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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here are the solutions for Question 1, Question 2, and Question 3. SECTION A: QUESTION 1: GENERAL 1.1 Resultant force The resultant force is the single force that produces the same effect on a body as all the individual forces acting on it. It is the vector sum of all forces. 1.2 Elastic limit The elastic limit is the maximum stress that a material can withstand without undergoing permanent deformation. If the stress exceeds this limit, the material will not return to its original shape once the stress is removed. 1.3 Strain Strain is a measure of the deformation of a material, defined as the ratio of the change in dimension (e.g., length, volume) to the original dimension. It is a dimensionless quantity. 1.4 Hydrostatic pressure Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth and is calculated as P = gh, where is the fluid density, g is the acceleration due to gravity, and h is the depth. 1.5 Second moment of area The second moment of area, also known as the area moment of inertia, is a geometrical property of a cross-section that describes how its area is distributed with respect to an axis. It is crucial in beam bending calculations to determine a beam's resistance to bending and deflection. SECTION B: QUESTION 2: KINEMATICS 2.1 A delivery truck travels at a velocity of 90 km/h on a bearing of 045^ true. It has to deliver goods to a warehouse that is 120 km due East of its current position. A cross wind of 20 km/h from the South acts on the truck. 2.1.1 Determine the true course the driver must steer to reach the warehouse. Let V_TA be the velocity of the truck relative to the air (what the driver steers), with magnitude 90 km/h and unknown bearing . Let V_W be the velocity of the wind. Since it's from the South, it blows North, so V_W = (0, 20) km/h. Let V_G be the resultant velocity of the truck relative to the ground. Since the destination is due East, V_G must be purely in the East direction, so V_G = (V_G, 0). The vector equation is V_G = V_TA + V_W. In components, using standard Cartesian coordinates (East as positive x, North as positive y): V_TA,x = 90 (East component, using bearing from North) V_TA,y = 90 (North component, using bearing from North) Step 1: Set up the vector equation in components. (V_G, 0) = (90 , 90 ) + (0, 20) (V_G, 0) = (90 , 90 + 20) Step 2: Equate the y-components to find the steering angle . 0 = 90 + 20 90 = -20 = -(20)/(90) = -(2)/(9) = (-(2)/(9)) ≈ 102.84^ The driver must steer on a true course (bearing) of 102.84^. 2.1.2 Calculate how long the journey will take. Step 1: Calculate the magnitude of the resultant velocity V_G (the effective speed towards East). From the x-components: V_G = 90 Substitute ≈ 102.84^: V_G = 90 (102.84^) km/h V_G ≈ 90 × 0.975 km/h V_G ≈ 87.75 km/h Step 2: Calculate the time taken for the journey. The distance to the warehouse is D = 120 km (East). t = (D)/(V_G) t = 120 km87.75 km/h t ≈ 1.3676 hours t ≈ 1 hour and (0.3676 × 60) minutes t ≈ 1 hour and 22.06 minutes 2.2 A steel ball is projected from the top of a building 48 m high with an initial velocity of 18 m/s at an angle of 25^ above the horizontal. Given: Initial velocity u = 18 m/s Angle of projection = 25^ Height of building H = 48 m Acceleration due to gravity g = 9.81 m/s^2 (downwards) 2.2.1 The maximum height above the ground reached by the ball. Step 1: Resolve initial velocity into vertical component. u_y = u = 18 (25^) m/s u_y ≈ 18 × 0.4226 ≈ 7.607 m/s Step 2: Calculate the maximum height reached above the launch point (h_max). At maximum height, the vertical velocity v_y = 0. Using the kinematic equation v_y^2 = u_y^2 + 2gh_max: 0^2 = (7.607 m/s)^2 + 2(-9.81 m/s^2)h_max 0 = 57.866 - 19.62 h_max 19.62 h_max = 57.866 h_max = (57.866)/(19.62) ≈ 2.949 m Step 3: Calculate the maximum height above the ground. Total maximum height = H + h_max = 48 m + 2.949 m = 50.95 m 2.2.2 The horizontal distance from the foot of the building where the ball hits the ground. 2.2.3 The time taken for the ball to hit the ground. Step 1: Calculate the time taken for the ball to hit the ground. The total vertical displacement from the launch point to the ground is y = -48 m. Using the kinematic equation y = u_y t + (1)/(2)gt^2: -48 = (7.607)t + (1)/(2)(-9.81)t^2 -48 = 7.607t - 4.905t^2 Rearrange into a quadratic equation: 4.905t^2 - 7.607t - 48 = 0 Using the quadratic formula t = -b ± sqrt(b^2 - 4ac)2a: t = -(-7.607) ± sqrt((-7.607)^2 - 4(4.905)(-48))2(4.905) t = 7.607 ± sqrt(57.866 + 941.76)9.81 t = 7.607 ± sqrt(999.626)9.81 t = (7.607 ± 31.617)/(9.81) Since time cannot be negative, we take the positive root: t = (7.607 + 31.617)/(9.81) = (39.224)/(9.81) ≈ 3.998 s Step 2: Calculate the horizontal distance. The horizontal velocity u_x is constant. u_x = u = 18 (25^) m/s u_x ≈ 18 × 0.9063 ≈ 16.313 m/s Horizontal distance x = u_x t: x = 16.313 m/s × 3.998 s x ≈ 65.22 m 2.3 A train accelerates uniformly from rest to a speed of 72 km/h in 45 seconds. It then travels at this constant speed for 3 minutes before applying the brakes and coming to rest uniformly in 60 seconds. 2.3.1 Calculate the acceleration of the train. This refers to the first phase of motion. Initial velocity u_1 = 0 m/s (from rest) Final velocity v_1 = 72 km/h Convert v_1 to m/s: v_1 = 72 km/h × 1000 m1 km × 1 h3600 s = 20 m/s Time t_1 = 45 s Step 1: Calculate the acceleration a_1. a_1 = (v_1 - u_1)/(t_1) a_1 = 20 m/s - 0 m/s45 s a_1 = (20)/(45) m/s^2 = (4)/(9) m/s^2 ≈ 0.444 m/s^2 2.3.2 Determine the total distance travelled by the train from the start until it comes to rest. This involves three phases of motion: Phase 1: Acceleration Initial velocity u_1 = 0 m/s Final velocity v_1 = 20 m/s Time t_1 = 45 s Step 1: Calculate distance s_1. s_1 = ((u_1 + v_1)/(2))t_1 s_1 = ((0 + 20)/(2)) × 45 = 10 × 45 = 450 m Phase 2: Constant speed Speed v_2 = 20 m/s Time t_2 = 3 minutes = 3 × 60 s = 180 s Step 2: Calculate distance s_2. s_2 = v_2 t_2 s_2 = 20 m/s × 180 s = 3600 m Phase 3: Deceleration Initial velocity u_3 = 20 m/s Final velocity v_3 = 0 m/s (comes to rest) Time t_3 = 60 s Step 3: Calculate distance s_3. s_3 = ((u_3 + v_3)/(2))t_3 s_3 = ((20 + 0)/(2)) × 60 = 10 × 60 = 600 m Step 4: Calculate the total distance travelled. S_total = s_1 + s_2 + s_3 S_total = 450 m + 3600 m + 600 m = 4650 m Got more? Send 'em!