A car, mass 1600 kg , travelling east at \( 30 \mathrm{~m} \cdot \mathrm{~s}^{-1} \), collides head-on with a minibus, mass 3000 kg , travelling west at \( 20 \mathrm{~m} \cdot \mathrm{~s}^{-1} \). The vehicles move as one after the collision. (i) What is the velocity of the vehicles after collision, and (ii) Proof whether the collision elastic or not?

To solve this problem, we will use the principles of conservation of momentum and the definition of elastic collisions. ### Known Conditions: - Mass of the car, \( m_1 = 1600 \, \text{kg} \) - Velocity of the car before collision, \( v_1 = 30 \, \text{m/s} \) (east) - Mass of the minibus, \( m_2 = 3000 \, \text{kg} \) - Velocity of the minibus before collision, \( v_2 = -20 \, \text{m/s} \) (west, hence negative) ### Step 1: Calculate the velocity after the collision Using the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. The formula for momentum is: \[ p = mv \] The total momentum before the collision is: \[ p_{\text{initial}} = m_1 v_1 + m_2 v_2 \] Substituting the values: \[ p_{\text{initial}} = (1600 \, \text{kg} \cdot 30 \, \text{m/s}) + (3000 \, \text{kg} \cdot -20 \, \text{m/s}) \] Calculating this: \[ p_{\text{initial}} = 48000 \, \text{kg m/s} - 60000 \, \text{kg m/s} = -12000 \, \text{kg m/s} \] Let \( v_f \) be the final velocity of both vehicles after the collision. The total mass after the collision is \( m_1 + m_2 \). The total momentum after the collision is: \[ p_{\text{final}} = (m_1 + m_2) v_f \] Setting the initial momentum equal to the final momentum: \[ -12000 \, \text{kg m/s} = (1600 \, \text{kg} + 3000 \, \text{kg}) v_f \] This simplifies to: \[ -12000 \, \text{kg m/s} = 4600 \, \text{kg} \cdot v_f \] Now, solving for \( v_f \): \[ v_f = \frac{-12000 \, \text{kg m/s}}{4600 \, \text{kg}} \approx -2.61 \, \text{m/s} \] ### Step 2: Determine if the collision is elastic A collision is elastic if both momentum and kinetic energy are conserved. We already have the momentum conservation. Now, we will check the kinetic energy before and after the collision. The kinetic energy before the collision is: \[ KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \] Calculating this: \[ KE_{\text{initial}} = \frac{1}{2} (1600 \, \text{kg}) (30 \, \text{m/s})^2 + \frac{1}{2} (3000 \, \text{kg}) (-20 \, \text{m/s})^2 \] \[ = \frac{1}{2} (1600) (900) + \frac{1}{2} (3000) (400) \] \[ = 720000 \, \text{J} + 600000 \, \text{J} = 1320000 \, \text{J} \] The kinetic energy after the collision is: \[ KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v_f^2 \] \[ = \frac{1}{2} (4600 \, \text{kg}) (-2.61 \, \text{m/s})^2 \] \[ = \frac{1}{2} (4600) (6.8121) \approx 15666.69 \, \text{J} \] ### Conclusion: 1. The velocity of the vehicles after the collision is approximately \( -2.61 \, \text{m/s} \) (indicating they move west). 2. Since the initial kinetic energy \( 1320000 \, \text{J} \) is not equal to the final kinetic energy \( 15666.69 \, \text{J} \), the collision is not elastic. Thus, the collision is inelastic.