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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2008 June Exam Higher Grade Paper 1 Q 7 7.1 The Principle of Conservation of Momentum states that: The total linear momentum of an isolated system remains constant in both magnitude and direction. 7.2 Percy's statement is FALSE. Explanation: The system (Percy and the bike) is not isolated because an external frictional force acts on it, causing it to slow down and stop. Therefore, the total momentum of the system is not conserved. Percy kept moving forward due to inertia. 7.3 Step 1: Identify the given values. Mass of Percy m_P = 75 kg Mass of quad bike m_QB = 100 kg Total mass of the system M = m_P + m_QB = 75 kg + 100 kg = 175 kg Initial velocity of the system u = 20 m/s Final velocity of the system v = 0 m/s Step 2: Calculate the change in momentum. p = M(v - u) p = (175 kg)(0 m/s - 20 m/s) p = (175 kg)(-20 m/s) p = -3500 kg·m/s The change in momentum of Percy and the bike is 3500 kg·m/s in the opposite direction of motion 7.4 Step 1: Use the impulse-momentum theorem. Given: Change in momentum p = -3500 kg·m/s (from 7.3) Time taken t = 8 s Average frictional force F_friction = ? Step 2: Calculate the average frictional force. F_net t = p In this case, the net force is the frictional force. F_friction (8 s) = -3500 kg·m/s F_friction = -3500 kg·m/s8 s F_friction = -437.5 N The magnitude of the average frictional force is 437.5 N 7.5 Step 1: Calculate the distance travelled while stopping. We need the acceleration first. a = (v - u)/( t) a = 0 m/s - 20 m/s8 s a = -20 m/s8 s a = -2.5 m/s^2 Now, calculate the distance s using a kinematic equation: v^2 = u^2 + 2as (0 m/s)^2 = (20 m/s)^2 + 2(-2.5 m/s^2)s 0 = 400 - 5s 5s = 400 s = (400)/(5) s = 80 m Step 2: Calculate the work done by the frictional force. Work done W = F · s · . The frictional force acts opposite to the direction of motion, so = 180^ and (180^) = -1. W = (437.5 N)(80 m)(-1) W = -35000 J Alternatively, using the work-energy theorem: W_net = KE = KE_f - KE_i KE_i = (1)/(2) M u^2 = (1)/(2) (175 kg)(20 m/s)^2 = (1)/(2) (175)(400) = 35000 J KE_f = (1)/(2) M v^2 = (1)/(2) (175 kg)(0 m/s)^2 = 0 J W_net = 0 J - 35000 J = -35000 J The work done by the frictional force is 35000 J (The negative sign indicates work done against motion). --- 2015 Free State Prelim Paper 1 Q4 4.1 The Law of Conservation of Momentum states that: The total linear momentum of an isolated system remains constant in both magnitude and direction. 4.2 The condition for an elastic collision is that both momentum and kinetic energy are conserved. 4.3 Step 1: Identify the given values. Mass of trolley m_T = 5 kg Initial velocity of trolley u_T = 4 m/s (east, let's take east as positive) Mass of brick m_B = 1.5 kg Initial velocity of brick u_B = 0 m/s (dropped onto the trolley) Step 2: Apply the principle of conservation of linear momentum to find the final velocity of the combined system. m_T u_T + m_B u_B = (m_T + m_B) v_f (5 kg)(4 m/s) + (1.5 kg)(0 m/s) = (5 kg + 1.5 kg) v_f 20 + 0 = (6.5) v_f 20 = 6.5 v_f v_f = (20)/(6.5) v_f ≈ 3.0769 m/s Step 3: Calculate the change in momentum of the 5 kg trolley. p_T = m_T v_f - m_T u_T p_T = (5 kg)(3.0769 m/s) - (5 kg)(4 m/s) p_T = 15.3845 kg·m/s - 20 kg·m/s p_T = -4.6155 kg·m/s The change in momentum of the 5 kg trolley is 4.62 kg·m/s west (rounded to two decimal places). --- 2008 Exemplar Paper 1 Q 2 2.1 Step 1: Identify the given values and define a positive direction. Let right be the positive direction. Mass of car m_C = 1600 kg Initial velocity of car u_C = -30 m/s (to the left) Mass of minibus m_M = 3000 kg Initial velocity of minibus u_M = 20 m/s (to the right) The vehicles move together after the collision, so they have a common final velocity v_f. Step 2: Apply the principle of conservation of linear momentum. m_C u_C + m_M u_M = (m_C + m_M) v_f (1600 kg)(-30 m/s) + (3000 kg)(20 m/s) = (1600 kg + 3000 kg) v_f -48000 + 60000 = (4600) v_f 12000 = 4600 v_f v_f = (12000)/(4600) v_f ≈ 2.61 m/s The velocity of the two vehicles after the collision is 2.61 m/s to the right 2.2 Step 1: Calculate the total initial kinetic energy. KE_i = (1)/(2) m_C u_C^2 + (1)/(2) m_M u_M^2 KE_i = (1)/(2) (1600 kg)(-30 m/s)^2 + (1)/(2) (3000 kg)(20 m/s)^2 KE_i = (1)/(2) (1600)(900) + (1)/(2) (3000)(400) KE_i = 800(900) + 1500(400) KE_i = 720000 J + 600000 J KE_i = 1320000 J Step 2: Calculate the total final kinetic energy. KE_f = (1)/(2) (m_C + m_M) v_f^2 KE_f = (1)/(2) (1600 kg + 3000 kg)(2.6087 m/s)^2 KE_f = (1)/(2) (4600)(6.8053) KE_f = 2300(6.8053) KE_f ≈ 15652.19 J Step 3: Compare initial and final kinetic energies. Since KE_i = 1320000 J and KE_f ≈ 15652.19 J, KE_i ≠ KE_f Therefore, kinetic energy is not conserved, and the collision was inelastic