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On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) is moving toward puck \(B\) (with mass \(0.350 \mathrm{kg} ),\) which is initially at rest. After the collision, puck \(A\) has a velocity of 0.120 \(\mathrm{m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of 0.650 \(\mathrm{m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Short Answer

Expert verified
(a) 0.790 m/s (b) -0.0023 Joules

Step by step solution

01

Understanding the Scenario

We have two pucks, A and B. Puck A with mass of 0.250 kg is initially moving towards Puck B with mass 0.350 kg, which is initially at rest. After the collision, Puck A moves with a speed of 0.120 m/s to the left and Puck B with 0.650 m/s to the right.
02

Conservation of Momentum

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. If we let the initial speed of A be \(v_{Ai}\), the equation is: \[m_A imes v_{Ai} + m_B imes 0 = m_A imes (-0.120) + m_B imes 0.650\] Plug in the masses: \[0.250 imes v_{Ai} = 0.250 imes (-0.120) + 0.350 imes 0.650\]
03

Solving for Initial Velocity of Puck A

Rearrange the equation to solve for \(v_{Ai}\): \[0.250 imes v_{Ai} = -0.030 + 0.2275\]\[0.250 imes v_{Ai} = 0.1975\]\[v_{Ai} = \frac{0.1975}{0.250} = 0.790 \, \text{m/s}\]
04

Calculating Initial and Final Kinetic Energies

The initial kinetic energy of the system is only due to puck A, since puck B is at rest:\[KE_{initial} = \frac{1}{2} imes 0.250 \times 0.790^2\]Calculate this value:\[KE_{initial} = 0.5 \times 0.250 \times 0.6241 = 0.0780 \, \text{Joules.}\]The final kinetic energy of puck A is:\[KE_{A,final} = \frac{1}{2} imes 0.250 \times (0.120)^2 = 0.0018 \, \text{Joules.}\]The final kinetic energy of puck B is:\[KE_{B,final} = \frac{1}{2} \times 0.350 \times (0.650)^2 = 0.0739 \, \text{Joules.}\]Thus, the total final kinetic energy is:\[KE_{final} = 0.0018 + 0.0739 = 0.0757 \, \text{Joules.}\]
05

Determining the Change in Kinetic Energy

Calculate the change in kinetic energy:\[\Delta KE = KE_{final} - KE_{initial}\]Plug in the values:\[\Delta KE = 0.0757 - 0.0780 = -0.0023 \, \text{Joules.}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It depends on both the mass of the object and its velocity. The formula to calculate kinetic energy is given by:
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass and \( v \) is the velocity.
In this exercise, the initial kinetic energy is only associated with puck A since puck B is at rest. After the collision, kinetic energies are calculated separately for both pucks.

It is calculated as:
Initial Kinetic Energy of puck A: \( KE_{A,initial} = \frac{1}{2} \times 0.250 \times (0.790)^2 = 0.0780 \) Joules.
Final Kinetic Energy of puck A: \( KE_{A,final} = \frac{1}{2} \times 0.250 \times (0.120)^2 = 0.0018 \) Joules.
Final Kinetic Energy of puck B: \( KE_{B,final} = \frac{1}{2} \times 0.350 \times (0.650)^2 = 0.0739 \) Joules.
The total kinetic energy change helps understand whether energy has been converted into other forms or lost due to factors like heat or sound, although in an ideal physics problem, these are ignored.
Elastic Collision
An elastic collision is one where both momentum and kinetic energy are conserved. This conservation reflects that no kinetic energy is lost in the form of heat or deformation. However, in practical scenarios, achieving a perfectly elastic collision is impossible as some amount of energy is usually lost.
In this problem, we calculated the change in kinetic energy to determine if the collision was perfectly elastic.

While momentum was conserved, a small change in kinetic energy (\( \Delta KE = -0.0023 \) Joules) indicated that the collision wasn't perfectly elastic, highlighting a realistic aspect of physics problems in idealized environments.
Physics Problem Solving
Physics problem solving requires a systematic approach to apply theoretical concepts practically. Here, the concepts of conservation of momentum and kinetic energy are central.
  • First, identify what is given in the problem and what needs to be solved.
  • Apply the conservation laws: momentum and, if applicable, energy.
  • Solve equations systematically, carefully ensuring that all variables are accounted for.
  • Verify answers by checking units and ensuring logical consistency in the context of the problem.
For this problem:
  • Given mass and velocity information, initial velocities and kinetic changes were derived using conservation laws.
  • Equations were systematically solved to find the desired initial speed and energy changes.
Working through these steps reinforces comfort with applying physics principles to real-world situations and simulations.
Frictionless Surface
A frictionless surface is a theoretical concept used in physics to study motion without the interference of friction. It allows us to focus solely on forces and energy changes without considering energy loss due to friction.
In this exercise, the frictionless air table lets us analyze the conservation laws without worrying about frictional forces acting on the pucks.
  • This simplification helps accurately apply the conservation of momentum and energy in idealized conditions.
Consider this:
The pucks slide without resistance, meaning all their energy changes result solely from collisions, not from external forces.
This assumption allows for a clean analysis of mechanical phenomena, crucial for grasping primary physics concepts before adding the complexity of friction in more advanced problems.

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