Required information In a lab experiment, two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of \( 0.460 \mathrm{~m} / \mathrm{s} \), one student holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at \( 1.300 \mathrm{~m} / \mathrm{s} \). (the string and spring both have negligible mass). What velocity does the other glider have? Enter a positive value if the velocity is toward right and enter a negative value if the velocity is toward left. \( \mathrm{m} / \mathrm{s} \)

In this scenario, we can use the principle of conservation of momentum. Before the string is burned, the total momentum of the system (both gliders) is simply the combined mass of the two gliders times their common velocity. After the string is burned, the momentum of the system is the sum of the individual momenta of the two gliders. Let’s denote the mass of each glider as \(m\). The initial momentum \(p_i\) before the string is burned is: \[ p_i = 2m \cdot 0.460 \, \mathrm{m/s} = 0.920m \, \mathrm{kg \cdot m/s} \] After the spring is released, one glider moves to the right at \(1.300 \, \mathrm{m/s}\), so its momentum \(p_1\) is: \[ p_1 = m \cdot 1.300 \, \mathrm{m/s} = 1.300m \, \mathrm{kg \cdot m/s} \] Let the velocity of the other glider be \(v_2\). Therefore, its momentum \(p_2\) will be: \[ p_2 = m \cdot v_2 \] According to the conservation of momentum: \[ p_i = p_1 + p_2 \] Substituting the values we have: \[ 0.920m = 1.300m + mv_2 \] Dividing the entire equation by \(m\) (and cancelling) gives us: \[ 0.920 = 1.300 + v_2 \] Solving for \(v_2\): \[ v_2 = 0.920 - 1.300 = -0.380 \, \mathrm{m/s} \] So the velocity of the other glider is \(-0.380 \, \mathrm{m/s}\), indicating that it moves toward the left.