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Here are the solutions to your questions: a) i) Acceleration is defined as the rate of change of velocity. Its unit of measurement is meters per second squared (m/s^2). a) ii) The term 'free fall' refers to the motion of an object solely under the influence of gravity, where the effects of air resistance are considered negligible. a) iii) Experiment to measure the acceleration of free fall 'g' A labelled diagram: (It is not possible to draw a diagram here, but the setup typically involves:) An electromagnet holding a steel ball* at a certain height. A timer (electronic stopwatch) connected to the electromagnet and a trap door* below. A metre rule* to measure the height. A power supply* for the electromagnet. The procedure you would follow: 1. Set up the apparatus with the electromagnet holding the steel ball above the trap door. 2. Measure the height (h) from the bottom of the steel ball to the top of the trap door using the metre rule. 3. Switch off the electromagnet. The ball will fall, and the timer will start automatically when the ball leaves the electromagnet. 4. The ball will hit the trap door, which will stop the timer. Record the time (t). 5. Repeat the experiment several times for the same height to obtain an average time. 6. Vary the height (h) and repeat steps 2-5 to obtain several pairs of (h, t) values. The measurements you would make: The height* (h) from which the ball is dropped, measured in meters (m). The time* (t) taken for the ball to fall through that height, measured in seconds (s). How you would use the measurements to obtain a value of 'g': 1. For an object in free fall starting from rest, the equation of motion is h = ut + (1)/(2)gt^2. Since the initial velocity (u) is 0\,m/s, the equation simplifies to h = (1)/(2)gt^2. 2. Rearrange the equation to g = (2h)/(t^2). 3. Alternatively, calculate t^2 for each measured time t. 4. Plot a graph of h (y-axis) against t^2 (x-axis). 5. The graph should be a straight line passing through the origin. The gradient (slope) of this line will be equal to (1)/(2)g. 6. Therefore, g = 2 × gradient. A precaution you would take to obtain accurate results: Ensure the timer starts and stops precisely* at the moment the ball is released and hits the trap door, respectively. Repeat each measurement multiple times and calculate the average* to minimize random errors. Ensure the apparatus is vertically aligned* to prevent the ball from hitting the sides. Minimize air currents* in the experimental area. b) i) The law of conservation of momentum states that for an isolated system, the total momentum before an interaction (like a collision or explosion) is equal to the total momentum after the interaction. In other words, momentum is neither created nor destroyed. b) ii) To calculate the recoil velocity of the rifle, we apply the law of conservation of momentum. Let: m_b = mass of the bullet = 0.02\,kg v_b = velocity of the bullet = 500\,m/s m_r = mass of the rifle = 0.8\,kg v_r = recoil velocity of the rifle (what we need to find) Step 1: State the initial momentum of the system (bullet + rifle). Before firing, both the bullet and the rifle are at rest, so the initial total momentum is zero. P_initial = (m_b + m_r) × 0 = 0 Step 2: State the final momentum of the system. After firing, the bullet moves forward, and the rifle recoils backward. P_final = m_b v_b + m_r v_r Step 3: Apply the law of conservation of momentum (P_initial = P_final). 0 = m_b v_b + m_r v_r 0 = (0.02\,kg)(500\,m/s) + (0.8\,kg)v_r Step 4: Solve for v_r. 0 = 10\,kg · m/s + (0.8\,kg)v_r -10\,kg · m/s = (0.8\,kg)v_r v_r = (-10\,kg · m/s)/(0.8\,kg) v_r = -12.5\,m/s The negative sign indicates that the recoil velocity is in the opposite direction to the bullet's velocity. The recoil velocity of the rifle is -12.5\,m/s. Send me the next one 📸