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
Step 1: Calculate the man's speed just before he grabs the boy. This is an inelastic collision where the man and boy stick together. The principle of conservation of linear momentum applies. Given: Mass of man, m_m = 87 kg Mass of boy, m_b = 22 kg Initial velocity of boy, v_bi = 0 m/s (standing still) Final velocity of man and boy combined, v_f = 2.4 m/s Let v_mi be the man's speed just before he grabs the boy. According to the principle of conservation of linear momentum: m_m v_mi + m_b v_bi = (m_m + m_b) v_f Substitute the known values: (87 kg) v_mi + (22 kg)(0 m/s) = (87 kg + 22 kg)(2.4 m/s) 87 v_mi + 0 = (109)(2.4) 87 v_mi = 261.6 v_mi = (261.6)/(87) v_mi = 3.00689... m/s v_mi ≈ 3.01 m/s The man's speed just before he grabs the boy is 3.01 m/s. Step 2: Determine if the collision is elastic and support the answer with a calculation. To determine if the collision is elastic, we compare the total kinetic energy before and after the collision. An elastic collision conserves kinetic energy, while an inelastic collision does not. Calculate the total initial kinetic energy (KE_i): KE_i = (1)/(2) m_m v_mi^2 + (1)/(2) m_b v_bi^2 Using v_mi = 3.00689... m/s: KE_i = (1)/(2) (87 kg) (3.00689... m/s)^2 + (1)/(2) (22 kg) (0 m/s)^2 KE_i = (1)/(2) (87)(9.0414) + 0 KE_i = 393.299... J KE_i ≈ 393.30 J Calculate the total final kinetic energy (KE_f): KE_f = (1)/(2) (m_m + m_b) v_f^2 KE_f = (1)/(2) (87 kg + 22 kg) (2.4 m/s)^2 KE_f = (1)/(2) (109)(5.76) KE_f = 313.92 J Compare KE_i and KE_f: Since KE_i = 393.30 J and KE_f = 313.92 J, KE_i ≠ KE_f. Therefore, the collision is not elastic. The collision is inelastic because the total kinetic energy before the collision (393.30 J) is not equal to the total kinetic energy after the collision (313.92 J). Step 3: Calculate the magnitude of the force exerted by the gravel surface on the man and the boy. After the collision, the combined mass moves at 2.4 m/s. They then decelerate to rest over a distance of 2 m. Given: Combined mass, M = m_m + m_b = 87 kg + 22 kg = 109 kg Initial velocity on gravel, u = 2.4 m/s Final velocity on gravel, v = 0 m/s Distance travelled on gravel, s = 2 m First, calculate the acceleration (a) using a kinematic equation: v^2 = u^2 + 2as (0 m/s)^2 = (2.4 m/s)^2 + 2a(2 m) 0 = 5.76 + 4a -5.76 = 4a a = (-5.76)/(4) a = -1.44 m/s^2 The negative sign indicates deceleration. Next, calculate the magnitude of the force (F) using Newton's second law: F = Ma F = (109 kg)(-1.44 m/s^2) F = -156.96 N The magnitude of the force is the absolute value of this result. The magnitude of the force exerted by the gravel surface on the man and the boy is 157 N.