Problem 13 - College Physics

Question

\(\bullet\) On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) ) is moving to the right 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 veloc- ity 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.

Step-by-Step Solution

Answer

(a) 0.29 m/s (to the right); (b) Increase of 0.055 J in kinetic energy.

Step 1: Identify known variables

We have two pucks involved in a collision. Puck A with mass \(m_A = 0.250\, \text{kg}\) and puck B with mass \(m_B = 0.350\, \text{kg}\). After the collision, puck A moves left with velocity \(v'_A = -0.120\, \text{m/s}\) and puck B moves right with velocity \(v'_B = 0.650\, \text{m/s}\). We need to find the initial velocity of puck A before the collision and the change in kinetic energy.

Step 2: Apply conservation of momentum

Since there is no external force acting on the system, the total linear momentum before the collision is equal to the total linear momentum after the collision. Thus,\[m_A \cdot v_{A0} + m_B \cdot v_{B0} = m_A \cdot v'_A + m_B \cdot v'_B\]Substitute the known values, noting that \(v_{B0} = 0\) since puck B is initially at rest:\[0.250 \cdot v_{A0} + 0.350 \cdot 0 = 0.250 \cdot (-0.120) + 0.350 \cdot 0.650\]Solve for \(v_{A0}\):\[0.250 \cdot v_{A0} = -0.30 + 0.2275\]\[0.250 \cdot v_{A0} = -0.30 + 0.2275 = -0.0725\]\[v_{A0} = \frac{-0.0725}{0.250} = -0.29\, \text{m/s (to the right)} \]

Step 3: Calculate initial kinetic energy

The initial kinetic energy \(KE_{initial}\) of the system is only due to puck A because puck B is initially at rest:\[KE_{initial} = \frac{1}{2} m_A v_{A0}^2 + \frac{1}{2} m_B v_{B0}^2 = \frac{1}{2} (0.250)(-0.29)^2 + 0\] \[= 0.5 \times 0.250 \times 0.0841 = 0.0105125 \] Joules.

Step 4: Calculate final kinetic energy

The final kinetic energy \(KE_{final}\) of the system is the sum of the kinetic energies of both pucks after the collision:\[KE_{final} = \frac{1}{2} m_A {v'_A}^2 + \frac{1}{2} m_B {v'_B}^2 \]\[= \frac{1}{2} (0.250)(0.120)^2 + \frac{1}{2} (0.350)(0.650)^2\]\[= 0.5 \cdot 0.250 \cdot 0.0144 + 0.5 \cdot 0.350 \cdot 0.4225\]\[= 0.0018 + 0.0739375 = 0.0757375 \]Joules.

Step 5: Calculate change in kinetic energy

The change in kinetic energy \(\Delta KE\) is the difference between the final and initial kinetic energies:\[\Delta KE = KE_{final} - KE_{initial}\]\[= 0.0757375 - 0.0205125\]\[= 0.055225 \, \text{J}\]

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that refers to the energy an object possesses due to its motion. In the case of the pucks on the frictionless air table, their kinetic energy depends on both their mass and velocity. The equation for kinetic energy (\( KE \)) is given by:

• \( KE = \frac{1}{2}mv^2 \)

where \( m \) is the mass of the object and \( v \) is its velocity. Before the collision, only puck A has kinetic energy since puck B is at rest. After the collision, both pucks possess kinetic energy due to their velocities. It is important to measure kinetic energy to assess how much of it is conserved or lost during interactions such as collisions, providing insights into energy transformations.

Elastic Collisions

An elastic collision is a type of collision in which there is no net loss of kinetic energy in the system. Conversely, inelastic collisions involve a loss of kinetic energy. In our puck experiment on the frictionless surface, we analyze whether the collision is perfectly elastic by checking if the kinetic energy before and after the collision is equal.

• In an elastic collision, both momentum and kinetic energy are conserved.
• For our scenario:
  • The calculated initial kinetic energy is \( 0.0105125 \, \text{J} \).
  • The final kinetic energy is found to be \( 0.0757375 \, \text{J} \).
• The discrepancy indicates that the collision was not perfectly elastic.

Understanding elastic collisions helps us in systems where restoring energy to original kinetic forms is paramount, like in atomic collisions or toy balls that bounce off surfaces.

Momentum

Momentum is the quantity of motion an object has, defined as the product of its mass and velocity. It plays a critical role in predicting the outcomes of collisions. The principle of conservation of momentum states that if no external forces act on a system, the total momentum remains constant.In the exercise, conservation of momentum is expressed mathematically as:

• \( m_A \cdot v_{A0} + m_B \cdot v_{B0} = m_A \cdot v'_A + m_B \cdot v'_B \)

This principle allows us to solve for the unknown initial velocity of puck A. By substituting the known velocities and masses, we found \( v_{A0} \) to be \( -0.29 \, \text{m/s} \) to the right. This demonstrates how momentum conservation is used in collision scenarios to analyze and determine unknown quantities based on known parameters.

Frictionless Surface

A frictionless surface is an idealized concept where no frictional force opposes the motion of objects. It allows for a clear demonstration of principles like conservation of momentum. In the exercise, the frictionless air table ensures that no external forces, such as friction, affect the momentum of the pucks. Key attributes of a frictionless surface include:

• No energy lost to heat or sound due to friction.
• Simplifies calculations by focusing solely on the interaction between the pucks.
• Ideal for studying pure collision dynamics without interference from additional forces.

Using a frictionless surface, like in our example, ensures the analysis of motion adheres closely to theoretical models, providing an accurate assessment of momentum and kinetic energy changes during the collision.