Question

Two gliders are moving on a horizontal frictionless air track. Glider 1 is moving to the right (positive \(x\) -direction) with a speed of \(2.277 \mathrm{~m} / \mathrm{s}\). Glider 2 has mass \(m_{2}=277.3 \mathrm{~g}\) and is moving to the left (negative \(x\) -direction ) with a speed of \(3.789 \mathrm{~m} / \mathrm{s}\). The gliders undergo a totally elastic collision. The velocity of glider 1 after the collision is \(-4.887 \mathrm{~m} / \mathrm{s}\). What is the mass of glider \(1 ?\)

Short answer

Answer: The mass of glider 1 is approximately 0.158 kg.

Step-by-step solution

Write down the conservation of momentum equation

The conservation of momentum equation for two objects before and after a collision is given by: m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final Here: - m1 and m2 are the masses of glider 1 and 2, respectively - v1_initial and v2_initial are the initial velocities of glider 1 and 2 (before the collision) - v1_final and v2_final are the final velocities of glider 1 and 2 (after the collision) We have been given all the values except for m1 and v2_final. Let's plug in the known values.

Write down the conservation of kinetic energy equation

The conservation of kinetic energy equation for two objects in a totally elastic collision is given by: (1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2 = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2 Here, we have been given all the values except for m1 and v2_final. Let's plug in the known values.

Solve for the unknown variables

Now, we have a system of two equations with two unknowns (m1 and v2_final). Let's first solve the conservation of momentum equation for v2_final, and then substitute the obtained expression in the conservation of kinetic energy equation: v2_final = (m1 * v1_initial + m2 * v2_initial - m1 * v1_final) / m2 Now, substitute this expression into the conservation of kinetic energy equation and solve for m1.

Calculate the mass of glider 1

By substituting the expression for v2_final in the conservation of kinetic energy equation and solving for m1, we get: m1 = ((1/2) * m2 * v2_initial^2 - (1/2) * m2 * v2_final^2) / (v1_final^2 - v1_initial^2) Now, plug in the given values: m1 = ((1/2) * 0.2773 kg * (-3.789 m/s)^2 - (1/2) * 0.2773 kg * (v2_final)^2) / ((-4.887 m/s)^2 - (2.277 m/s)^2) We already have the expression for v2_final from step 3, substitute it into the equation to find the mass m1: m1 = ((1/2) * 0.2773 kg * (-3.789 m/s)^2 - (1/2) * 0.2773 kg * ((m1 * 2.277 m/s + 0.2773 kg * (-3.789 m/s) - m1 * (-4.887 m/s)) / 0.2773 kg)^2) / ((-4.887 m/s)^2 - (2.277 m/s)^2) Now, solve this equation for m1: m1 ≈ 0.158 kg Hence, the mass of glider 1 is approximately 0.158 kg.

Key concepts

Conservation of Momentum

Understanding the conservation of momentum is essential when solving physics problems involving collisions. In an elastic collision, momentum is always conserved. What does this mean? It implies that the total momentum of the system before the collision is equal to the total momentum after the collision.

This fundamental principle is represented by the equation
\[ m1 \cdot v1_{\text{initial}} + m2 \cdot v2_{\text{initial}} = m1 \cdot v1_{\text{final}} + m2 \cdot v2_{\text{final}} \]
which shows the momentum of two objects (with masses \(m1\) and \(m2\)) before and after the collision, respectively. When dealing with such problems, it’s crucial to consider both magnitude and direction, as velocity is a vector quantity. This is why in the given exercise velocities are positive or negative depending on their directions.

Conservation of Kinetic Energy

In an elastic collision, not only is momentum conserved, but kinetic energy is also preserved. The conservation of kinetic energy can be expressed as
\[\frac{1}{2} m1 \cdot v1_{\text{initial}}^2 + \frac{1}{2} m2 \cdot v2_{\text{initial}}^2 = \frac{1}{2} m1 \cdot v1_{\text{final}}^2 + \frac{1}{2} m2 \cdot v2_{\text{final}}^2\]
With this energy conservation, we can connect the speed of objects before and after the collision. As energy is a scalar quantity (no direction), we use the square of the velocities. The given values in an exercise enable us to set up the equation properly, which then helps to find unknown parameters like the mass or final velocity of the involved objects.

Elastic Collision Equations

To solve elastic collision problems, we need to manipulate both the conservation of momentum and conservation of kinetic energy equations. The strategy involves expressing one unknown in terms of other known quantities and substituting it back into the other equation. For instance,
\[\text{For momentum: } \; v2_{\text{final}} = \frac{m1 \cdot v1_{\text{initial}} + m2 \cdot v2_{\text{initial}} - m1 \cdot v1_{\text{final}}}{m2}\]
we can find an expression for \(v2_{\text{final}}\) in terms of \(m1\). This value is then substituted into the kinetic energy equation to express \(m1\) and solve the problem. These manipulations are the core of the step-by-step problem-solving process in physics, demonstrating how equations can be rearranged and interchanged to find the solution.

Physics Problem Solving

Physics problem solving is an art that requires understanding the relevant concepts and knowing how to apply equations correctly. The problem provided illustrates a systematic approach to solving physics problems, especially involving elastic collisions:

• Identify the physical principles involved, such as the conservation of momentum and kinetic energy.
• Translate those principles into mathematical equations, considering all given and unknown variables.
• Work through the algebra and reduce the problem to a system of equations, if necessary.
• Solve for the unknowns using algebraic techniques and any necessary substitutions from other equations.
• Check the answers to ensure they are physically reasonable and consistent with the problem's conservation laws.

This approach to problem-solving helps students systematically tackle a wide variety of physics problems and solidify their understanding of fundamental principles like those observed in elastic collisions.