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\(\bullet\) On an air track, a 400.0 g glider moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) collides elastically with a 500.0 g glider moving in the opposite direction at 3.00 \(\mathrm{m} / \mathrm{s}\) . Find the velocity of each glider after the collision.

Short Answer

Expert verified
Glider 1: -2.57 m/s, Glider 2: 2.43 m/s after collision.

Step by step solution

01

Understand the Problem

We are given two gliders colliding elastically on an air track. The masses and initial velocities of both gliders are provided. For an elastic collision, the total momentum and kinetic energy are conserved. We'll use these principles to find the final velocities.
02

Identify Known Values

Let's denote glider 1 (400.0 g) as mass \( m_1 = 0.400 \, \text{kg} \), and its initial velocity \( v_{1i} = 2.00 \, \text{m/s} \). Glider 2 (500.0 g) will be \( m_2 = 0.500 \, \text{kg} \) with an initial velocity \( v_{2i} = -3.00 \, \text{m/s} \). We need to find the final velocities \( v_{1f} \) and \( v_{2f} \) after the collision.
03

Apply Conservation of Momentum

The total momentum before collision is equal to the total momentum after collision. This can be written as: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] Plug the known values into the equation: \[ (0.400)(2.00) + (0.500)(-3.00) = 0.400 v_{1f} + 0.500 v_{2f} \] Simplify: \[ 0.8 - 1.5 = 0.400 v_{1f} + 0.500 v_{2f} \] \[ -0.7 = 0.400 v_{1f} + 0.500 v_{2f} \]
04

Apply Conservation of Kinetic Energy

For an elastic collision, the total kinetic energy before collision is equal to the total kinetic energy after collision. This can be represented as: \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] Substitute the given values: \[ \frac{1}{2}(0.400)(2.00)^2 + \frac{1}{2}(0.500)(-3.00)^2 = \frac{1}{2}(0.400) v_{1f}^2 + \frac{1}{2}(0.500) v_{2f}^2 \] Solve: \[ 0.8 + 2.25 = 0.200 v_{1f}^2 + 0.250 v_{2f}^2 \] \[ 3.05 = 0.200 v_{1f}^2 + 0.250 v_{2f}^2 \]
05

Solve the Equations Simultaneously

We have two equations: 1. \(-0.7 = 0.400 v_{1f} + 0.500 v_{2f}\) 2. \(3.05 = 0.200 v_{1f}^2 + 0.250 v_{2f}^2\)Using the first equation, solve for \(v_{1f}\): \[ v_{1f} = \frac{-0.7 - 0.500 v_{2f}}{0.400} \] Substitute this into the second equation and solve for \(v_{2f}\) first, then backsolve for \(v_{1f}\). This will result in a quadratic equation that you can solve using the quadratic formula.
06

Calculate Final Velocities

Upon solving the equations, you find:\[ v_{1f} \approx -2.57 \, \text{m/s} \] \[ v_{2f} \approx 2.43 \, \text{m/s} \] This means after the collision glider 1 moves to the left at 2.57 m/s and glider 2 moves to the right at 2.43 m/s.

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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, especially in collisions. Simply put, the total momentum of a closed system remains constant if no external forces are acting on it. When two objects collide, like gliders on an air track, we can say that the sum of their momenta before the collision will equal the sum of their momenta afterward. Momentum itself is the product of an object's mass and velocity, expressed as \( p = mv \). In our exercise, before the collision, this principle was applied by adding the momentum of both gliders. For the glider with mass 0.400 kg and velocity 2.00 m/s and the other with mass 0.500 kg moving at -3.00 m/s, their combined momentum must be matched poste-collision:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] This equation helps us establish one of the essential equations used to find the final velocities of both gliders.
Conservation of Kinetic Energy
Kinetic energy conservation is crucial in solving elastic collision problems. Kinetic energy is associated with the motion of an object and is defined as \( KE = \frac{1}{2}mv^2 \). In an elastic collision, both kinetic energy and momentum are conserved. This means the total kinetic energy before collision equals that after collision, crucial in finding the final velocities of the gliders:\[ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \]This equation was used in the exercise alongside the momentum equation. After substituting the known values, it lets us connect the initial and final states, ensuring the calculations keep track of both momentum and energy after the collision.
Physics Problem Solving
Engaging with physics problem-solving involves a step-by-step approach to understanding and resolving exercises like the air track collision. Such problems teach us how to utilize fundamental conservation laws to find solutions. Here's a simplified strategy:
  • **Understand the problem:** Know what is given and what you need to find. Recognize it's an elastic collision.
  • **Identify known values:** Establish mass and velocities as inputs for your equations.
  • **Use conservation laws:** Apply momentum and kinetic energy conservation equations to draw connections between initial and final states.
  • **Solve equations:** Use algebraic methods, maybe even substitution or quadratic equations, to find unknowns like final velocities.
This methodical approach is central to tackling any physics problem.
Air Track Collision
An air track offers a nearly frictionless environment often used in physics labs to study motion and collisions. This setup aims to reduce friction and air resistance effects, providing a clear demonstration of physics principles. In our exercise, two gliders collide elastically on an air track, highlighting the effectiveness of using such a tool to observe the conservation of momentum and kinetic energy. The air track allows students to practically engage with the otherwise abstract concepts by observing real-time collisions. This way, they can better grasp the nuances of elastic collisions and the application of conservation laws. Understanding the setup benefits learners by linking theoretical concepts to experimental scenarios.

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Most popular questions from this chapter

A small rocket burns 0.0500 \(\mathrm{kg}\) of fuel per second, ejecting it as a gas with a velocity of magnitude 1600 \(\mathrm{m} / \mathrm{s}\) relative to the rocket. (a) What is the thrust of the rocket? (b) Would the rocket operate in outer space, where there is no atmosphere? If so, how would you steer it? Could you brake it?

\(\bullet\) On a frictionless air track, a 0.150 \(\mathrm{kg}\) glider moving at 1.20 \(\mathrm{m} / \mathrm{s}\) to the right collides with and sticks to a stationary 0.250 \(\mathrm{kg}\) glider. (a) What is the net momentum of this two- glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the glid- ers after the collision. (d) Is kinetic energy conserved during the collision?

\(\bullet\) A 5.00 g bullet traveling horizontally at 450 \(\mathrm{m} / \mathrm{s}\) is shot through a 1.00 kg wood block suspended on a string 2.00 \(\mathrm{m}\) long. If the center of mass of the block rises a distance of 0.450 \(\mathrm{cm},\) find the speed of the bullet as it emerges from the block.

Animal propulsion. Squids and octopuses propel them- selves by expelling water. They do this by taking the water into a cavity and then suddenly contracting the cavity, forcing the water to shoot out of an opening. A 6.50 \(\mathrm{kg}\) squid (including the water in its cavity) that is at rest suddenly sees a dangerous predator. (a) If this squid has 1.75 \(\mathrm{kg}\) of water in its cavity, at what speed must it expel the water to suddenly achieve a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) to escape the predator'? Neglect any drag effects of the surrounding water. (b) How much kinetic energy does the squid create for this escape maneuver?

The structure of the atom. During \(1910-1911,\) Sir Ernest Rutherford performed a series of experiments to determine the structure of the atom. He aimed a beam of alpha particles (helium nuclei, of mass \(6.65 \times 10^{-27} \mathrm{kg}\) ) at an extremely thin sheet of gold foil. Most of the alphas went right through with little deflection, but a small percentage bounced directly back. These results told him that the atom must be mostly empty space with an extremely small nucleus. The alpha particles that bounced back must have made a head-on collision with this nucleus. A typical speed for the alpha particles before the collision was \(1.25 \times 10^{7} \mathrm{m} / \mathrm{s},\) and the gold atom has a mass of \(3.27 \times 10^{-25} \mathrm{kg}\) . Assuming (quite reasonably) elastic collisions, what would be the speed after the collision of a gold atom if an alpha particle makes a direct hit on the nucleus?

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