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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2013 November Paper 1 Q4 4.1 At the instant the package leaves the boy's hand, the package moves east. By the principle of conservation of momentum, the boy must move in the opposite direction to conserve the total momentum of the boy-package system, which was initially zero. The boy moves WEST 4.2 The law that explains the direction in which the boy experiences a force is Newton's Third Law of Motion. Statement: When object A exerts a force on object B, object B simultaneously exerts an equal in magnitude and opposite in direction force on object A. 4.3 Step 1: Identify the given values. Mass of boy m_b = 60 kg Mass of package m_p = 5 kg Initial velocity of boy u_b = 0 m/s Initial velocity of package u_p = 0 m/s Final velocity of package v_p = 4 m/s (east, let's take east as positive) Final velocity of boy v_b = ? Step 2: Apply the principle of conservation of linear momentum. m_b u_b + m_p u_p = m_b v_b + m_p v_p (60 kg)(0 m/s) + (5 kg)(0 m/s) = (60 kg) v_b + (5 kg)(4 m/s) 0 + 0 = 60 v_b + 20 0 = 60 v_b + 20 60 v_b = -20 v_b = (-20)/(60) v_b = -(1)/(3) m/s The magnitude of the velocity of the boy is 0.33 m/s (rounded to two decimal places). 4.4 4.4.1 If the boy throws the same package at a higher velocity in the same direction, the momentum of the package (m_p v_p) will increase. To conserve the total momentum of the system (which remains zero), the magnitude of the boy's momentum (m_b v_b) must also increase. Since the boy's mass (m_b) is constant, his velocity (v_b) must increase in magnitude. The answer to QUESTION 4.3 will INCREASES 4.4.2 If the boy throws a package of double the mass at the same velocity as in QUESTION 4.3, the momentum of the package (m_p v_p) will double. To conserve the total momentum of the system, the magnitude of the boy's momentum (m_b v_b) must also double. Since the boy's mass (m_b) is constant, his velocity (v_b) must double in magnitude. The answer to QUESTION 4.3 will INCREASES Explanation: The momentum of the package (m_p v_p) will double. By the principle of conservation of linear momentum, the magnitude of the boy's momentum (m_b v_b) must also double. Since the boy's mass (m_b) remains constant, his velocity (v_b) must increase. --- 2015 Metro South Prelim Paper Q4 4.1 The principle of conservation of linear momentum states that: The total linear momentum of an isolated system remains constant in both magnitude and direction. 4.2 Step 1: Identify the given values. Mass of wooden block m_w = 2 kg Initial velocity of wooden block u_w = 5 m/s (to the right, let's take right as positive) Mass of crate m_c = 9 kg Initial velocity of crate u_c = 0 m/s Final velocity of crate v_c = 1 m/s (to the right) Final velocity of wooden block v_w = ? Step 2: Apply the principle of conservation of linear momentum. m_w u_w + m_c u_c = m_w v_w + m_c v_c (2 kg)(5 m/s) + (9 kg)(0 m/s) = (2 kg) v_w + (9 kg)(1 m/s) 10 + 0 = 2 v_w + 9 10 = 2 v_w + 9 2 v_w = 10 - 9 2 v_w = 1 v_w = (1)/(2) m/s v_w = 0.5 m/s The magnitude of the velocity of the wooden block immediately after the collision is 0.5 m/s 4.3 Step 1: Identify the given values for the crate. Mass of crate m_c = 9 kg Initial velocity of crate u_c = 0 m/s Final velocity of crate v_c = 1 m/s Time of collision t = 0.6 s Force exerted on the crate F = ? Step 2: Apply the impulse-momentum theorem to the crate. F t = p_c F t = m_c v_c - m_c u_c F (0.6 s) = (9 kg)(1 m/s) - (9 kg)(0 m/s) F (0.6) = 9 - 0 F (0.6) = 9 F = (9)/(0.6) F = 15 N The force the wooden block exerts on the crate during the collision is 15 N