Problem 58 - College Physics

Question

A \(4.0-\mathrm{kg}\) ball with a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction collides head-on elastically with a stationary \(2.0-\mathrm{kg}\) ball. What are the velocities of the balls after the collision?

Step-by-Step Solution

Answer

After the collision, the 4 kg ball has a velocity of 2.5 m/s, and the 2 kg ball has a velocity of 3.0 m/s, both in the +x-direction.

Step 1: Identify the Variables

The mass of the first ball is given as \( m_1 = 4.0 \, \text{kg} \), and its velocity is \( v_1 = 4.0 \, \text{m/s} \) in the positive x-direction. The second ball has a mass \( m_2 = 2.0 \, \text{kg} \) and is initially stationary, so its velocity \( v_2 = 0 \, \text{m/s} \). We need to find the velocities \( v_1' \) and \( v_2' \) of the balls after the collision.

Step 2: Apply Conservation of Momentum

Since momentum is conserved in an elastic collision, the total momentum before the collision equals the total momentum after the collision. Set up the equation:\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]Substitute the given values:\[ 4.0 \, \text{kg} \cdot 4.0 \, \text{m/s} + 2.0 \, \text{kg} \cdot 0 \, \text{m/s} = 4.0 \, \text{kg} \cdot v_1' + 2.0 \, \text{kg} \cdot v_2' \]\[ 16.0 = 4.0 v_1' + 2.0 v_2' \] (Equation 1)

Step 3: Apply Conservation of Kinetic Energy

In elastic collisions, kinetic energy is also conserved. Set up the equation:\[ \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2 \]Substitute the values:\[ \frac{1}{2} \times 4.0 \times 4.0^2 + \frac{1}{2} \times 2.0 \times 0^2 = \frac{1}{2} \times 4.0 \times v_1'^2 + \frac{1}{2} \times 2.0 \times v_2'^2 \]\[ 32.0 = 2.0 v_1'^2 + 1.0 v_2'^2 \] (Equation 2)

Step 4: Solve Simultaneous Equations

Solve Equation 1 and Equation 2 together to find \( v_1' \) and \( v_2' \). Using substitution or elimination methods:1. From Equation 1: \( 4.0 v_1' + 2.0 v_2' = 16.0 \) Simplify to \( 2.0 v_1' + v_2' = 8.0 \) (Equation 3)2. Express \( v_1' = (8.0 - v_2')/2.0 \)Substitute \( v_1' \) in Equation 2:\[ 32.0 = 2.0 \left(\frac{8 - v_2'}{2}\right)^2 + v_2'^2 \]Simplify and solve for \( v_2' \):\[ 32.0 = 2.0 \left(16 - 8v_2' + v_2'^2\right)/4 + v_2'^2 \]\[ 32.0 = 8.0 - 4.0v_2' + v_2'^2 + v_2'^2 \]\[ 32.0 = 8.0 + 2v_2'^2 - 4.0v_2' \]\[ 24.0 = 2v_2'^2 - 4.0v_2' \]Divide through by 2:\[ 12.0 = v_2'^2 - 2.0v_2' \]Solve the quadratic equation to find \( v_2' \), then use it to find \( v_1' \).

Step 5: Calculate the Velocities After Collision

Using the quadratic formula or by completing the square, solve the equation \( v_2'^2 - 2v_2' - 12 = 0 \). The solutions are \( v_2' = 3.0 \text{ m/s} \) and \( v_2' = -4.0 \text{ m/s} \). However, physically only one solution is valid based on conservation laws and directions.Substitute \( v_2' = 3.0 \text{ m/s} \) back into the expression for \( v_1' = (8.0 - v_2')/2.0 \):\[ v_1' = (8.0 - 3.0)/2.0 = 2.5 \, \text{m/s} \]

Step 6: Conclude Velocities

After evaluating potential solutions and ensuring physical consistency with the problem's setup, we have:- The 4.0-kg ball moves with a velocity of \( v_1' = 2.5 \text{ m/s} \) in the +x-direction.- The 2.0-kg ball moves with a velocity of \( v_2' = 3.0 \text{ m/s} \) in the +x-direction.

Key Concepts

Conservation of Momentum

One central principle in physics, especially during collisions, is the conservation of momentum. Momentum, a vector quantity, is expressed as the product of an object's mass and velocity, given by the formula: \( p = mv \). During an elastic collision, like the one in our exercise, the total momentum before the collision equals the total momentum after the collision.

This means that any change in the momentum of one object in the system is exactly balanced by a change in the momentum of another object. The total system momentum is conserved.

• For our problem, the equation for momentum conservation is set up as follows: \( m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \).
• Substituting known values, we have \( 4.0 \times 4.0 + 2.0 \times 0 = 4.0 \times v_1' + 2.0 \times v_2' \).
• Simplifying gives: \( 16 = 4v_1' + 2v_2' \).

This equation plays a crucial role in understanding how velocities change after the collision.

Conservation of Kinetic Energy

In addition to conserving momentum, an elastic collision also conserves kinetic energy. Kinetic energy, associated with the motion of an object, is given by the formula: \( KE = \frac{1}{2}mv^2 \).

During an elastic collision, the total kinetic energy of the system remains the same before and after the collision. Unlike inelastic collisions, where kinetic energy is lost (usually converted to heat or sound), no kinetic energy is lost here.

• For our problem, the kinetic energy conservation equation becomes: \( \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2 \).
• Inputting known values, it simplifies to \( \frac{1}{2} \times 4.0 \times 4.0^2 + 0 = \frac{1}{2} \times 4.0 \times v_1'^2 + \frac{1}{2} \times 2.0 \times v_2'^2 \).
• This simplifies further to \( 32.0 = 2.0v_1'^2 + 1.0v_2'^2 \).

Ensuring that this condition holds helps us determine which solutions are physically possible.

Quadratic Equation

Solving simultaneous equations often involves quadratic equations, a staple in solving physics problems. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \) and can be solved by several methods including factoring, completing the square, or using the quadratic formula.

For our exercise, we rearranged the system's kinetic energy conservation equation to form a quadratic equation for \( v_2' \). This is crucial for determining the final velocities after collision.

• Our equation was simplified to \( v_2'^2 - 2v_2' - 12 = 0 \).
• Applying the quadratic formula, \( v_2' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), gave us potential solutions for the velocity \( v_2' \).
• Through this, we found \( v_2' = 3.0 \text{ m/s} \), the physically valid solution.

Recognizing and correctly solving quadratic equations is important in handling similar physics problems.

Physics Problem Solving

Solving physics problems is like following a recipe. It involves systematic steps and understanding fundamental principles such as momentum and energy conservation.

Understanding what the problem asks and breaking it into parts can greatly help. Consider these problem-solving strategies:

• Identify and list down all known variables from the problem statement.
• Apply the principles of physics relevant to the problem, like conservation laws.
• Set up equations derived from these principles and solved them systematically.
• Check each solution's physical validity through consistency with known facts about motion and collisions.

Such a structured approach helps one not only solve the problem but deeply understand the concepts behind the calculations.