Question

1. a train carriage a of mass 500 kg is moving horizontally at 6.0ms⁻¹. it collides with another train carriage b of mass 700kg that is initially at rest, as shown in the diagram below. the graph below shows the variation with time t of the velocities of the two train carriages before, during and after the collision.

Explanation:

1. Explanation:

• Step 1: Recall the principle of conservation of momentum
• The law of conservation of momentum states that the total initial - momentum \(p_i\) of a system is equal to the total final - momentum \(p_f\) in the absence of external forces. The formula for momentum is \(p = mv\), where \(m\) is the mass and \(v\) is the velocity.
• The initial momentum of the system is \(p_i=m_Av_A + m_Bv_B\). Given \(m_A = 500\ kg\), \(v_A=6.0\ m/s\), and \(m_B = 700\ kg\), \(v_B = 0\ m/s\). So, \(p_i=500\times6.0+700\times0=3000\ kg\cdot m/s\).
• Step 2: Determine the final - momentum expression
• Let the final velocity of carriage A be \(v_{A_f}\) and of carriage B be \(v_{B_f}\). The final momentum of the system is \(p_f=m_Av_{A_f}+m_Bv_{B_f}\).
• From the graph, we can read the final velocities. After the collision, assume the time is \(t = 6.0\ s\). The velocity of carriage A, \(v_{A_f}=- 1.0\ m/s\) and the velocity of carriage B, \(v_{B_f}=5.0\ m/s\).
• Then \(p_f=500\times(-1.0)+700\times5.0\).
• Calculate \(p_f=-500 + 3500=3000\ kg\cdot m/s\).

1. Answer:

• The total initial momentum of the two - carriage system is \(3000\ kg\cdot m/s\) and the total final momentum is also \(3000\ kg\cdot m/s\), which verifies the conservation of momentum.

Answer:

1. Explanation:

• Step 1: Recall the principle of conservation of momentum
• The law of conservation of momentum states that the total initial - momentum \(p_i\) of a system is equal to the total final - momentum \(p_f\) in the absence of external forces. The formula for momentum is \(p = mv\), where \(m\) is the mass and \(v\) is the velocity.
• The initial momentum of the system is \(p_i=m_Av_A + m_Bv_B\). Given \(m_A = 500\ kg\), \(v_A=6.0\ m/s\), and \(m_B = 700\ kg\), \(v_B = 0\ m/s\). So, \(p_i=500\times6.0+700\times0=3000\ kg\cdot m/s\).
• Step 2: Determine the final - momentum expression
• Let the final velocity of carriage A be \(v_{A_f}\) and of carriage B be \(v_{B_f}\). The final momentum of the system is \(p_f=m_Av_{A_f}+m_Bv_{B_f}\).
• From the graph, we can read the final velocities. After the collision, assume the time is \(t = 6.0\ s\). The velocity of carriage A, \(v_{A_f}=- 1.0\ m/s\) and the velocity of carriage B, \(v_{B_f}=5.0\ m/s\).
• Then \(p_f=500\times(-1.0)+700\times5.0\).
• Calculate \(p_f=-500 + 3500=3000\ kg\cdot m/s\).

1. Answer:

• The total initial momentum of the two - carriage system is \(3000\ kg\cdot m/s\) and the total final momentum is also \(3000\ kg\cdot m/s\), which verifies the conservation of momentum.