An object of mass \(m_1\) and an object of mass \(m_2\) are initially attached to each other and at rest. At the instant \(t=0\) they are pushed apart by a spring that has been compressed between them. After \(t=0\), \(v_1=2 \mathrm{~m} / \mathrm{s}\) and \(v_2=-3 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) \(m_1 / m_2=3 / 2\) (B) \(m_1=2 \mathrm{~kg}\) and \(m_2=3 \mathrm{~kg}\) (C) \(m_1=3 \mathrm{~kg}\) and \(m_2=2 \mathrm{~kg}\) (D) \(m_1 / m_2=2 / 3\) (E) \(m_1 / m_2=-3 / 2\)

When we talk about the principle of conservation of linear momentum, we are referring to a fundamental law of physics, stating that the momentum of a system remains constant if no external force acts upon it.

The momentum of an object is given by the product of its mass \( m \) and its velocity \( v \)—expressed in the formula \( p = mv \). In a situation where two objects interact with each other without external influence, the total momentum before the interaction is equal to the total momentum after it.

In our exercise, we used this principle to solve for the mass ratio of two objects pushed apart by a spring. Since the force exerted by the spring is internal to the system, the total momentum must remain constant. The scenario perfectly illustrates momentum conservation, as there is zero net external force acting on the objects.

The mass ratio calculation is crucial in this scenario for figuring out the relationship between the masses of the two objects. The mass ratio, written as \( m_1 / m_2 \) or the mass of object 1 divided by the mass of object 2, can give us insight into relative masses without needing absolute values.

In the problem, we are looking for this ratio based on given velocities post-separation. Using the conservation of momentum, we set up an equation that, once the velocities are plugged in, allows us to solve for the mass ratio directly. It demonstrates how understanding a concept can lead to solutions even when not all data is given explicitly.

A common mistake when calculating mass ratios is to confuse the direction of velocities or to impose external force assumptions on an isolated system. It's crucial to treat the velocities sign with care, as it indicates direction and will affect the mass ratio.

Problems involving concepts like momentum conservation often appear on standardized tests like the Physics SAT. These questions are built to test students' understanding of fundamental physics concepts and their ability to apply these concepts in different contexts.

Physics SAT questions are designed to be clear and precise, though sometimes they are presented in a way that makes them appear more complicated than they are. It's important for students to have strong problem-solving strategies, such as breaking the problem down into known and unknown variables and using relevant physics laws or formulas.

This kind of problem is typical of what might be found on the Physics SAT: it's a single concept check (momentum conservation), but it requires the student to consider the implications of the concept for mass and velocity.

For students preparing for AP Physics, it's essential to review key topics that are frequently tested, such as momentum conservation. In-depth understanding and the ability to solve complex problems around these topics are critical for success on the AP exam.

During an AP Physics review, students should practice various problems that require application of the conservation laws—energy, momentum, and charge. Problem-solving techniques such as isolation of variables and drawing free-body diagrams can be very helpful.

The problem we've tackled would be a prime candidate for material in an AP Physics review session, encompassing both conceptual understanding of momentum conservation and practical application through mass ratio calculation. Students should note how initial conditions are used to simplify equations and lead to a solution, a strategy often required in AP Physics exams.