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 is the ratio between the total kinetic energy of the two gliders after the string is burned to the total kinetic energy of the gliders before the string is burned?

Before we crunch those numbers, let’s consider the total kinetic energy before the string was burned! Both gliders were moving together at \( 0.460 \, \text{m/s} \), so their combined initial kinetic energy can be expressed as \( KE_{initial} = 2 \times \frac{1}{2} m (0.460)^2 = m (0.460)^2 \). Now, when the string is burned, we see one glider shoot off to the right at \( 1.300 \, \text{m/s} \). We can assume conservation of momentum to find the velocity of the second glider. Using the equation \( m \cdot 0.460 + m \cdot 0.460 = m \cdot 1.300 + m \cdot v_2 \), we get that \( v_2 = 0.460 - 1.300 = -0.840 \, \text{m/s} \) (the negative sign indicates it's moving in the opposite direction). So, the total kinetic energy after the string is burned becomes \( KE_{final} = \frac{1}{2} m (1.300)^2 + \frac{1}{2} m (0.840)^2 \). When we calculate this, we get: \[ KE_{final} = 0.5m(1.300^2) + 0.5m(0.840^2) = 0.5m(1.690 + 0.7056) = 0.5 m (2.3956) \] Now, to find the ratio of total kinetic energies, we take \( \frac{KE_{final}}{KE_{initial}} \): \[ \text{Ratio} = \frac{0.5 m (2.3956)}{m \cdot (0.460)^2} = \frac{2.3956}{0.2116} \approx 11.31 \] Thus, the final calculation gives a ratio of approximately **11.31**. Kinetic energy boost, anyone?