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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8. (a) (i) The law of conservation of linear momentum states that for an isolated system, the total linear momentum remains constant in the absence of external forces. 8. (a) (ii) Given: Mass of ball 1, m_1 = 1.0 kg Initial velocity of ball 1, u_1 = 4.0 m/s Mass of ball 2, m_2 = 2.0 kg Initial velocity of ball 2, u_2 = 0 m/s (stationary) The collision is perfectly elastic. Step 1: Apply the principle of conservation of linear momentum. The total momentum before collision equals the total momentum after collision. m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 Substitute the given values: (1.0 kg)(4.0 m/s) + (2.0 kg)(0 m/s) = (1.0 kg)v_1 + (2.0 kg)v_2 4.0 = v_1 + 2v_2 (*) Step 2: Apply the principle for a perfectly elastic collision using relative velocities. For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: u_1 - u_2 = v_2 - v_1 Substitute the given values: 4.0 m/s - 0 m/s = v_2 - v_1 4.0 = v_2 - v_1 (**) Step 3: Solve the system of equations. From equation (**), express v_2 in terms of v_1: v_2 = v_1 + 4.0 Substitute this expression for v_2 into equation (*): 4.0 = v_1 + 2(v_1 + 4.0) 4.0 = v_1 + 2v_1 + 8.0 4.0 = 3v_1 + 8.0 3v_1 = 4.0 - 8.0 3v_1 = -4.0 v_1 = -(4.0)/(3) m/s v_1 ≈ -1.33 m/s Step 4: Calculate v_2 using the value of v_1. v_2 = v_1 + 4.0 v_2 = -(4.0)/(3) + 4.0 v_2 = -(4.0)/(3) + (12.0)/(3) v_2 = (8.0)/(3) m/s v_2 ≈ 2.67 m/s The velocities of the balls after collision are: Velocity of the first ball: -1.33 m/s Velocity of the second ball: 2.67 m/s Send me the next one 📸