the area of the body wave, and we know the power required to develop these waves. Using the formula for the intensity of the wave, we can find the intensity of the body wave.

A similar procedure has to be used to find the intensity of the surface waves.

Formula:

The intensity of a wave,

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$ …(i)

The power exerted by the body,

$\mathit{P}\mathbf{=}\frac{\mathbf{Kinetic}\mathbf{}\mathbf{Energy}}{\mathbf{\Delta t}}$ …(ii)

The potential energy of the body,

…(iii)

$\mathit{P}\mathit{E}\mathbf{=}\mathit{m}\mathit{g}\mathit{h}$

a) Calculation of block’s kinetic energy

As the block is falling from a certain height and no other forces are acting on it, according to conservation of energy, potential energy at that height will be converted into kinetic energy just before it comes to rest.

$PE=KE$

Hence, the kinetic energy of the block using equation (iii) is given as:

$KE=mgh$

We take,$m=7.3\times {10}^7\text{\hspace{0.17em}kg}$

$\begin{array}{l}KE=(7.3\times {10}^7\text{\hspace{0.17em}kg})\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 500\text{\hspace{0.17em}m}\\ =3.6\times {10}^{11}\text{\hspace{0.17em}J}\end{array}$

We take,$m=1.7\times {10}^8\text{\hspace{0.17em}kg}$

$\begin{array}{rcl}KE& =& (1.7\times {10}^8\text{\hspace{0.17em}kg})\times 9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 500\text{\hspace{0.17em}m}\\ & =& 8.3\times {10}^{11}\text{\hspace{0.17em}J}\end{array}$

So, the range of the kinetic energy will be from$3.6\times {10}^{11}\text{\hspace{0.17em}J}$to.$8.3\times {10}^{11}\text{\hspace{0.17em}J}$

b) Calculation of intensity of the block wave

Now$20\%$of the kinetic energy is converted into the seismic wave.

When,$KE=3.6\times {10}^{11}\text{\hspace{0.17em}J}$

$\begin{array}{c}K{E}_{wave}=\frac{20}{100}\times (3.6\times {10}^{11}\text{\hspace{0.17em}J})\\ =0.72\times {10}^{11}\text{\hspace{0.17em}J}\end{array}$

The power exerted by the body using equation (ii) is given as:

$\begin{array}{c}P=\frac{0.72\times {10}^{11}\text{\hspace{0.17em}J}}{0.5\text{\hspace{0.17em}s}}\\ =1.44\times {10}^{11}\text{\hspace{0.17em}W}\end{array}$

The area of the surface of the body is given as:

$\begin{array}{l}{A}_{body}=\frac{1}{2}\times 4\pi r^2\\ =2\pi \times {(200\times {10}^3\text{\hspace{0.17em}m})}^2\\ A_{body}=2.5\times {10}^{11}{\text{\hspace{0.17em}m}}^{\text{2}}\end{array}$

Intensity of the body wave using equation (i) can be given as:

$\begin{array}{c}{I}_{body}=\frac{1.44\times {10}^{11}\text{\hspace{0.17em}W}}{2.5\times {10}^{11}{\text{\hspace{0.17em}m}}^{\text{2}}}\\ I_{body}=0.58\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}$

When,

$\begin{array}{l}K{E}_{wave}=\frac{20}{100}\times (8.3\times {10}^{11}\text{\hspace{0.17em}J})\\ K{E}_{wave}=1.66\times {10}^{10}\text{\hspace{0.17em}J}\end{array}$

The power exerted by the body using equation (ii) is given as:

$\begin{array}{c}P=\frac{1.66\times {10}^{11}\text{\hspace{0.17em}J}}{0.5\text{\hspace{0.17em}s}}\\ P=3.32\times {10}^{11}\text{\hspace{0.17em}W}\end{array}$

The area of the block is given as:

$A_{body}=2.5\times {10}^{11}{\text{\hspace{0.17em}m}}^{\text{2}}$

Intensity of the body wave using equation (i) can be given as:

$\begin{array}{l}{I}_{body}=\frac{3.32\times {10}^{11}\text{\hspace{0.17em}W}}{2.5\times {10}^{11}{\text{\hspace{0.17em}m}}^{\text{2}}}\\ I_{body}=1.33\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}$

So, the intensity of the body wave ranges from$0.58\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$ to.$1.33\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$

c) Calculation of intensity of the surface wave

Now$20\%$of the kinetic energy is converted into the seismic wave.

When$KE=3.6\times {10}^{11}\text{\hspace{0.17em}J}$

$\begin{array}{l}K{E}_{wave}=\frac{20}{100}\times (3.6\times {10}^{11}\text{\hspace{0.17em}J})\\ K{E}_{wave}=0.72\times {10}^{11}\text{\hspace{0.17em}J}\end{array}$

The power radiated by the body wave using equation (ii) is given as:

$\begin{array}{c}P=\frac{0.72\times {10}^{11}\text{\hspace{0.17em}J}}{0.5\text{\hspace{0.17em}s}}\\ =1.44\times {10}^{11}\text{\hspace{0.17em}W}\end{array}$

The area of the seismic surface is given as:

$\begin{array}{l}{A}_{surface}=2\pi rd\\ =2\pi \times (200\times {10}^3\text{\hspace{0.17em}m})\times 5.0\text{\hspace{0.17em}m}\\ A_{surface}=6.28\times {10}^6{\text{\hspace{0.17em}m}}^{\text{2}}\end{array}$

Intensity of the surface wave using equation (i) can be given as:

$\begin{array}{l}{I}_{surface}=\frac{1.44\times {10}^{11}\text{\hspace{0.17em}W}}{6.28\times {10}^6{m}^2}\\ =22.92\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ I_{surface}\approx 23\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}$

When $KE=8.3\times {10}^{11}\text{\hspace{0.17em}J}$

$\begin{array}{l}K{E}_{wave}=\frac{20}{100}\times (8.3\times {10}^{11}\text{\hspace{0.17em}J})\\ K{E}_{wave}=1.66\times {10}^{10}\text{\hspace{0.17em}J}\end{array}$

The power radiated by the body wave using equation (ii) is given as:

$\begin{array}{c}P=\frac{1.66\times {10}^{11}\text{\hspace{0.17em}J}}{0.5\text{\hspace{0.17em}s}}\\ =3.32\times {10}^{11}\text{\hspace{0.17em}W}\end{array}$

The area of the seismic surface is given as:

$A_{surface}=6.28\times {10}^6{\text{\hspace{0.17em}m}}^{\text{2}}$

Intensity of the surface wave using equation (i) can be given as:

$\begin{array}{l}{I}_{surface}=\frac{3.32\times {10}^{11}\text{\hspace{0.17em}J}}{6.28\times {10}^6{m}^2}\\ =52.86\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ I_{surface}\approx 53\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}$

So, the intensity of the surface wave ranges from $23\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}to53\times {10}^3\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

d) Identifying the wave on the seismograph

Several factors are involved in determining which seismic waves are being detected by thedistant seismograph. In this situation, the waves which have higher intensity will be detected by the distant seismograph, and in this case, they are the surface waves.