Answer any three numerical questions. a. A disoriented physics professor drives 3.25\ km north, then 4.75\ km west and then 1.50\ km south. Find the magnitude and direction of the resultant…

a. Resultant Displacement of the Professor’s Motion

Step 1: Break down the motion into components

The professor’s motion consists of three legs:

1. 3.25 km north → $\vec{A}=3.25\text{km}\hat{j}$
2. 4.75 km west → $\vec{B}=-4.75\text{km}\hat{i}$
3. 1.50 km south → $\vec{C}=-1.50\text{km}\hat{j}$

Step 2: Compute the resultant displacement vector

The total displacement $\vec{R}$ is the vector sum of $\vec{A}$, $\vec{B}$, and $\vec{C}$: $$\vec{R}=\vec{A}+\vec{B}+\vec{C}=(0\hat{i}-4.75\hat{i})+(3.25\hat{j}-1.50\hat{j})=-4.75\hat{i}+1.75\hat{j}$$

Step 3: Calculate the magnitude of the resultant displacement

Using the Pythagorean theorem: $$|\vec{R}|=\sqrt{(-4.75{)}^2+(1.75{)}^2}=\sqrt{22.5625+3.0625}=\sqrt{25.625}=5.06\text{km}$$

Step 4: Determine the direction of the resultant displacement

The angle $\theta$ with respect to the west direction (negative x-axis) is: $$\tan \theta =\frac{1.75}{4.75}⟹\theta ={\tan }^{-1}\left(\frac{1.75}{4.75}\right)=20.3^{\circ }$$ Thus, the resultant displacement is 5.06 km at 20.3° north of west.

---

b. Tension in the String and Speed of the Bob

Given:

• Mass of bob, $m=200\text{g}=0.2\text{kg}$
• Radius of circle, $r=50\text{cm}=0.5\text{m}$
• Angle of string with vertical, $\theta =30^{\circ }$

Step 1: Resolve forces in the vertical direction

The tension $T$ in the string has two components:

• Vertical: $T\cos 30^{\circ }$
• Horizontal: $T\sin 30^{\circ }$

Since the bob is in equilibrium in the vertical direction: $$T\cos 30^{\circ }=mg$$ $$T=\frac{mg}{\cos 30^{\circ }}=\frac{0.2\times 9.8}{\cos 30^{\circ }}=\frac{1.96}{0.866}=2.26\text{N}$$

Step 2: Calculate the centripetal force (horizontal component)

The horizontal component of tension provides the centripetal force: $$T\sin 30^{\circ }=\frac{m{v}^2}{r}$$ $$2.26\times \sin 30^{\circ }=\frac{0.2\times v^2}{0.5}$$ $$1.13=0.4v^2⟹v^2=\frac{1.13}{0.4}=2.825⟹v=\sqrt{2.825}=1.68\text{m/s}$$

Final Results:

• Tension in the string = 2.26 N
• Speed of the bob = 1.68 m/s

---

c. Elongation of Copper and Steel Wires

Given:

• Young’s modulus of copper, $Y_{\text{Cu}}=1.1\times 10^{11}{\text{N/m}}^2$
• Young’s modulus of steel, $Y_{\text{St}}=2.0\times 10^{11}{\text{N/m}}^2$
• Length of copper wire, $L_{\text{Cu}}=1\text{m}$
• Length of steel wire, $L_{\text{St}}=2\text{m}$
• Total elongation, $\Delta L=1\text{cm}=0.01\text{m}$

Step 1: Define elongation for each wire

Let $\Delta L_{\text{Cu}}$ and $\Delta L_{\text{St}}$ be the elongations of copper and steel wires, respectively. $$\Delta L_{\text{Cu}}+\Delta L_{\text{St}}=0.01\text{m}$$

Step 2: Relate elongation to stress and strain

The stress $\sigma$ is the same in both wires (same force $F$ and cross-sectional area $A$): $$\sigma =\frac{F}{A}=Y_{\text{Cu}}{ϵ}_{\text{Cu}}=Y_{\text{St}}{ϵ}_{\text{St}}$$ where $ϵ=\frac{\Delta L}{L}$.

Thus: $$Y_{\text{Cu}}\frac{\Delta L_{\text{Cu}}}{L_{\text{Cu}}}=Y_{\text{St}}\frac{\Delta L_{\text{St}}}{L_{\text{St}}}$$ $$1.1\times 10^{11}\frac{\Delta L_{\text{Cu}}}{1}=2.0\times 10^{11}\frac{\Delta L_{\text{St}}}{2}$$ $$1.1\Delta L_{\text{Cu}}=1.0\Delta L_{\text{St}}⟹\Delta L_{\text{St}}=1.1\Delta L_{\text{Cu}}$$

Step 3: Solve for elongations

Substitute into the total elongation equation: $$\Delta L_{\text{Cu}}+1.1\Delta L_{\text{Cu}}=0.01$$ $$2.1\Delta L_{\text{Cu}}=0.01⟹\Delta L_{\text{Cu}}=\frac{0.01}{2.1}=0.00476\text{m}=0.476\text{cm}$$ $$\Delta L_{\text{St}}=1.1\times 0.476=0.524\text{cm}$$

Final Results:

• Copper wire elongates by 0.476 cm
• Steel wire elongates by 0.524 cm

---

d. Terminal Speed of the Coalesced Raindrop

Given:

• Eight identical drops, each with terminal speed $v=5\text{cm/s}$
• Radius of each drop = $r$
• Mass of each drop = $m$

Step 1: Relate terminal speed to radius

Terminal speed $v$ of a spherical drop is given by Stokes’ law: $$v=\frac{2}{9}\frac{(\rho -{\rho }_{\text{air}})g{r}^2}{\eta }$$ where:

• $\rho$ = density of water
• $\eta$ = viscosity of air
• $g$ = acceleration due to gravity

Since all drops are identical, $v\propto r^2$.

Step 2: Determine the radius of the coalesced drop

When eight identical drops coalesce, the volume remains constant: $$8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3⟹8r^3=R^3⟹R=2r$$

Step 3: Calculate the new terminal speed

Since $v\propto r^2$, the new terminal speed $V$ of the coalesced drop (radius $2r$) is: $$\frac{V}{v}={\left(\frac{2r}{r}\right)}^2=4⟹V=4v=4\times 5\text{cm/s}=20\text{cm/s}$$

Final Result:

• Terminal speed of the coalesced drop = 20 cm/s