A minibus taxi with an unknown number of passengers travels at \( 25 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) when it collides with a car with a mass of 1200 kg (passengers included), travelling at \( 15 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) in the opposite direction, as shown. During the collision the vehicles stick together and travel at \( 14 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) immediately after the collision in the direction of the original motion of the taxi. 6.1 Ignore friction. Use momentum principles to determine whether the minibus taxi was overloaded, that is, above the legal combined mass of 3500 kg . 6.2 Is the collision betiveen the vehicles elastic or inelastic? Support your answer with an appropriate calculation.
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Beyond the Answer
To solve this problem, we first need to apply the law of conservation of momentum and then analyze the type of collision that occurred. ### 6.1 Determine if the minibus taxi was overloaded: Let's denote the mass of the minibus taxi as \( m_t \) and the combined mass of the car as \( m_c = 1200 \, \text{kg} \). We know the velocities before and after the collision: - Velocity of the minibus taxi, \( v_t = 25 \, \text{m/s} \) - Velocity of the car, \( v_c = -15 \, \text{m/s} \) (negative because it's in the opposite direction) - Velocity after the collision, \( v_f = 14 \, \text{m/s} \) Using the conservation of momentum: \[ m_t \cdot v_t + m_c \cdot v_c = (m_t + m_c) \cdot v_f \] Plugging in the known values: \[ m_t \cdot 25 + 1200 \cdot (-15) = (m_t + 1200) \cdot 14 \] This simplifies to: \[ 25m_t - 18000 = 14m_t + 16800 \] Rearranging gives: \[ 25m_t - 14m_t = 18000 + 16800 \] \[ 11m_t = 34800 \] \[ m_t = \frac{34800}{11} \approx 3163.64 \, \text{kg} \] Now, calculating the total mass: \[ m_{total} = m_t + m_c \approx 3163.64 + 1200 \approx 4363.64 \, \text{kg} \] Since \( 4363.64 \, \text{kg} > 3500 \, \text{kg} \), it appears that the minibus taxi was indeed overloaded. ### 6.2 Type of collision (elastic or inelastic): For a collision to be elastic, both momentum and kinetic energy must be conserved. We already know momentum was conserved, so we check kinetic energy before and after the collision. **Kinetic Energy Before Collision:** \[ KE_{initial} = \frac{1}{2} m_t v_t^2 + \frac{1}{2} m_c v_c^2 = \frac{1}{2}(m_t)(25^2) + \frac{1}{2}(1200)(15^2) \] \[ = \frac{1}{2} m_t (625) + \frac{1}{2}(1200)(225) \] \[ = 312.5 m_t + 135000 \] **Kinetic Energy After Collision:** \[ KE_{final} = \frac{1}{2} (m_t + m_c) v_f^2 \] \[ = \frac{1}{2} (m_t + 1200)(14^2) = \frac{1}{2} (m_t + 1200)(196) \] \[ = 98(m_t + 1200) \] To analyze if the kinetic energy is conserved, we'd compare \( KE_{initial} \) and \( KE_{final} \). Since the kinetic energy is not likely conserved given the collision involved two objects sticking together, we can already conclude the collision is inelastic. Thus, the collision between the vehicles is **inelastic** since kinetic energy is not conserved, even though momentum is conserved.