What are vector and scalar quantities? To which class do: momentum, current, electrical energy, acceleration due to gravity and mass belong?

Welcome back Blessing — missed you this week. I will solve question 11a (i), (ii), (iii) and 11b (i), (ii) from the image. 11a (i) What are vector and scalar quantities? To which class do: momentum, current, electrical energy, acceleration due to gravity and mass belong? 11a (i) Definition of Vector and Scalar Quantities: • A scalar quantity is a physical quantity that has magnitude only. Examples include mass, temperature, and time. • A vector quantity is a physical quantity that has both magnitude and direction. Examples include force, velocity, and displacement. Classification: • Momentum: Vector (has magnitude and direction) • Current: Scalar (though it has direction of flow, it does not follow vector addition rules) • Electrical energy: Scalar (has magnitude only) • Acceleration due to gravity: Vector (has magnitude and direction, always towards the center of the Earth) • Mass: Scalar (has magnitude only) 11a (ii) Explain the term 'relative velocity'. Relative velocity is the velocity of an object or observer B in the reference frame of another object or observer A. It is the velocity at which B appears to move to A. Mathematically, the relative velocity of B with respect to A is given by v_BA = v_B - v_A. 11a (iii) At the instant the traffic lights turn green, an auto mobile that has been waiting at the intersection of roads accelerates with 3 ms^-2. At the same instant, a truck travelling with a constant velocity of 12 ms^-1 overtakes and passes the automobile. How far beyond its starting point will the automobile overtake the truck and how fast will it be travelling at that time? Step 1: Define variables and set up equations for the automobile and the truck. For the automobile (starting from rest): Initial velocity, u_A = 0 ms^-1 Acceleration, a_A = 3 ms^-2 Distance covered, s_A = u_A t + (1)/(2) a_A t^2 = 0 · t + (1)/(2) (3) t^2 = (3)/(2) t^2 Final velocity, v_A = u_A + a_A t = 0 + 3t = 3t For the truck (constant velocity): Velocity, v_T = 12 ms^-1 Acceleration, a_T = 0 ms^-2 Distance covered, s_T = v_T t = 12t Step 2: Determine the time when the automobile overtakes the truck. The automobile overtakes the truck when their distances covered are equal: s_A = s_T (3)/(2) t^2 = 12t Since t ≠ 0 (as they start at the same time and the automobile needs time to catch up), we can divide by t: (3)/(2) t = 12 3t = 24 t = (24)/(3) = 8 s Step 3: Calculate the distance covered when the automobile overtakes the truck. Using the truck's distance formula (or automobile's): s_T = 12t = 12 × 8 = 96 m So, the automobile will overtake the truck 96 m beyond its starting point. Step 4: Calculate the automobile's speed at the overtaking point. Using the automobile's final velocity formula: v_A = 3t = 3 × 8 = 24 ms^-1 So, the automobile will be travelling at 24 ms⁻¹ at that time. The automobile will overtake the truck 96 m beyond its starting point and will be travelling at 24 ms^-1 at that time. 11b (i) Define centripetal force and derive its expression. Centripetal force is the force that acts on an object moving in a circular path and is directed towards the center of the circle. It is responsible for continuously changing the direction of the object's velocity, thus keeping it in circular motion. Derivation of Centripetal Force Expression: Consider an object of mass m moving with a constant speed v in a circular path of radius r. The centripetal acceleration a_c is given by: a_c = (v^2)/(r) According to Newton's second law of motion, force F is equal to mass m times acceleration a: F = ma Substituting the expression for centripetal acceleration into Newton's second law, we get the expression for centripetal force F_c: F_c = m a_c F_c = (mv^2)/(r) 11b (ii) A bend in a level road has a radius of 100 m. Find the maximum speed which a car turning this bend may have without skidding if the coefficient of friction between the road and the car is 0.25. Step 1: Identify the forces involved. For a car to turn a bend without skidding, the centripetal force required for circular motion must be provided by the static friction force between the tires and the road. Centripetal force, F_c = (mv^2)/(r) Maximum static friction force, F_f = _s N, where _s is the coefficient of static friction and N is the normal force. Step 2: Relate normal force to gravitational force. On a level road, the normal force N is equal to the gravitational force mg: N = mg So, F_f = _s mg Step 3: Set the centripetal force equal to the maximum static friction force to find the maximum speed. For no skidding, F_c F_f. For maximum speed, F_c = F_f: (mv^2)/(r) = _s mg Step 4: Solve for the maximum speed v. Cancel out the mass m from both sides: (v^2)/(r) = _s g v^2 = _s g r v = sqrt(_s g r) Step 5: Substitute the given values. Given: Radius, r = 100 m Coefficient of friction, _s = 0.25 Acceleration due to gravity, g = 9.8 ms^-2 (standard value) v = sqrt(0.25 × 9.8 ms)^-2 × 100 m v = sqrt(245 m)^2s^-2 v ≈ 15.65 ms^-1 The maximum speed the car can have without skidding is 15.65 ms^-1. Drop the next question.