91Ó°ÊÓ

In an elastic collision, the total momentum before and after the collision is conserved. Initially, only particle 1 is moving. Thus, the initial momentum is given by:\[ p_i = m_1 v_{1,i} = 0.30 \, \text{kg} \times 2.0 \, \text{m/s} = 0.60 \, \text{kg} \cdot \text{m/s} \]After the collision, let the velocities of particles 1 and 2 be \(v_{1,f}\) and \(v_{2,f}\) respectively. The momentum conservation equation becomes:\[ m_1 v_{1,i} = m_1 v_{1,f} + m_2 v_{2,f} \]

In a one-dimensional elastic collision, the kinetic energy is also conserved. Hence:\[ \frac{1}{2} m_1 v_{1,i}^2 = \frac{1}{2} m_1 v_{1,f}^2 + \frac{1}{2} m_2 v_{2,f}^2 \]This can be simplified to:\[ m_1 v_{1,i}^2 = m_1 v_{1,f}^2 + m_2 v_{2,f}^2 \]

Using the equations from steps 1 and 2, solve for the final velocities:From momentum conservation:\[ v_{1,f} + \frac{m_2}{m_1} v_{2,f} = v_{1,i} \]Substitute the values:\[ v_{1,f} + \frac{0.40}{0.30} v_{2,f} = 2.0 \]From energy conservation, we can use the relation derived for one-dimensional elastic collisions:\[ v_{1,f} = \frac{(m_1 - m_2)}{(m_1 + m_2)}v_{1,i} \]\[ v_{1,f} = \frac{(0.30 - 0.40)}{(0.30 + 0.40)} \times 2.0 = -\frac{1}{7} \times 2.0 = -0.29 \, \text{m/s} \]And for particle 2:\[ v_{2,f} = \frac{2m_1}{(m_1 + m_2)}v_{1,i} \]\[ v_{2,f} = \frac{2 \times 0.30}{(0.30 + 0.40)} \times 2.0 = \frac{6}{7} \times 2.0 = 1.71 \, \text{m/s} \]

Particle 2 moves towards the wall, hits it, and bounces back with no loss of speed at \( x_n = 70 \, \text{cm} = 0.70 \, \text{m} \). The collision with the wall changes the direction of particle 2 but not its speed. Thus, particle 2 continues to move at \( v_{2,f} = 1.71 \, \text{m/s} \) back towards particle 1.

Particle 1 is now moving left with speed \( 0.29 \, \text{m/s} \) after the initial collision, while particle 2 moves right at \( 1.71 \, \text{m/s} \). The relative speed between particles is:\[ v_\text{relative} = 1.71 + 0.29 = 2.0 \, \text{m/s} \]Particle 2 must travel an additional 70 cm (0.70 m) back to meet particle 1 since it collides with the wall and bounces back. The time taken for this can be calculated as:\[ t = \frac{0.70 \, \text{m}}{2.0 \, \text{m/s}} = 0.35 \, \text{s} \]During this time, particle 1 moves:\[ x = v_{1,f} \times t = 0.29 \times 0.35 = 0.1015 \, \text{m} \]Initial position of particle 1 before moving left was 0. This implies the collision occurs at 0 - 0.1015 m = -0.1015 m (10.15 cm to the left of the original starting position of particle 1).