Chapter 7, Problem 91P

(a)

To determine

The change in momentum of the car due to the fly.

Answer to Problem 91P

The change in momentum of the car due to the fly is $-0.01\text{kg}\cdot \text{km/h}$ opposite to the car’s motion.

Explanation of Solution

Write the expression for conservation of momentum.

  $m_1{v}_1=m_2{v}_2$        (I)

Here, $m_1$ is the mass of the automobile, $v_1$ is the speed of the automobile, $m_2$ is the mass of the fly, and $v_2$ is the speed of the fly.

Write the expression for car’s final momentum.

  $p_f=m_1{v}_2$        (II)

Here, $p_f$ is the final momentum of the car.

Write the expression for car’s initial momentum.

  $p_i=m_1{v}_1$        (III)

Here, $p_i$ is the initial momentum of the car.

Write the expression for change in momentum of the car due to the fly.

  $\Delta p=p_f-p_i$        (IV)

Here, $\Delta p$ is the change in momentum of the car due to the fly.

Conclusion:

Substitute $1000\text{kg}$ for $m_1$, $100\text{km/h}$ for $v_1$, and $0.1\text{g}+1000\text{kg}$ for $m_2$ in the equation (1) to find $v_2$.

  $\begin{array}{c}\left(1000\text{kg}\right)\left(100\text{km/h}\right)=\left[\left(0.1\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)+1000\text{kg}\right]v_2\\ 100,000\text{kg}\cdot \text{km/h}=\left(0.0001\text{kg}+1000\text{kg}\right)v_2\\ v_2=\frac{100,000\text{kg}\cdot \text{km/h}}{1000.0001\text{kg}}\\ =99.999990\text{km/h}\end{array}$

Substitute $1000\text{kg}$ for $m_1$ and $99.999990\text{km/h}$ for $v_2$ in the equation (1I) to find $P_f$.

  $\begin{array}{c}{p}_f=\left(1000\text{kg}\right)\left(99.999990\text{km/h}\right)\\ =99,999.990\text{kg}\cdot \text{km/h}\end{array}$

Substitute $1000\text{kg}$ for $m_1$ and $100\text{km/h}$ for $v_1$ in the equation (1II) to find $P_i$.

  $\begin{array}{c}{p}_i=\left(1000\text{kg}\right)\left(100\text{km/h}\right)\\ =100,000\text{kg}\cdot \text{km/h}\end{array}$

Substitute $99,999.990\text{kg}\cdot \text{km/h}$ for $P_f$ and $100,000\text{kg}\cdot \text{km/h}$ for $P_i$ in equation (V).

  $\begin{array}{c}\Delta p=99,999.990\text{kg}\cdot \text{km/h}-100,000\text{kg}\cdot \text{km/h}\\ \approx -0.01\text{kg}\cdot \text{km/h}\end{array}$

Therefore, the change in momentum of the car due to the fly is $-0.01\text{kg}\cdot \text{km/h}$ opposite to the car’s motion.

(b)

To determine

The change in momentum of the fly due to the car.

Answer to Problem 91P

The change in momentum of the fly due to the car is $0.01\text{kg}\cdot \text{km/h}$.

Explanation of Solution

Write the expression for change in momentum of the fly due to the car.

  $\Delta p=m_{\text{fly}}{v}_1$        (V)

Here, $\Delta p$ is the change in momentum of the fly due to the car, $m_{\text{fly}}$ is the mass of the fly, and $v_1$ is the speed of the automobile.

Conclusion:

Substitute $0.1\text{g}$ for $m_{\text{fly}}$ and $100\text{km/h}$ for $v_1$ in the equation (V) to find $\Delta P$.

  $\begin{array}{c}\Delta p=\left(0.1\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)\left(100\text{km/h}\right)\\ =0.01\text{kg}\cdot \text{km/h}\end{array}$

Therefore, the change in momentum of the fly due to the car is $0.01\text{kg}\cdot \text{km/h}$.

(c)

To determine

The number of flies takes to reduce the car’s speed into $1\text{km/h}$ approximately.

Answer to Problem 91P

The number of flies takes to reduce the car’s speed into $1\text{km/h}$ approximately is ${10}^5\text{flies}$.

Explanation of Solution

Write the expression for the number of flies takes to reduce the car’s speed.

  $m_1{v}_1=\left[m_1+n{m}_{\text{fly}}\right]99$        (VI)

Here, $n$ is the number of flies takes to reduce the car’s speed.

Conclusion:

Substitute $0.1\text{g}$ for $m_{\text{fly}}$, $1000\text{kg}$ for $m_1$, and $100\text{km/h}$ for $v_1$ in the equation (VI) to find $n$.

  $\begin{array}{c}\left(1000\text{kg}\right)\left(100\text{km/h}\right)=\left[1000\text{kg}+n\left(0.0001\text{kg}\right)\right]99\text{km/h}\\ 100,000\text{kg}\cdot \text{km/h}=99000\text{kg}\cdot \text{km/h}+n\left(0.0099\text{kg}\right)\\ 1000\text{kg}\cdot \text{km/h}=n\left(0.0099\text{kg}\right)\\ n={10}^5\text{flies}\end{array}$

Therefore, the number of flies takes to reduce the car’s speed into $1\text{km/h}$ approximately is ${10}^5\text{flies}$.