Problem 29 - College Physics

Question

A \(0.300 \mathrm{~kg}\) glider is moving to the right on a frictionless, horizontal air track with a speed of \(0.80 \mathrm{~m} / \mathrm{s}\) when it makes a head-on collision with a stationary \(0.150 \mathrm{~kg}\) glider. (a) Find the magnitude and direction of the final velocity of each glider if the collision is elastic. (b) Find the final kinetic energy of each glider.

Step-by-Step Solution

Answer

The final velocities are -0.4 m/s for the 0.300 kg glider and 1.2 m/s for the 0.150 kg glider. The final kinetic energies are 0.024 J and 0.108 J, respectively.

Step 1: Understand the Problem

We have two gliders undergoing an elastic collision on a frictionless track. The gliders are of masses 0.300 kg and 0.150 kg. The first glider has an initial velocity of 0.80 m/s, while the second is stationary. We need to find their final velocities and kinetic energies after the collision.

Step 2: Apply the Conservation of Momentum

In an elastic collision, the total momentum before and after the collision is conserved. The equation for momentum conservation is:\[m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}\]where- \( m_1 = 0.300 \text{ kg} \), \( v_{1i} = 0.80 \text{ m/s} \)- \( m_2 = 0.150 \text{ kg} \), \( v_{2i} = 0 \text{ m/s} \)Substitute the values:\[0.300 \times 0.80 + 0.150 \times 0 = 0.300 v_{1f} + 0.150 v_{2f}\]\[0.24 = 0.300 v_{1f} + 0.150 v_{2f}\]

Step 3: Apply the Conservation of Kinetic Energy

For elastic collisions, the total kinetic energy is also conserved. The equation is:\[\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2\]Substituting the initial values:\[\frac{1}{2} \times 0.300 \times (0.80)^2 + \frac{1}{2} \times 0.150 \times (0)^2 = \frac{1}{2} \times 0.300 \times v_{1f}^2 + \frac{1}{2} \times 0.150 \times v_{2f}^2\]Simplifies to:\[0.096 = 0.150 v_{1f}^2 + 0.075 v_{2f}^2\]

Step 4: Solve the System of Equations

We have two equations:1. \(0.24 = 0.300 v_{1f} + 0.150 v_{2f}\)2. \(0.096 = 0.150 v_{1f}^2 + 0.075 v_{2f}^2\)First, solve equation (1) for \(v_{1f}\):\[v_{1f} = \frac{0.24 - 0.150 v_{2f}}{0.300}\]Substitute \(v_{1f}\) in equation (2) and solve for \(v_{2f}\). After simplification, you'll find:\(v_{2f} = 1.2 \text{ m/s}\),and substituting this back for \(v_{1f}\):\(v_{1f} = -0.4 \text{ m/s}\).

Step 5: Calculate Final Kinetic Energies

To find the final kinetic energies, use:\[KE_f = \frac{1}{2} m v^2\]For the 0.300 kg glider:\[KE_{1f} = \frac{1}{2} \times 0.300 \times (-0.4)^2 = 0.024 \text{ J}\]For the 0.150 kg glider:\[KE_{2f} = \frac{1}{2} \times 0.150 \times (1.2)^2 = 0.108 \text{ J}\]

Key Concepts

Conservation of Momentum

In any collision, the principle of conservation of momentum is crucial to understand how objects exchange forces and velocities. Momentum is defined as the product of mass and velocity and is a vector quantity; this means both magnitude and direction are important.

• Momentum conservation occurs because there are no external forces acting on the system.
• Before and after an elastic collision, the total momentum remains constant.
• For our glider example, the equation for momentum conservation is \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \).

In our scenario, with a 0.300 kg glider moving at 0.80 m/s and a stationary 0.150 kg glider, the total initial momentum is \(0.24 \text{ kg m/s}\). After the collision, solving the momentum equation gives us the final velocities of the gliders, maintaining the total momentum.

Conservation of Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In an elastic collision, both momentum and kinetic energy are conserved, meaning there is no loss of energy due to heat, sound, or deformation.

• Kinetic energy conservation can be described with the equation \( \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \).
• Along with momentum equations, this provides a system of equations necessary to solve for unknown variables, like final velocities.

For our gliders, since the collision is elastic, we find the initial kinetic energy of the system is equal to the final, distributing between the two gliders according to their speeds and masses post-collision.

Physics Problem Solving

Solving physics problems involving collisions can be simplified by using systematic approaches. Here's a helpful method:

• **Identify given data** – List out the known values such as masses, initial velocities, and collision type.
• **Write equations** – Use conservation laws like momentum and kinetic energy to create equations related to the problem.
• **Solve equations** – Use algebraic methods to find unknown values such as final velocities or final energies.
• **Interpret results** – Confirm that the results make sense in a real-world context and adhere to the principles of conservation.

In this example, once you have written down the conservation equations, solving involves basic algebra to find the final velocities and kinetic energies. Always check both mathematical accuracy and physical feasibility of your answers.

Frictionless Surface Dynamics

A frictionless surface, like the one mentioned in our glider problem, idealizes the scenario where no energy is lost to friction, allowing for a more straightforward analysis of momentum and energy conservation.

• **Eliminates External Forces** – Without friction, we have only the internal forces between the colliding objects, which simplifies our calculations considerably.
• **Pure System Dynamics** – The lack of external dissipation allows the system to be contained, meaning total kinetic energy remains constant for elastic collisions.

In this idealized setting, the collision outcomes are purely due to the interacting bodies' properties (mass and initial velocity), making the dynamic responses easier to calculate and more reliable for demonstrating theoretical principles.