Compare the momenta of both trolleys for each combination bofore and after explosion.

To compare the momenta of both trolleys before and after an explosion, we'll analyze the situation using the principle of **conservation of momentum**. This principle states that if no external forces act on a system, the total momentum of the system remains constant before and after an event, such as an explosion. Let's consider the following scenarios for clarity: 1. **Before the Explosion** 2. **After the Explosion** Assume we have two trolleys, **Trolley A** and **Trolley B**. We'll examine their momenta in both scenarios. --- ### 1. Before the Explosion **Assumptions:** - Both trolleys are part of a single system. - They may be moving together with a common velocity or at rest. **Momentum Calculations:** - **Total Momentum Before Explosion (\(p_{\text{total, before}}\))** The total momentum is the vector sum of the individual momenta of Trolley A and Trolley B. \[ p_{\text{total, before}} = p_A + p_B = m_A v_A + m_B v_B \] Where: - \(m_A\) and \(m_B\) are the masses of Trolley A and Trolley B, respectively. - \(v_A\) and \(v_B\) are the velocities of Trolley A and Trolley B, respectively. ### 2. After the Explosion **Assumptions:** - The explosion is internal to the system (no external forces). - The trolleys move apart with new velocities due to the explosion. **Momentum Calculations:** - **Total Momentum After Explosion (\(p_{\text{total, after}}\))** According to the conservation of momentum: \[ p_{\text{total, after}} = p_A' + p_B' = m_A v_A' + m_B v_B' \] Where: - \(v_A'\) and \(v_B'\) are the velocities of Trolley A and Trolley B after the explosion. --- ### **Comparison of Momenta** 1. **Total Momentum:** - **Before Explosion:** \(p_{\text{total, before}} = m_A v_A + m_B v_B\) - **After Explosion:** \(p_{\text{total, after}} = m_A v_A' + m_B v_B'\) **Conservation Principle:** \[ p_{\text{total, before}} = p_{\text{total, after}} \] This means that the combined momentum of both trolleys remains the same before and after the explosion. 2. **Individual Momenta:** - **Trolley A:** - **Before Explosion:** \(p_A = m_A v_A\) - **After Explosion:** \(p_A' = m_A v_A'\) - **Trolley B:** - **Before Explosion:** \(p_B = m_B v_B\) - **After Explosion:** \(p_B' = m_B v_B'\) While the **total** momentum of the system remains unchanged, the **individual** momenta of Trolley A and Trolley B can change due to the redistribution of energy and momentum during the explosion. --- ### **Possible Scenarios After Explosion** 1. **Trolleys Move in Opposite Directions:** If the trolleys move in opposite directions after the explosion, their momenta will have opposite signs (assuming direction is represented by the sign of velocity). The magnitudes of their momenta will adjust such that their vector sum equals the original total momentum. \[ m_A v_A' = -m_B v_B' \quad \text{(if initial total momentum was zero)} \] 2. **One Trolley Moves While the Other Remains Stationary:** If one trolley gains all the momentum, the other may remain stationary or move in the same direction with adjusted momentum to keep the total momentum conserved. 3. **Both Trolleys Move in the Same Direction:** Both trolleys may continue moving in the same direction as before the explosion but with different speeds. Their combined momentum will still equal the initial total momentum. --- ### **Key Takeaways** - **Total Momentum Conservation:** Regardless of the explosion, the **sum** of the momenta of both trolleys before and after the explosion remains **constant**, provided no external forces act on the system. - **Individual Momentum Changes:** The explosion redistributes momentum between the trolleys, changing their individual momenta. The direction and magnitude of each trolley's momentum after the explosion depend on the specifics of the explosion and the system's initial conditions. - **Vector Nature of Momentum:** Momentum is a vector quantity. Both magnitude and direction are essential. Changes in direction post-explosion play a crucial role in how individual momenta compare before and after the event. --- ### **Example Problem** **Situation:** - Two trolleys, A and B, are moving together on a frictionless track with velocities \(v_A = 2 \, \text{m/s}\) and \(v_B = 3 \, \text{m/s}\) respectively. - Masses are \(m_A = 2 \, \text{kg}\) and \(m_B = 3 \, \text{kg}\). - An explosion occurs, and their velocities change to \(v_A' = 4 \, \text{m/s}\) and \(v_B' = 1 \, \text{m/s}\). **Calculations:** - **Total Momentum Before Explosion:** \[ p_{\text{total, before}} = (2 \, \text{kg})(2 \, \text{m/s}) + (3 \, \text{kg})(3 \, \text{m/s}) = 4 + 9 = 13 \, \text{kg·m/s} \] - **Total Momentum After Explosion:** \[ p_{\text{total, after}} = (2 \, \text{kg})(4 \, \text{m/s}) + (3 \, \text{kg})(1 \, \text{m/s}) = 8 + 3 = 11 \, \text{kg·m/s} \] **Analysis:** Here, the total momentum before the explosion (13 kg·m/s) does not equal the total momentum after the explosion (11 kg·m/s). This discrepancy suggests that either external forces acted on the system during the explosion or there was an error in the measurements or calculations. **Correct Conservation:** To satisfy conservation of momentum, the total momentum after the explosion must equal 13 kg·m/s. If the example's numbers do not satisfy this, adjustments must be made to the velocities post-explosion to ensure that: \[ (2 \, \text{kg})v_A' + (3 \, \text{kg})v_B' = 13 \, \text{kg·m/s} \] --- ### **Conclusion** When analyzing explosions (or any interactions) involving multiple objects: - **Always start with the conservation of total momentum.** - **Examine how individual momenta change to maintain the total momentum.** - **Consider directions, as momentum is a vector quantity.** Understanding these principles allows for accurate comparisons of the momenta of objects before and after events like explosions.