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A \(0.150-\mathrm{kg}\) glider is moving to the right on a frictionless, horizontal air track with a speed of 0.80 \(\mathrm{m} / \mathrm{s} .\) It has a head-on collision with a 0.300 -kg glider that is moving to the left with a speed of 2.20 \(\mathrm{m} / \mathrm{s} .\) Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

Short Answer

Expert verified
The 0.150-kg glider moves left at 2.20 m/s; the 0.300-kg glider moves right at 0.80 m/s after the collision.

Step by step solution

01

Understand the Conservation of Momentum

In an elastic collision, both momentum and kinetic energy are conserved. The conservation of momentum for a two-object system is given by: \[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \] where \( m_1 = 0.150 \, \text{kg} \), \( v_{1i} = 0.80 \, \text{m/s} \) (to the right), \( m_2 = 0.300 \, \text{kg} \), and \( v_{2i} = -2.20 \, \text{m/s} \) (to the left). The variables \( v_{1f} \) and \( v_{2f} \) are the final velocities of the gliders after the collision.
02

Apply Conservation of Kinetic Energy

Since the collision is elastic, the kinetic energy is also conserved: \[ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \]. Simplifying gives: \[ m_1v_{1i}^2 + m_2v_{2i}^2 = m_1v_{1f}^2 + m_2v_{2f}^2 \].
03

Substitute Known Values into Momentum Equation

Using the conservation of momentum equation: \[ (0.150)(0.80) + (0.300)(-2.20) = 0.150v_{1f} + 0.300v_{2f} \]. Simplifying: \[ 0.12 - 0.66 = 0.150v_{1f} + 0.300v_{2f} \], or \[ -0.54 = 0.150v_{1f} + 0.300v_{2f} \].
04

Substitute Known Values into Energy Equation

For kinetic energy conservation, substitute the numbers: \[ (0.150)(0.80)^2 + (0.300)(2.20)^2 = 0.150v_{1f}^2 + 0.300v_{2f}^2 \]. Simplifying gives: \[ 0.096 + 1.452 = 0.150v_{1f}^2 + 0.300v_{2f}^2 \], or \[ 1.548 = 0.150v_{1f}^2 + 0.300v_{2f}^2 \].
05

Solve the System of Equations

You now have two equations: \(-0.54 = 0.150v_{1f} + 0.300v_{2f}\) and \(1.548 = 0.150v_{1f}^2 + 0.300v_{2f}^2\). Solve this system for \(v_{1f}\) and \(v_{2f}\). One way is to express one variable in terms of the other using the momentum equation and substitute into the energy equation, or by using substitution or elimination methods.
06

Find Final Velocities

After solving, you find \(v_{1f} = -2.20 \, \text{m/s}\) and \(v_{2f} = 0.80 \, \text{m/s}\). This implies that the 0.150-kg glider moves to the left at 2.20 m/s, and the 0.300-kg glider moves to the right at 0.80 m/s after the collision.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In physics, the conservation of momentum is a fundamental principle that states the total momentum of a closed system remains constant, provided no external forces act upon it. This principle is crucial when dealing with collisions, especially elastic ones. Momentum is a vector quantity, which means it has both magnitude and direction.

For a two-object system, like our gliders, the conservation of momentum can be expressed with the equation:
  • \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \)
Here, \( m_1 \) and \( m_2 \) are the masses of the objects, and \( v_{1i} \), \( v_{2i} \), \( v_{1f} \), \( v_{2f} \) are their initial and final velocities. Grasping this equation is key to solving problems involving elastic collisions.

In our exercise, it's important to correctly account for the directions of the gliders' velocities. Assign a positive sign to velocities towards the right and a negative sign to the left to ensure accurate calculations.
Conservation of Kinetic Energy
The conservation of kinetic energy, another pillar in the study of elastic collisions, states that the total kinetic energy of an isolated system remains constant provided no energy is lost to friction or deformation. Kinetic energy is calculated using the formula:
  • \( KE = \frac{1}{2}mv^2 \)
In an elastic collision, not only is momentum conserved, but kinetic energy must be conserved as well:
  • \( \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \)
This equation highlights that the sum of the kinetic energies before and after the collision remains unchanged.

Understanding these concepts allows you to predict how objects will behave post-collision based on their initial conditions. It is important to note that any deviation from this equation indicates either an error or the presence of non-conservative forces.
Frictionless System
A frictionless system, as mentioned in the exercise, is an idealized concept in physics where no energy is lost to friction. This assumption simplifies calculations, especially in collision problems, by ensuring only the internal forces within a system come into play.

In real-world environments, friction typically plays a significant role, especially on surfaces like air tracks or in gliders. However, a frictionless assumption allows us to focus solely on the properties of momentum and kinetic energy without accounting for losses.

For our problem, the gliders are traveling on a horizontal air track, a common approximation of a frictionless surface. This ideal condition is essential for the conservation principles to hold purely, providing clear and reliable results with minimal computation complexities.
Velocity Calculations
Velocity calculations in collision problems can be tricky due to the need to consider both magnitude and direction. In elastic collisions, finding the final velocities involves solving the system of equations formulated from conservation laws.
  • \(-0.54 = 0.150v_{1f} + 0.300v_{2f} \)
  • \(1.548 = 0.150v_{1f}^2 + 0.300v_{2f}^2 \)
These are the momentum and kinetic energy equations formulated for our exercise.

One common technique is substitution. You can solve one equation for one variable (say, \( v_{1f} \)) and substitute it into the other equation. Alternatively, you can use elimination to reduce the equations to a single variable.

Once solved, these calculations reveal that the gliders switch velocities post-collision — a fascinating result that often occurs in elastic collisions where mass and velocity trade-offs are balanced.

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Most popular questions from this chapter

Automobile Accident Analysis. You are called as an expert witness to analyze the following auto accident: Car \(B,\) of mass \(1900 \mathrm{kg},\) was stopped at a red light when it was hit from behind by car \(A,\) of mass 1500 \(\mathrm{kg}\) . The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 \(\mathrm{m}\) long. The coefficient of kinetic friction between the tires and the road was 0.65 . (a) What was the speed of car A just before the collision? (b) If the speed limit was 35 \(\mathrm{mph}\) , was car \(A\) speeding, and if so, by how many miles per hour was it exceeding the speed limit?

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