Define momentum. Distinguish between angular velocity and angular acceleration. State Hooke's law. Distinguish between elasticity and limit of proportionality.

Let the positive x-axis be East and the positive y-axis be North.
Vehicle A: $v_A = 105 \text{ km/h}$ northwest. This means $45^\circ$ north of west, or $135^\circ$ from the positive x-axis.

$$\vec{v}_A = (105 \cos 135^\circ) \hat{i} + (105 \sin 135^\circ) \hat{j}$$ $$\vec{v}_A = (105 \times -\frac{\sqrt{2}}{2}) \hat{i} + (105 \times \frac{\sqrt{2}}{2}) \hat{j}$$ $$\vec{v}_A \approx -74.246 \hat{i} + 74.246 \hat{j} \text{ km/h}$$

Vehicle B: $v_B = 115 \text{ km/h}$ directly east.

$$\vec{v}_B = 115 \hat{i} + 0 \hat{j} \text{ km/h}$$

$$\vec{v}_{A/B} = (-74.246 \hat{i} + 74.246 \hat{j}) - (115 \hat{i})$$ $$\vec{v}_{A/B} = (-74.246 - 115) \hat{i} + 74.246 \hat{j}$$ $$\vec{v}_{A/B} = -189.246 \hat{i} + 74.246 \hat{j} \text{ km/h}$$

$$|\vec{v}_{A/B}| = \sqrt{(-189.246)^2 + (74.246)^2}$$ $$|\vec{v}_{A/B}| = \sqrt{35813.9 + 5512.4}$$ $$|\vec{v}_{A/B}| = \sqrt{41326.3} \approx 203.289 \text{ km/h}$$

Let $\phi$ be the angle with the negative x-axis (West).

$$\tan \phi = \frac{|y\text{-component}|}{|x\text{-component}|} = \frac{74.246}{189.246} \approx 0.39233$$ $$\phi = \arctan(0.39233) \approx 21.42^\circ$$

The direction is $21.4^\circ$ North of West.
Rounding to three significant figures:
Magnitude: $\boxed{\text{203 km/h}}$
Direction: $\boxed{\text{21.4}^\circ \text{ North of West}}$

2.2 Calculate the resulting velocity in magnitude and direction.

Let the positive x-axis be East and the positive y-axis be North.
Ship velocity in still water: $v_S = 95 \text{ km/h}$ south-east. This means $45^\circ$ south of east, or $-45^\circ$ (or $315^\circ$) from the positive x-axis.

$$\vec{v}_S = (95 \cos(-45^\circ)) \hat{i} + (95 \sin(-45^\circ)) \hat{j}$$ $$\vec{v}_S = (95 \times \frac{\sqrt{2}}{2}) \hat{i} + (95 \times -\frac{\sqrt{2}}{2}) \hat{j}$$ $$\vec{v}_S \approx 67.175 \hat{i} - 67.175 \hat{j} \text{ km/h}$$

Current velocity: $v_C = 20 \text{ km/h}$ E $25^\circ$ N. This means $25^\circ$ north of east.

$$\vec{v}_C = (20 \cos 25^\circ) \hat{i} + (20 \sin 25^\circ) \hat{j}$$ $$\vec{v}_C \approx (20 \times 0.9063) \hat{i} + (20 \times 0.4226) \hat{j}$$ $$\vec{v}_C \approx 18.126 \hat{i} + 8.452 \hat{j} \text{ km/h}$$

$$\vec{v}_{resultant} = (67.175 \hat{i} - 67.175 \hat{j}) + (18.126 \hat{i} + 8.452 \hat{j})$$ $$\vec{v}_{resultant} = (67.175 + 18.126) \hat{i} + (-67.175 + 8.452) \hat{j}$$ $$\vec{v}_{resultant} = 85.301 \hat{i} - 58.723 \hat{j} \text{ km/h}$$

$$|\vec{v}_{resultant}| = \sqrt{(85.301)^2 + (-58.723)^2}$$ $$|\vec{v}_{resultant}| = \sqrt{7276.26 + 3448.39}$$ $$|\vec{v}_{resultant}| = \sqrt{10724.65} \approx 103.559 \text{ km/h}$$

Let $\phi$ be the angle with the positive x-axis (East).

$$\tan \phi = \frac{|y\text{-component}|}{|x\text{-component}|} = \frac{58.723}{85.301} \approx 0.68842$$ $$\phi = \arctan(0.68842) \approx 34.54^\circ$$

The direction is $34.5^\circ$ South of East.
Rounding to three significant figures:
Magnitude: $\boxed{\text{104 km/h}}$
Direction: $\boxed{\text{34.5}^\circ \text{ South of East}}$

2.3 Projectile motion: A boy throws a cricket ball at an angle of $20^\circ$ to the horizontal with an initial velocity of $45 \text{ m/s}$.

Given: $v_0 = 45 \text{ m/s}$, $\theta = 20^\circ$. Assume $g = 9.81 \text{ m/s}^2$.
Initial vertical velocity: $v_{0y} = v_0 \sin \theta = 45 \sin 20^\circ \approx 45 \times 0.3420 = 15.39 \text{ m/s}$.
Initial horizontal velocity: $v_{0x} = v_0 \cos \theta = 45 \cos 20^\circ \approx 45 \times 0.9397 = 42.2865 \text{ m/s}$.

2.3.1 The time it will take to reach maximum height.

At maximum height, the vertical velocity $v_y = 0$.

$$v_y = v_{0y} - gt$$ $$0 = 15.39 \text{ m/s} - (9.81 \text{ m/s}^2)t$$ $$t = \frac{15.39 \text{ m/s}}{9.81 \text{ m/s}^2}$$ $$t \approx 1.5688 \text{ s}$$

Rounding to three significant figures:

$$\boxed{\text{1.57 s}}$$

2.3.2 The maximum height reached by the ball.

$$y_{max} = v_{0y}t - \frac{1}{2}gt^2$$

Using $t \approx 1.5688 \text{ s}$ from 2.3.1:

$$y_{max} = (15.39 \text{ m/s})(1.5688 \text{ s}) - \frac{1}{2}(9.81 \text{ m/s}^2)(1.5688 \text{ s})^2$$ $$y_{max} = 24.148 \text{ m} - \frac{1}{2}(9.81)(2.4612) \text{ m}$$ $$y_{max} = 24.148 \text{ m} - 12.073 \text{ m}$$ $$y_{max} \approx 12.075 \text{ m}$$

Alternatively, using $v_y^2 = v_{0y}^2 - 2gy_{max}$:

$$0^2 = (15.39 \text{ m/s})^2 - 2(9.81 \text{ m/s}^2)y_{max}$$ $$y_{max} = \frac{(15.39)^2}{2 \times 9.81} = \frac{236.8521}{19.62} \text{ m}$$ $$y_{max} \approx 12.072 \text{ m}$$

Rounding to three significant figures:

$$\boxed{\text{12.1 m}}$$

2.3.3 The horizontal displacement of the ball.

The total time of flight is twice the time to reach maximum height.

$$T_{total} = 2 \times t_{max} = 2 \times 1.5688 \text{ s} = 3.1376 \text{ s}$$

$$R = v_{0x} \times T_{total}$$ $$R = 42.2865 \text{ m/s} \times 3.1376 \text{ s}$$ $$R \approx 132.65 \text{ m}$$

Rounding to three significant figures:

$$\boxed{\text{133 m}}$$

QUESTION 3: ANGULAR MOTION

3.1 An electric motor has a vee pulley with an effective diameter of $11 \text{ cm}$ on its shaft. If the motor, after being switched on, takes $6 \text{ seconds}$ to reach maximum speed of $1380 \text{ r/min}$.

Given:

• Diameter $D = 11 \text{ cm} = 0.11 \text{ m}$
• Radius $r = D/2 = 0.11/2 = 0.055 \text{ m}$
• Time $t = 6 \text{ s}$
• Initial angular velocity $\omega_i = 0 \text{ r/min}$ (starts from rest)
• Final angular velocity $\omega_f = 1380 \text{ r/min}$

3.1.1 The angular acceleration of the pulley in rad/s$^2$.

$$\omega_f = 1380 \text{ r/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}$$ $$\omega_f = \frac{1380 \times 2\pi}{60} \text{ rad/s}$$ $$\omega_f = 46\pi \text{ rad/s}$$ $$\omega_f \approx 144.513 \text{ rad/s}$$

$$\alpha = \frac{46\pi \text{ rad/s} - 0 \text{ rad/s}}{6 \text{ s}}$$ $$\alpha = \frac{46\pi}{6} \text{ rad/s}^2$$ $$\alpha = \frac{23\pi}{3} \text{ rad/s}^2$$ $$\alpha \approx 24.0855 \text{ rad/s}^2$$

Rounding to three significant figures:

$$\boxed{\text{24.1 rad/s}^2}$$

3.1.2 The linear acceleration of the vee belt on the pulley in m/s$^2$.

Using $\alpha = \frac{23\pi}{3} \text{ rad/s}^2$ and $r = 0.055 \text{ m}$.

$$a = \frac{23\pi}{3} \text{ rad/s}^2 \times 0.055 \text{ m}$$ $$a \approx 24.0855 \text{ rad/s}^2 \times 0.055 \text{ m}$$ $$a \approx 1.3247 \text{ m/s}^2$$

Rounding to three significant figures:

$$\boxed{\text{1.32 m/s}^2}$$

3.1.3 The maximum linear velocity of the vee belt in m/s.

Using $\omega_f = 46\pi \text{ rad/s}$ and $r = 0.055 \text{ m}$.

$$v = 46\pi \text{ rad/s} \times 0.055 \text{ m}$$ $$v \approx 144.513 \text{ rad/s} \times 0.055 \text{ m}$$ $$v \approx 7.9482 \text{ m/s}$$

Rounding to three significant figures:

$$\boxed{\text{7.95 m/s}}$$