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A 6000 kg railroad car moving at 5m/s collides into a stationary car with a mass of 4000 kg. If they couple together after the collision, what will be their combined velocity immediately after impact? 1m/s 2m/s 3m/s 4m/s

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A 6000 kg railroad car moving at 5m/s collides into a stationary car with a mass of 4000 kg. If they couple together after the collision, what will be their combined
velocity immediately after impact?
1m/s
2m/s
3m/s
4m/s

A 6000 kg railroad car moving at 5m/s collides into a stationary car with a mass of 4000 kg. If they couple together after the collision, what will be their combined velocity immediately after impact? 1m/s 2m/s 3m/s 4m/s

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OwenProfessional · Tutor for 6 years

Answer

To solve this problem, we will use the principle of conservation of momentum. According to this principle, if no external forces act on a system, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.Let's denote the mass of the moving railroad car as \(m_1 = 6000\) kg and its velocity as \(v_1 = 5\) m/s. The mass of the stationary car is \(m_2 = 4000\) kg and its velocity \(v_2 = 0\) m/s (since it's stationary).The total momentum before the collision (\(p_{\text{before}}\)) is the sum of the momenta of the two cars:\[p_{\text{before}} = m_1 \cdot v_1 + m_2 \cdot v_2\]\[p_{\text{before}} = (6000 \text{ kg}) \cdot (5 \text{ m/s}) + (4000 \text{ kg}) \cdot (0 \text{ m/s})\]\[p_{\text{before}} = 30000 \text{ kg} \cdot \text{m/s} + 0\]\[p_{\text{before}} = 30000 \text{ kg} \cdot \text{m/s}\]After the collision, the two cars couple together and move with a common velocity (\(v_{\text{after}}\)). The combined mass of the two cars is \(m_1 + m_2 = 6000 \text{ kg} + 4000 \text{ kg} = 10000 \text{ kg}\).The total momentum after the collision (\(p_{\text{after}}\)) is:\[p_{\text{after}} = (m_1 + m_2) \cdot v_{\text{after}}\]Since momentum is conserved:\[p_{\text{before}} = p_{\text{after}}\]\[30000 \text{ kg} \cdot \text{m/s} = (10000 \text{ kg}) \cdot v_{\text{after}}\]Now we can solve for \(v_{\text{after}}\):\[v_{\text{after}} = \frac{30000 \text{ kg} \cdot \text{m/s}}{10000 \text{ kg}}\]\[v_{\text{after}} = 3 \text{ m/s}\]Therefore, the combined velocity of the two railroad cars immediately after the collision is \(3 \text{ m/s}\).
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