Chapter 17, Problem 17.57AP

Review. A 150-g glider moves at v₁ = 2.30 m/s on an air track toward an originally stationary 200-g glider as shown in Figure P16.53. The gliders undergo a completely inelastic collision and latch together over a time interval of 7.00 ms. A student suggests roughly half the decrease in mechanical energy of the two-glider system is transferred to the environment by sound. Is this suggestion reasonable? To evaluate the idea, find the implied sound level at a position 0.800 m from the gliders. If the student’s idea is unreasonable, suggest a better idea.

Figure P16 53

To determine

The idea and find the implied sound level at the position $0.800\text{m}$ from the gliders. If the student’s idea is unreasonable then suggest better idea.

Answer to Problem 17.57AP

It is unreasonable, implying a sound level $123\text{dB}$. Almost all of the mechanical energy is transformed into internal energy.

Explanation of Solution

The mass of first glider is $150.0\text{g}$, the mass of the second glider is $200.0\text{g}$, the latched time interval is $7.00\text{ms}$, the position of the sound level from the glider is $0.800\text{m}$ and the speed of the first glider system is $2.30\text{m}/\text{s}$.

The glider stick together and moved with final speed momentum conservation for the two sliders system.

Formula to calculate the momentum of the two glider systems before collision is,

    $P_1=m_1{v}_1+m_2{v}_2$

Here, $P_1$ is the momentum of the two glider systems before collision, $m_1$ is the mass of the first glider, $m_2$ is the mass of the second glider, $v_1$ is the speed of the first glider and $v_2$ is the second glider.

The velocity of the second slider systemis zero after collision so, formula to calculte the momentum of the two glider systems after collision is,

    $P_2=m_1{v}_1$

Here, $P_2$ is the momentum of the two glider systems after collision.

From momentum conservation system, the momentum of the two sliders system before collision and after collision is equal.

    $P_1=P_2$        (1)

Substitute $m_1{v}_1+m_2{v}_2$ for $P_1$ and $m_1{v}_1$ for $P_2$ in equation (1).

    $m_1{v}_1+m_2{v}_2=m_1{v}_1$        (2)

Since, the velocities of the two sliders systems are equal before collision. Hence, the equation (2) can be written as,

    $\begin{array}{c}{m}_1{v}_1+m_2{v}_2=m_1{v}_1\\ \left(m_1+m_2\right)v=m_1{v}_1\\ v=\frac{m_1{v}_1}{\left(m_1+m_2\right)}\end{array}$        (3)

Substitute $150.0\text{g}$ for $m_1$, $200.0\text{g}$ for $m_2$ and $2.30\text{m}/\text{s}$ for $v_1$ in equation (3) to find the $v$.

    $\begin{array}{c}v=\frac{\left(150.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{kg}}\right)\right)2.30\text{m}/\text{s}}{\left(150.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{kg}}\right)+150.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{kg}}\right)\right)}\\ =0.986\text{m}/\text{s}\end{array}$

Formula to calculate the missing kinetic energy is,

    $E=\frac{1}{2}{m}_1{v}_1{}^2-\frac{1}{2}\left(m_1+m_2\right)v^2$        (4)

Substitute $150.0\text{g}$ for $m_1$, $200.0\text{g}$ for $m_2$ and $0.986\text{m}/\text{s}$ for $v$ in equation (4) to find $E$.

    $\begin{array}{c}E=\frac{1}{2}150.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{kg}}\right)v_1{}^2-\frac{1}{2}\left(150.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{kg}}\right)+200.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{kg}}\right)\right){\left(0.986\text{m}/\text{s}\right)}^2\\ =0.277\text{J}\end{array}$

Imagine that one half of $0.277\text{J}$ goes into the internal energy and half into sound radiated isotropically in $7.00\text{ms}$ and its intensity is $0.800\text{m}$ away.

Formula to calculate the intensity is,

    $I=\frac{E}{At}$        (5)

Here, $E$ is the missing kinetic energy and $t$ is the latched time interval.

Formula to calculate the area of the area of the cross section is,

    $A=4\pi r^2$

Here, $r$ is the position of the sound level from the glider and $A$ is the cross section area of the glider.

Substitute $4\pi r^2$  for $A$ in equation (5).

    $I=\frac{E}{4\pi r^{2}t}$        (6)

Substitute $7.00\text{ms}$ for $t$, $0.277\text{J}$ for $E$ and $0.800\text{m}$ for $r$ in equation (6) to find the $I$.

    $\begin{array}{c}I=\frac{0.277\text{J}}{4\pi {\left(0.800\text{m}\right)}^2\left(7.00\text{ms}\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)\right)}\\ =2.01{\text{W}/\text{m}}^2\end{array}$

Formula to calculate the intensity level is,

    $\beta =10\text{dB}\log \left(\frac{I}{I_1}\right)$        (7)

Here, $I_1\left(1\times {10}^{-12}{\text{W}/\text{m}}^2\right)$ is the threshold intensity of hearing, $I$ is the intensity of the sound and $\beta$ is the intensity level.

Substitute $2.01{\text{W}/\text{m}}^2$ for $I$ and $1\times {10}^{-12}{\text{W}/\text{m}}^2$ for $I_1$ to find the $\beta$.

    $\begin{array}{c}\beta =10\text{dB}\log \left(\frac{2.01{\text{W}/\text{m}}^2}{1\times {10}^{-12}{\text{W}/\text{m}}^2}\right)\\ =123\text{dB}\end{array}$

It is unreasonable, implying a sound level $123\text{dB}$. Nearly all of the decreases in the mechanical energy becomes internal energy in the latch

Conclusion:

Therefore, it is unreasonable, implying a sound level $123\text{dB}$. Almost all of the mechanical energy is transformed into internal energy.