Chapter 15, Problem 76PQ

(a)

To determine

Time of flight of the medicine from the syringe into the child’s mouth.

Answer to Problem 76PQ

Time of flight of the medicine from the syringe into the child’s mouth is $\underset{\_}{0.101\text{s}}$.

Explanation of Solution

Write the equation to find the distance travelled by medicine into the mouth of child.

$\Delta y=v_{0}t+\frac{1}{2}{a}_y{t}^2$ (I)

Here, $\Delta y$ is the vertical distance travelled by medicine to reach the mouth of child, $v_0$ is the initial velocity of the medicine, $t$ is the time take to reach the mouth, and $a_y$ is the acceleration.

The syringe is fixed horizontally, so the initial components of velocities are zero. That is $v_0$ becomes zero. Also $a_y$ is equal to $-g$.

Put $0\text{m}/\text{s}$ for $v_0$ and $-g$ for $a_y$ in equation (I) and rewrite to get $t$.

$\begin{array}{c}\Delta y=-\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2\Delta y}{-g}}\end{array}$ (II)

Conclusion:

Substitute $-5.00\text{cm}$ for $\Delta y$, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (II) to get $t$.

$\begin{array}{c}t=\sqrt{\frac{2\left(-5.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{-9.81\text{m}/{\text{s}}^{\text{2}}}}\\ =0.101\text{s}\end{array}$

Therefore, time of flight of the medicine from the syringe into the child’s mouth is $\underset{\_}{0.101\text{s}}$.

(b)

To determine

Speed of medicine while leaving the syringe to the mouth of the child.

Answer to Problem 76PQ

The speed of the medicine while leaving he syringe to the mouth of the child is $\underset{\_}{0.693\text{m}/\text{s}}$.

Explanation of Solution

Write the equation to find the horizontal displacement of medicine.

$\Delta x=v_{\text{opening}}t$ (III)

Here, $\Delta x$ is the horizontal displacement of the medicine, $v_{\text{opening}}$ is the speed of medicine at the opening of syringe, and $t$ is the time.

Rewrite equation (III) to get $v_{\text{opening}}$.

$v_{\text{opening}}=\frac{\Delta x}{t}$ (IV)

Conclusion:

Substitute $7.00\text{cm}$ for $\Delta x$ and $0.101\text{s}$ for $t$ in equation (IV) to get $v_{\text{opening}}$.

$\begin{array}{c}{v}_{\text{opening}}=\frac{7.00\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{0.101\text{s}}\\ =0.693\text{m}/\text{s}\end{array}$

Therefore, speed of the medicine while leaving he syringe to the mouth of the child is $\underset{\_}{0.693\text{m}/\text{s}}$.

(c)

To determine

Speed with which mother should push the plunger of the syringe.

Answer to Problem 76PQ

The speed with which mother should push the plunger of syringe is $\underset{\_}{0.0251\text{m}/\text{s}}$.

Explanation of Solution

Here the medicine pushed by a plunger from a barrel to opening of the syringe. Therefore apply continuity equation to connect the area and velocity of both the region.

Write the continuity equation to the barrel of medicine and the opening of the syringe.

$A_1{v}_1=A_2{v}_2$ (V)

Here, $A_1$ is the area of the barrel, $v_1$ is the velocity of medicine in barrel, $A_2$ is the area of the opening of the syringe, and $v_2$ is the velocity of medicine in the opening of syringe or it is equal to $v_{\text{opening}}$.

Replace $v_2$ by $v_{\text{opening}}$ in equation (V).

$A_1{v}_1=A_2{v}_{\text{opening}}$ (VI)

Area of the opening of syringe is not given, but the radius is provided.

Write the equation to find the area of the opening of the syringe.

$A_2=\pi r_2{}^2$ (VII)

Here, $r_2$ is the radius of the opening of syringe.

Rewrite equation (VI) to get $v_1$.

$v_1=\frac{A_2{v}_{\text{opening}}}{A_1}$ (VIII)

Conclusion:

Substitute $1.20\text{mm}$ for $r_2$ in equation (VII) to get $A_2$.

$\begin{array}{c}{A}_2=\pi {\left(1.20\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\\ =4.52\times {10}^{-6}{\text{m}}^2\end{array}$

Substitute $4.52\times {10}^{-6}{\text{m}}^2$ for $A_2$, $0.693\text{m}/\text{s}$ for $v_{\text{opening}}$, and $1.25{\text{cm}}^2$ for $A_1$ in equation (VIII) to get $v_1$.

$\begin{array}{c}{v}_1=\left(\frac{4.52\times {10}^{-6}{\text{m}}^2}{1.25{\text{cm}}^2\times \frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}}\right)\left(0.693\text{m}/\text{s}\right)\\ =0.0251\text{m}/\text{s}\end{array}$

Therefore, the speed with which mother should push the plunger of syringe is $\underset{\_}{0.0251\text{m}/\text{s}}$.

(d)

To determine

Pressure of medicine in the opening of the syringe.

Answer to Problem 76PQ

Pressure of medicine at the opening of the syringe is $\underset{\_}{1.013\times {10}^5\text{Pa}}$.

Explanation of Solution

The pressure experienced by the medicine at the opening of syringe is nothing but the atmospheric pressure. The magnitude of atmospheric pressure is $1.000\text{atm}$. The value of atmospheric pressure at the nozzle in pascals unit correct to four significant figure is $\underset{\_}{1.013\times {10}^5\text{Pa}}$.

Conclusion:

Therefore, pressure of medicine at the opening of the syringe is $\underset{\_}{1.013\times {10}^5\text{Pa}}$.

(d)

To determine

Pressure in the barrel of medicine.

Answer to Problem 76PQ

Pressure of medicine at the barrel is $\underset{\_}{1.015\times {10}^5\text{Pa}}$.

Explanation of Solution

Here the flow of medicine through is assumed to be an ideal one. Hence we can apply Bernoulli’s law. Bernoulli’s law states that the total mechanical energy stored in the system is a constant.

Write the Bernoulli’s equation.

$P_1+\frac{1}{2}{\rho }_w{v}_1{}^2+{\rho }_{w}g{y}_1=P_2+\frac{1}{2}{\rho }_w{v}_2{}^2+{\rho }_{w}g{y}_2$ (IX)

Here, $P_1$ is the pressure at barrel, ${\rho }_w$ is the density of medicine which is equal to density of water, $v_1$ is the velocity of medicine at barrel, $g$ is acceleration due to gravity, $P_2$ is the pressure of medicine at the opening of syringe, $v_2$ is the velocity of medicine at the opening, $y_1$ is the height of barrel from the ground, and $y_2$ is the height of opening of syringe from the ground.

Since the medicine is flowing through a horizontal syringe, $y_1=y_2$. So the gravity terms in equation (IX) cancels out.

Rewrite equation (IX).

$P_1+\frac{1}{2}{\rho }_w{v}_1{}^2=P_2+\frac{1}{2}{\rho }_w{v}_2{}^2$ (X)

Rearrange equation (X) to get $P_1$.

$P_1=P_2+\frac{{\rho }_w}{2}\left(v_2{}^2-v_1{}^2\right)$ (XI)

Conclusion:

Substitute $1.013\times {10}^5\text{Pa}$ for $P_2$, $1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $0.693\text{m}/\text{s}$ for $v_2$, and $0.0251\text{m}/\text{s}$ for $v_1$ in equation (XI) to get $P_1$.

$\begin{array}{c}{P}_1=1.013\times {10}^5\text{Pa}+\frac{1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}{2}\left[{\left(0.693\text{m}/\text{s}\right)}^2-{\left(0.0251\text{m}/\text{s}\right)}^2\right]\\ =1.015\times {10}^5\text{Pa}\end{array}$

Therefore, pressure of medicine at the opening of the syringe is $\underset{\_}{1.015\times {10}^5\text{Pa}}$.