Explain how car safety features like airbags and crumple zones reduce passenger injuries during a collision, referencing the impulse-momentum theorem and Newton's Third Law.

Car Safety Features The safety features, such as airbags and crumple zones, contribute to passenger safety by increasing the time of impact during a collision. According to the impulse-momentum theorem (F_net t = p), for a given change in momentum ( p), increasing the time interval ( t) over which the momentum changes results in a decrease in the average net force (F_net) exerted on the passengers. A reduced force significantly lowers the risk and severity of injuries. Airbags also spread the impact force over a larger area of the body, further reducing pressure and potential harm. --- 2008 Prelim Paper 1 Q6 6.1 The weight of truck A is the gravitational force exerted by the Earth on truck A. According to Newton's Third Law, the reaction force to the weight of truck A is the gravitational force exerted by truck A on the Earth. Step 1: Calculate the magnitude of the weight of truck A. W_A = m_A g W_A = (2000 kg)(9.8 m/s^2) W_A = 19600 N The magnitude of the reaction force is equal to the magnitude of the weight. The direction of the reaction force (force of truck A on Earth) is opposite to the direction of the weight (force of Earth on truck A). The magnitude and direction of the 'reaction force' to the weight of truck A is 19600 N UPWARDS 6.2 Step 1: Identify the given values and define a positive direction. Let west be the positive direction. Mass of truck A m_A = 2000 kg Initial velocity of truck A u_A = 3 m/s Mass of truck B m_B = 1200 kg + 300 kg = 1500 kg Initial velocity of truck B u_B = 0 m/s After the collision, the trucks combine and move with a common final velocity v_f. Step 2: Apply the principle of conservation of linear momentum. m_A u_A + m_B u_B = (m_A + m_B) v_f (2000 kg)(3 m/s) + (1500 kg)(0 m/s) = (2000 kg + 1500 kg) v_f 6000 kg·m/s + 0 = (3500 kg) v_f 6000 = 3500 v_f v_f = (6000)/(3500) m/s v_f = (12)/(7) m/s v_f ≈ 1.71 m/s The velocity of truck B after the collision is 1.71 m/s WESTWARDS 6.3 Step 1: Identify the given values for truck B and the collision time. Mass of truck B m_B = 1500 kg Initial velocity of truck B u_B = 0 m/s Final velocity of truck B v_B = 1.714 m/s (using the unrounded value (12)/(7) m/s) Time of collision t = 0.5 s Force exerted by truck A on truck B F = ? Step 2: Apply the impulse-momentum theorem to truck B. F t = p_B F t = m_B v_B - m_B u_B F (0.5 s) = (1500 kg)((12)/(7) m/s) - (1500 kg)(0 m/s) F (0.5) = (18000)/(7) kg·m/s F = (18000)/(7 × 0.5) N F = (18000)/(3.5) N F ≈ 5142.86 N The magnitude of the force that truck A exerts on truck B is 5143 N 6.4 Bubble plastic is preferred because it increases the time of impact ( t) during a collision. According to the impulse-momentum theorem (F_net t = p), for a given change in momentum ( p), a longer impact time ( t) results in a smaller average force (F_net) exerted on the electronic equipment. This reduced force minimizes damage to the equipment.