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

QUESTION 1: GENERAL

1.1 Define momentum.

Momentum is a measure of the mass in motion. It is a vector quantity, calculated as the product of an object's mass and its velocity.

1.2 Distinguish between angular velocity and angular acceleration.

• Angular velocity ($\omega$) is the rate at which an object rotates or revolves relative to another point, measured in radians per second (rad/s). It describes how fast the angular position changes.
• Angular acceleration ($\alpha$) is the rate of change of angular velocity, measured in radians per second squared (rad/s$^2$). It describes how fast the angular velocity changes.

1.3 State Hooke's law.

Hooke's law states that the force ($F$) needed to extend or compress a spring by some distance ($x$) is directly proportional to that distance, provided the elastic limit is not exceeded. Mathematically, $F = kx$, where $k$ is the spring constant.

1.4 Distinguish between elasticity and limit of proportionality.

• Elasticity is the property of a material to return to its original shape and size after the deforming force has been removed.
• Limit of proportionality is the point on a stress-strain curve beyond which stress is no longer directly proportional to strain. Up to this limit, Hooke's law is obeyed.

QUESTION 2: KINEMATICS

2.1 Calculate the velocity of vehicle A relative to vehicle B.

Step 1: Define the velocity vectors for Vehicle A and Vehicle B. 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. $\vec{v}_A = (105 \cos 135^\circ) \hat{i} + (105 \sin 135^\circ) \hat{j}$ $\vec{v}_A = (105 \times -\frac{\sqrt{2}}{2}) \hat{i} + (105 \times \frac{\sqrt{2}}{2}) \hat{j}$ $\vec{v}_A \approx -74.246 \hat{i} + 74.246 \hat{j} km/h$ Vehicle B: $v_B = 115 \text{ km/h}$ directly east. $\vec{v}_B = 115 \hat{i} + 0 \hat{j} km/h$

Step 2: Calculate the relative velocity $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. $\vec{v}_{A/B} = (-74.246 \hat{i} + 74.246 \hat{j}) - (115 \hat{i})$ $\vec{v}_{A/B} = (-74.246 - 115) \hat{i} + 74.246 \hat{j}$ $\vec{v}_{A/B} = -189.246 \hat{i} + 74.246 \hat{j} km/h$

Step 3: Calculate the magnitude of the relative velocity. $|\vec{v}_{A/B}| = \sqrt{(-189.246)^2 + (74.246)^2}$ $|\vec{v}_{A/B}| = \sqrt{35813.9 + 5512.4}$ $|\vec{v}_{A/B}| = \sqrt{41326.3} \approx 203.289 km/h$

Step 4: Calculate the direction of the relative velocity. Let $\phi$ be the angle with the negative x-axis (West). $\tan \phi = \frac{|y-component|}{|x-component|} = \frac{74.246}{189.246} \approx 0.39233$ $\phi = \arctan(0.39233) \approx 21.42^\circ$ The direction is $21.4^\circ$ North of West. Rounding to three significant figures: Magnitude: $\boxed{203 km/h}$ Direction: $\boxed{21.4^\circ North of West}$

2.2 Calculate the resulting velocity in magnitude and direction.

Step 1: Define the velocity vectors for the ship and the current. 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. $\vec{v}_S = (95 \cos(-45^\circ)) \hat{i} + (95 \sin(-45^\circ)) \hat{j}$ $\vec{v}_S = (95 \times \frac{\sqrt{2}}{2}) \hat{i} + (95 \times -\frac{\sqrt{2}}{2}) \hat{j}$ $\vec{v}_S \approx 67.175 \hat{i} - 67.175 \hat{j} km/h$ Current velocity: $v_C = 20 \text{ km/h}$ E $25^\circ$ N. This means $25^\circ$ north of east. $\vec{v}_C = (20 \cos 25^\circ) \hat{i} + (20 \sin 25^\circ) \hat{j}$ $\vec{v}_C \approx (20 \times 0.9063) \hat{i} + (20 \times 0.4226) \hat{j}$ $\vec{v}_C \approx 18.126 \hat{i} + 8.452 \hat{j} km/h$

Step 2: Calculate the resulting velocity $\vec{v}_{resultant} = \vec{v}_S + \vec{v}_C$. $\vec{v}_{resultant} = (67.175 \hat{i} - 67.175 \hat{j}) + (18.126 \hat{i} + 8.452 \hat{j})$ $\vec{v}_{resultant} = (67.175 + 18.126) \hat{i} + (-67.175 + 8.452) \hat{j}$ $\vec{v}_{resultant} = 85.301 \hat{i} - 58.723 \hat{j} km/h$

Step 3: Calculate the magnitude of the resulting velocity. $|\vec{v}_{resultant}| = \sqrt{(85.301)^2 + (-58.723)^2}$ $|\vec{v}_{resultant}| = \sqrt{7276.26 + 3448.39}$ $|\vec{v}_{resultant}| = \sqrt{10724.65} \approx 103.559 km/h$

Step 4: Calculate the direction of the resulting velocity. Let $\phi$ be the angle with the positive x-axis (East). $\tan \phi = \frac{|y-component|}{|x-component|} = \frac{58.723}{85.301} \approx 0.68842$ $\phi = \arctan(0.68842) \approx 34.54^\circ$ The direction is $34.5^\circ$ South of East. Rounding to three significant figures: Magnitude: $\boxed{104 km/h}$ Direction: $\boxed{34.5^\circ South of East}$

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 $g = 9.81 m/s^2$. 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.

Step 1: Use the kinematic equation for vertical motion. At maximum height, the vertical velocity $v_y = 0$. $v_y = v_{0y} - gt$ $0 = 15.39 m/s - (9.81 m/s^2)t$ $t = \frac{15.39 m/s}{9.81 m/s^2}$ $t \approx 1.5688 s$ Rounding to three significant figures: $\boxed{1.57 s}$

2.3.2 The maximum height reached by the ball.

Step 1: Use the kinematic equation for vertical displacement. $y_{max} = v_{0y}t - \frac{1}{2}gt^2$ Using $t \approx 1.5688 \text{ s}$ from 2.3.1: $y_{max} = (15.39 m/s)(1.5688 s) - \frac{1}{2}(9.81 m/s^2)(1.5688 s)^2$ $y_{max} = 24.148 m - \frac{1}{2}(9.81)(2.4612) m$ $y_{max} = 24.148 m - 12.073 m$ $y_{max} \approx 12.075 m$ Alternatively, using $v_y^2 = v_{0y}^2 - 2gy_{max}$: $0^2 = (15.39 m/s)^2 - 2(9.81 m/s^2)y_{max}$ $y_{max} = \frac{(15.39)^2}{2 \times 9.81} = \frac{236.8521}{19.62} m$ $y_{max} \approx 12.072 m$ Rounding to three significant figures: $\boxed{12.1 m}$

2.3.3 The horizontal displacement of the ball.

Step 1: Calculate the total time of flight. The total time of flight is twice the time to reach maximum height. $T_{total} = 2 \times t_{max} = 2 \times 1.5688 s = 3.1376 s$

Step 2: Calculate the horizontal displacement (range) using the horizontal velocity and total time. $R = v_{0x} \times T_{total}$ $R = 42.2865 m/s \times 3.1376 s$ $R \approx 132.65 m$ Rounding to three significant figures: $\boxed{133 m}$

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 $D = 11 cm = 0.11 \text{ m}$
• 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$^2$.

Step 1: Convert the final angular velocity from revolutions per minute (r/min) to radians per second (rad/s). $\omega_f = 1380 r/min \times \frac{2\pi rad}{1 rev} \times \frac{1 min}{60 s}$ $\omega_f = \frac{1380 \times 2\pi}{60} \text{ rad/s}$ $\omega_f = 46\pi \text{ rad/s}$ $\omega_f \approx 144.513 \text{ rad/s}$

Step 2: Calculate the angular acceleration ($\alpha$) using the formula $\alpha = \frac{\omega_f - \omega_i}{t}$. $\alpha = \frac{46\pi rad/s - 0 rad/s}{6 s}$ $\alpha = \frac{46\pi}{6} rad/s^2$ $\alpha = \frac{23\pi}{3} rad/s^2$ $\alpha \approx 24.0855 rad/s^2$ Rounding to three significant figures: $\boxed{24.1 rad/s^2}$

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

Step 1: Calculate the linear acceleration ($a$) using the relationship $a = \alpha r$. Using $\alpha = \frac{23\pi}{3} rad/s^2$ and $r = 0.055 \text{ m}$. $a = \frac{23\pi}{3} rad/s^2 \times 0.055 \text{ m}$ $a \approx 24.0855 rad/s^2 \times 0.055 \text{ m}$ $a \approx 1.3247 m/s^2$ Rounding to three significant figures: $\boxed{1.32 m/s^2}$

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

Step 1: Calculate the maximum linear velocity ($v$) using the relationship $v = \omega_f r$. Using $\omega_f = 46\pi \text{ rad/s}$ and $r = 0.055 \text{ m}$. $v = 46\pi rad/s \times 0.055 \text{ m}$ $v \approx 144.513 rad/s \times 0.055 \text{ m}$ $v \approx 7.9482 \text{ m/s}$ Rounding to three significant figures: $\boxed{7.95 m/s}$

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