Problem 17 - College Physics

Question

A ball with a mass of \(0.600 \mathrm{~kg}\) is initially at rest. It is struck by a second ball having a mass of \(0.400 \mathrm{~kg}\), initially moving with a velocity of \(0.250 \mathrm{~m} / \mathrm{s}\) toward the right along the \(x\) axis. After the collision, the \(0.400 \mathrm{~kg}\) ball has a velocity of \(0.200 \mathrm{~m} / \mathrm{s}\) at an angle of \(36.9^{\circ}\) above the \(x\) axis in the first quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the \(0.600 \mathrm{~kg}\) ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

Step-by-Step Solution

Answer

(a) 0.100 m/s at -53.1° below the x-axis; (b) -0.0015 J change in kinetic energy.

Step 1: Define the Problem

We need to determine the velocity of the 0.600 kg ball after collision and the change in kinetic energy of the system before and after the collision. We can use the laws of conservation of momentum for this problem.

Step 2: Set Up Conservation of Momentum Equations

The law of conservation of momentum says the total momentum of a system is conserved if no external forces act on it. We will set up equations for both the x and y components of momentum before and after the collision.Initial momentum:- The 0.600 kg ball is at rest, so its initial momentum is zero.- The 0.400 kg ball has a momentum of \( p_{x, \text{initial}} = 0.400 \text{ kg} \times 0.250 \text{ m/s} = 0.100 \text{ kg m/s} \).Final momentum:- For the 0.400 kg ball: - \( p_{x, \text{final}} = 0.400 \times 0.200 \times \cos(36.9^\circ) \) - \( p_{y, \text{final}} = 0.400 \times 0.200 \times \sin(36.9^\circ) \)Now equate these with the total momentum after collision for both components.

Step 3: Calculate the Momentum Components

Calculate the momentum components for the 0.400 kg ball after the collision:- \( p_{x, \text{final}} = 0.400 \times 0.200 \times \cos(36.9^\circ) = 0.400 \times 0.200 \times 0.8 = 0.064 \text{ kg m/s} \)- \( p_{y, \text{final}} = 0.400 \times 0.200 \times \sin(36.9^\circ) = 0.400 \times 0.200 \times 0.6 = 0.048 \text{ kg m/s} \)

Step 4: Apply Conservation of Momentum

Using conservation of momentum:For the x component:\[ 0.100 = 0.064 + 0.600 \cdot v_{x, \text{final}} \]For the y component:\[ 0 = 0.048 + 0.600 \cdot v_{y, \text{final}} \]

Step 5: Solve for Velocities

From the x component:\[ v_{x, \text{final}} = \frac{0.100 - 0.064}{0.600} = \frac{0.036}{0.600} = 0.060 \text{ m/s} \]From the y component:\[ v_{y, \text{final}} = \frac{-0.048}{0.600} = -0.080 \text{ m/s} \]

Step 6: Calculate Magnitude and Direction of 0.600 kg Ball's Velocity

The magnitude of velocity is given by the expression:\[ v = \sqrt{v_{x, \text{final}}^2 + v_{y, \text{final}}^2} = \sqrt{(0.060)^2 + (-0.080)^2} \approx 0.100 \text{ m/s} \]The direction (angle \(\theta\)) can be calculated using the tangent of the angle:\[ \theta = \tan^{-1}\left(\frac{v_{y, \text{final}}}{v_{x, \text{final}}}\right) = \tan^{-1}\left(\frac{-0.080}{0.060}\right) \approx -53.1^\circ \]

Step 7: Determine the Change in Kinetic Energy

Calculate the initial and final kinetic energy and find the difference.Initial kinetic energy:\[ KE_{\text{initial}} = \frac{1}{2} \times 0.400 \times (0.250)^2 = 0.0125 \text{ J} \]Final kinetic energy:\[ KE_{0.400 \text{ ball}} = \frac{1}{2} \times 0.400 \times (0.200)^2 = 0.008 \text{ J} \]\[ KE_{0.600 \text{ ball}} = \frac{1}{2} \times 0.600 \times (0.100)^2 = 0.003 \text{ J} \]\[ KE_{\text{final}} = 0.008 + 0.003 = 0.011 \text{ J} \]Change in kinetic energy:\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 0.011 - 0.0125 = -0.0015 \text{ J} \]

Step 8: Conclusion: Answer to Questions

The magnitude and direction of the velocity of the 0.600 kg ball are 0.100 m/s and \(-53.1^\circ\) below the x-axis, respectively. The change in the total kinetic energy of the two balls is -0.0015 J.

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It can be calculated using the formula: \( KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.
The initial kinetic energy in our exercise comes from the second ball which was moving before the collision occurred. It had a mass of 0.400 kg and a velocity of 0.250 m/s. Calculating, the initial kinetic energy is \( KE_{\text{initial}} = \frac{1}{2} \times 0.400 \times (0.250)^2 = 0.0125 \text{ J} \).
After the collision, the kinetic energies of the two balls combined amounted to less than this initial kinetic energy, indicating a loss. Such a change in kinetic energy is important to understand in collision problems, as it can signify energy transformations, such as sound or heat, during the collision process.

Elastic Collision

Elastic collisions are an important concept in physics, where both momentum and kinetic energy are conserved.
This exercise asks us to look for any changes in kinetic energy, which tells us that the collision may not be perfectly elastic. In perfectly elastic collisions, no kinetic energy is lost. Therefore, the kinetic energy before and after the collision would be the same.
In our example, we observe a change in the kinetic energy of -0.0015 J. This negative change indicates that not all energy was conserved as kinetic energy, affirming that the collision isn’t perfectly elastic. Some energy likely transformed into other forms, like heat or sound, showing the interplay between different energy forms during collisions.

Momentum Components

Momentum is a vector quantity, meaning it has both magnitude and direction. To solve problems like this, we need to separate momentum into its components, usually along the x and y axes.
Initially, the momentum is only in the x-direction because the 0.400 kg ball is moving straight. Its x-component is \( p_{x, \text{initial}} = 0.400 \times 0.250 = 0.100 \text{ kg m/s} \).
Post-collision, momentum divides into two components for each ball. For the 0.400 kg ball, the x-component is \( 0.064 \text{ kg m/s} \) and the y-component is \( 0.048 \text{ kg m/s} \), calculated using trigonometric functions based on the given angle. By using the conservation of momentum separately for each axis, we can find the velocity components of the 0.600 kg ball after the collision.

Physics Problem Solving

Solving physics problems involves a clear understanding and application of physical principles like conservation laws.
For momentum and collision-related problems, breaking down the events step-by-step and verifying with known equations helps in finding a solution.
This problem-solving process uses systematic approaches. The conservation of momentum is first considered, setting up equations for both x and y components separately. Each component must be conserved independently. Tracking changes in kinetic energy helps assess whether the collision is elastic or inelastic.
Once the system's equations are laid out, we solve for unknowns systematically, using basic algebra and trigonometry. This disciplined approach clarifies complex interactions in each problem, empowering us to solve diverse physics puzzles confidently.