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In an elastic collision, the total momentum before and after the collision remains constant. This principle is known as the conservation of momentum. Let's illustrate this concept with our specific problem involving two balls. Imagine a small ball of mass \(m\) and a larger ball of mass \(M\). When they are falling, both have the same downward velocity right before they hit the floor because they were dropped from the same height. This velocity can be calculated using the formula: \[ v = \sqrt{2gh} \] Here, \(v\) is the velocity just before impact, \(g\) is the acceleration due to gravity, and \(h\) is the height. The moment the larger ball hits the floor, it rebounds elastically with a velocity of \(\sqrt{2gh}\) upward. When the larger ball and smaller ball collide, the conservation of momentum tells us: \[ M \cdot \sqrt{2gh} + m \cdot (- \sqrt{2gh}) = M \cdot v_M + m \cdot v_m\] \(v_M\) and \(v_m\) represent the velocities of the larger and smaller balls after the collision. This equation ensures that the total momentum before and after the collision is the same. It essentially balances out the momentum each ball has gained or lost.

The conservation of energy principle states that the total kinetic energy before and after an elastic collision remains constant. Kinetic energy in our problem is calculated using the formula: \[ \frac{1}{2} mv^2 \] In this case, both balls fall from a height of \(h\), hence their initial kinetic energies just before hitting the ground can be calculated. When the larger ball rebounds, it has the same kinetic energy as when it fell: \[ \frac{1}{2} Mv^2 \] When both balls collide, the conservation of energy principle gives us: \[ \frac{1}{2} M(\sqrt{2gh})^2 + \frac{1}{2} m(\sqrt{2gh})^2 = \frac{1}{2}Mv_M^2 + \frac{1}{2}mv_m^2 \] This equation means that the total kinetic energy before and after the collision remains the same. Every energy transformation happens without losses. This principle is crucial because it helps us form the necessary equations to solve for the velocities post-collision.

The mass ratio in our elastic collision problem is key to solving the exercise. We need to find the ratio \( \frac{m}{M} \) that causes the larger ball to stop after colliding with the smaller ball. Using both the conservation of momentum and energy equations, we solve for these conditions. When the larger ball stops, its final velocity \(v_M\) is zero. Plugging this into our momentum equation, we get: \[ M \sqrt{2gh} = m v_m \] Simplifying, we find: \[ v_m = \sqrt{2gh} \frac{M}{m} \] We then use this result in our energy conservation equation, which simplifies into the following formulation: \[ Mgh + mgh = \frac{1}{2} m (\sqrt{2gh} \frac{M}{m} )^2 \] This leads us to: \[ M + m = \frac{M^2}{m} \] Solving this quadratic equation, we discover: \[ m / M \approx 1 / 3 \] This is the ratio of the masses which ensures the desired outcome. To determine how high the small ball will bounce, we look at the velocity after the collision. Given that: \[ v_m = 2 \sqrt{2gh} \] Using the energy conservation principle for upward motion, the small ball reaches a height: \[ h' = 4h \] This final step shows us that the small ball reaches a height four times the original drop height.