Problem 3 - Applied Physics

Question

A \(0.600-\mathrm{kg}\) ball traveling \(4.00 \mathrm{~m} / \mathrm{s}\) to the right collides with a \(1.00-\mathrm{kg}\) ball traveling \(5.00 \mathrm{~m} / \mathrm{s}\) to the left. After the collision, the lighter ball is traveling \(7.25 \mathrm{~m} / \mathrm{s}\) to the left. What is the velocity of the heavier ball after the collision?

Step-by-Step Solution

Answer

The velocity of the heavier ball after the collision is 1.75 m/s to the right.

Step 1: Identify Given Data

We have two balls colliding: \( m_1 = 0.600 \, \mathrm{kg} \) with an initial velocity \( v_{1i} = 4.00 \, \mathrm{m/s} \) to the right and \( m_2 = 1.00 \, \mathrm{kg} \) with an initial velocity \( v_{2i} = -5.00 \, \mathrm{m/s} \) to the left. After the collision, the velocity of the lighter ball is \( v_{1f} = -7.25 \, \mathrm{m/s} \). We need to find the final velocity of the heavier ball, \( v_{2f} \).

Step 2: Apply Conservation of Momentum

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Thus, we write the equation: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] Substitute the known values into the equation:\[ 0.600 \, \mathrm{kg} \times 4.00 \, \mathrm{m/s} + 1.00 \, \mathrm{kg} \times (-5.00 \, \mathrm{m/s}) = 0.600 \, \mathrm{kg} \times (-7.25 \, \mathrm{m/s}) + 1.00 \, \mathrm{kg} \times v_{2f} \]

Step 3: Simplify the Equation

Perform the multiplication and simplify the left side and right side of the equation:\[ 2.40 - 5.00 = -4.35 + v_{2f} \]Simplify further:\[ -2.60 = -4.35 + v_{2f} \]

Step 4: Solve for \( v_{2f} \)

Isolate \( v_{2f} \) by adding \( 4.35 \) to both sides:\[ -2.60 + 4.35 = v_{2f} \]\[ v_{2f} = 1.75 \, \mathrm{m/s} \]

Key Concepts

Collision Physics

In the world of physics, collisions are fascinating phenomena that occur when two or more objects come into contact with each other, influencing their subsequent motion. We categorize most collisions as either elastic or inelastic. In an

• elastic collision, both momentum and kinetic energy are conserved. Objects may bounce off one another, retaining their energy.
• inelastic collision, momentum is conserved, but kinetic energy isn't. Objects might stick together or deform.

Understanding the type of collision can be essential in predicting how objects behave post-impact.
In the given exercise, the balls did not stick together, indicating a combination of elastic and inelastic behaviors. The key takeaway here is the concept of momentum conservation, which is foundational regardless of collision type.

Momentum Calculation

Momentum is a measure of the motion of an object and is calculated as the product of an object's mass and velocity, represented by the formula:
\[ p = mv \]
where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. It's a vector quantity, meaning it has both magnitude and direction.
In collisions, the momentum before the event should equal the momentum after, provided no external forces act on the system. This is called the conservation of momentum, expressed by:

• Initial Momentum: \( m_1 v_{1i} + m_2 v_{2i} \)
• Final Momentum: \( m_1 v_{1f} + m_2 v_{2f} \)

By substituting known values into these formulas, we can solve for unknown variables. In our case, we successfully found the final velocity of the heavier ball \( v_{2f} \) using this principle.

Physics Problem Solving

Problem-solving in physics involves a systematic application of theories and equations to arrive at answers. Here are some general steps to approach such problems:

• Understand the Problem: Clearly determine what you need to find and list all given data.
• Choose the Right Physics Principle: Identify which laws or formulas apply, such as the conservation of momentum for collisions.
• Set Up Equations: Convert the real-world problem into mathematical equations.
• Solve the Equations: Use algebraic manipulation to isolate unknowns and solve.
• Check Your Work: Verify the answer makes sense both mathematically and conceptually.

By following these steps, we effectively determined the final velocity of the heavier ball. This approach not only aids in solving textbook problems but also builds a solid foundation for tackling more complex real-world physics issues.