This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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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* () 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* () 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 km/h northwest. This means 45^ north of west, or 135^ from the positive x-axis. v_A = (105 135^) i + (105 135^) j v_A = (105 × -sqrt(2)2) i + (105 × sqrt(2)2) j v_A ≈ -74.246 i + 74.246 j km/h Vehicle B: v_B = 115 km/h directly east. v_B = 115 i + 0 j km/h Step 2: Calculate the relative velocity v_A/B = v_A - v_B. v_A/B = (-74.246 i + 74.246 j) - (115 i) v_A/B = (-74.246 - 115) i + 74.246 j v_A/B = -189.246 i + 74.246 j km/h Step 3: Calculate the magnitude of the relative velocity. |v_A/B| = sqrt((-189.246)^2 + (74.246)^2) |v_A/B| = sqrt(35813.9 + 5512.4) |v_A/B| = sqrt(41326.3) ≈ 203.289 km/h Step 4: Calculate the direction of the relative velocity. Let be the angle with the negative x-axis (West). = |y-component||x-component| = (74.246)/(189.246) ≈ 0.39233 = (0.39233) ≈ 21.42^ The direction is 21.4^ North of West. Rounding to three significant figures: Magnitude: 203 km/h Direction: 21.4^ 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 km/h south-east. This means 45^ south of east, or -45^ (or 315^) from the positive x-axis. v_S = (95 (-45^)) i + (95 (-45^)) j v_S = (95 × sqrt(2)2) i + (95 × -sqrt(2)2) j v_S ≈ 67.175 i - 67.175 j km/h Current velocity: v_C = 20 km/h E 25^ N. This means 25^ north of east. v_C = (20 25^) i + (20 25^) j v_C ≈ (20 × 0.9063) i + (20 × 0.4226) j v_C ≈ 18.126 i + 8.452 j km/h Step 2: Calculate the resulting velocity v_resultant = v_S + v_C. v_resultant = (67.175 i - 67.175 j) + (18.126 i + 8.452 j) v_resultant = (67.175 + 18.126) i + (-67.175 + 8.452) j v_resultant = 85.301 i - 58.723 j km/h Step 3: Calculate the magnitude of the resulting velocity. |v_resultant| = sqrt((85.301)^2 + (-58.723)^2) |v_resultant| = sqrt(7276.26 + 3448.39) |v_resultant| = sqrt(10724.65) ≈ 103.559 km/h Step 4: Calculate the direction of the resulting velocity. Let be the angle with the positive x-axis (East). = |y-component||x-component| = (58.723)/(85.301) ≈ 0.68842 = (0.68842) ≈ 34.54^ The direction is 34.5^ South of East. Rounding to three significant figures: Magnitude: 104 km/h Direction: 34.5^ South of East 2.3 Projectile motion: A boy throws a cricket ball at an angle of 20^ to the horizontal with an initial velocity of 45 m/s. Given: v_0 = 45 m/s, = 20^. Assume g = 9.81 m/s^2. Initial vertical velocity: v_0y = v_0 = 45 20^ ≈ 45 × 0.3420 = 15.39 m/s. Initial horizontal velocity: v_0x = v_0 = 45 20^ ≈ 45 × 0.9397 = 42.2865 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 = 15.39 m/s9.81 m/s^2 t ≈ 1.5688 s Rounding to three significant figures: 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_0yt - (1)/(2)gt^2 Using t ≈ 1.5688 s from 2.3.1: y_max = (15.39 m/s)(1.5688 s) - (1)/(2)(9.81 m/s^2)(1.5688 s)^2 y_max = 24.148 m - (1)/(2)(9.81)(2.4612) m y_max = 24.148 m - 12.073 m y_max ≈ 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 = ((15.39)^2)/(2 × 9.81) = (236.8521)/(19.62) m y_max ≈ 12.072 m Rounding to three significant figures: 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 × t_max = 2 × 1.5688 s = 3.1376 s Step 2: Calculate the horizontal displacement (range) using the horizontal velocity and total time. R = v_0x × T_total R = 42.2865 m/s × 3.1376 s R ≈ 132.65 m Rounding to three significant figures: 133 m QUESTION 3: ANGULAR MOTION 3.1 An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min. Given: Diameter D = 11 cm = 0.11 m Radius r = D/2 = 0.11/2 = 0.055 m Time t = 6 s Initial angular velocity _i = 0 r/min (starts from rest) Final angular velocity _f = 1380 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). _f = 1380 r/min × 2 rad1 rev × 1 min60 s _f = (1380 × 2)/(60) rad/s _f = 46 rad/s _f ≈ 144.513 rad/s Step 2: Calculate the angular acceleration () using the formula = (_f - _i)/(t). = 46 rad/s - 0 rad/s6 s = (46)/(6) rad/s^2 = (23)/(3) rad/s^2 ≈ 24.0855 rad/s^2 Rounding to three significant figures: 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 = r. Using = (23)/(3) rad/s^2 and r = 0.055 m. a = (23)/(3) rad/s^2 × 0.055 m a ≈ 24.0855 rad/s^2 × 0.055 m a ≈ 1.3247 m/s^2 Rounding to three significant figures: 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 = _f r. Using _f = 46 rad/s and r = 0.055 m. v = 46 rad/s × 0.055 m v ≈ 144.513 rad/s × 0.055 m v ≈ 7.9482 m/s Rounding to three significant figures: 7.95 m/s Drop the next question.