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.
ScanSolve AI Answer
Welcome back Candice — missed you this week. Here are the solutions to your questions: 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, provided no external forces act on the system. 4.2 Step 1: Identify the known values and the principle to be used. The system consists of the block and bullet. After the collision, the combined system swings upwards. We use the conservation of mechanical energy between the lowest point (immediately after impact) and the highest point of the swing. Given: Mass of bullet, m_b = 15 \, g = 0.015 \, kg Mass of block, m_B = 5 \, kg Height reached, h = 15 \, cm = 0.15 \, m Acceleration due to gravity, g = 9.8 \, m/s^2 Step 2: Calculate the total mass of the block-bullet system. M = m_b + m_B = 0.015 \, kg + 5 \, kg = 5.015 \, kg Step 3: Apply the principle of conservation of mechanical energy. The kinetic energy of the system immediately after impact is converted into gravitational potential energy at the maximum height. (1)/(2) M (v')^2 = Mgh Where v' is the velocity of the block-bullet system immediately after impact. Step 4: Solve for v'. (1)/(2) (v')^2 = gh (v')^2 = 2gh v' = sqrt(2gh) Substitute the values: v' = sqrt(2 × 9.8 \, m/s)^2 × 0.15 \, m v' = sqrt(2.94 \, m)^2/s^2 v' ≈ 1.7146 \, m/s Rounding to two decimal places, as given in the question: v' ≈ 1.71 \, m/s Thus, the magnitude of the velocity of the block-bullet system immediately after the bullet struck the block is 1.71 m/s. 4.3 Step 1: Identify the known values and the principle to be used. We use the conservation of linear momentum for the inelastic collision between the bullet and the block. Initial velocity of bullet = v_b (unknown) Initial velocity of block = v_B = 0 \, m/s Final velocity of block-bullet system = v' = 1.7146 \, m/s (from 4.2) Step 2: Apply the principle of conservation of linear momentum. Total momentum before collision = Total momentum after collision m_b v_b + m_B v_B = (m_b + m_B) v' Since v_B = 0: m_b v_b = (m_b + m_B) v' Step 3: Solve for v_b. v_b = ((m_b + m_B) v')/(m_b) Substitute the values: v_b = (0.015 \, kg + 5 \, kg) × 1.7146 \, m/s0.015 \, kg v_b = 5.015 \, kg × 1.7146 \, m/s0.015 \, kg v_b = 8.598979 \, kg·m/s0.015 \, kg v_b ≈ 573.265 \, m/s Rounding to one decimal place: The magnitude of the velocity of the bullet just before it strikes the block is 573.3 m/s. 4.4 This phenomenon is explained by Newton's Third Law of Motion. When the rifle fires, it exerts a forward force on the bullet (action). According to Newton's Third Law, the bullet exerts an equal and opposite backward force on the rifle (reaction). This backward force causes the rifle to recoil, pushing the police officer slightly backwards. Send me the next one 📸