Problem 1 - Applied Physics

Question

One ball of mass \(0.500 \mathrm{~kg}\) traveling \(6.00 \mathrm{~m} / \mathrm{s}\) to the right collides with a ball of mass \(0.200 \mathrm{~kg}\) initially at rest. After the collision, the heavier ball is traveling \(2.57 \mathrm{~m} / \mathrm{s}\) to the right. What is the velocity of the lighter ball after the collision?

Step-by-Step Solution

Answer

The lighter ball's velocity is 8.575 m/s to the right.

Step 1: Understand the problem

We have two balls colliding. Ball 1 has a mass of 0.500 kg and initially moves at a velocity of 6.00 m/s to the right. Ball 2 has a mass of 0.200 kg and is initially at rest. After the collision, Ball 1 moves to the right at 2.57 m/s, and we need to find the velocity of Ball 2 after the collision.

Step 2: Apply conservation of momentum

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. The equation for this is:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]Where:- \( v_{1i} = 6.00 \) m/s is the initial velocity of Ball 1- \( v_{2i} = 0 \) m/s as Ball 2 is at rest initially- \( v_{1f} = 2.57 \) m/s is the final velocity of Ball 1- \( v_{2f} \) is the final velocity of Ball 2 (what we are solving for)

Step 3: Insert known values into the equation

Substitute the known values into the conservation of momentum equation:\[ (0.500 \, \text{kg})(6.00 \, \text{m/s}) + (0.200 \, \text{kg})(0 \, \text{m/s}) = (0.500 \, \text{kg})(2.57 \, \text{m/s}) + (0.200 \, \text{kg})v_{2f} \]

Step 4: Solve for the unknown velocity

Calculate the left-hand side and simplify:\[ 3.00 \, \text{kg} \, \text{m/s} = 1.285 \, \text{kg} \, \text{m/s} + 0.200 \, \text{kg} \cdot v_{2f} \]Rearrange to solve for \(v_{2f}\):\[ 0.200 \, \text{kg} \cdot v_{2f} = 3.00 \, \text{kg} \, \text{m/s} - 1.285 \, \text{kg} \, \text{m/s} \]\[ 0.200 \, \text{kg} \cdot v_{2f} = 1.715 \, \text{kg} \, \text{m/s} \]Divide both sides by 0.200 kg:\[ v_{2f} = \frac{1.715 \, \text{kg} \, \text{m/s}}{0.200 \, \text{kg}} = 8.575 \, \text{m/s} \]

Step 5: Conclusion

The velocity of the lighter ball after the collision is 8.575 m/s to the right.

Key Concepts

Elastic Collision

In physics, an elastic collision is a collision in which there is no net loss in kinetic energy. This means that the total kinetic energy of the system (both objects involved in the collision) remains the same before and after the impact. During such collisions, both momentum and kinetic energy are conserved.
In the given exercise, we observe an elastic collision between two balls: one heavier ball and one lighter ball. The heavier ball is initially in motion, while the lighter ball is at rest. After the collision, both balls move in the same direction.
Understanding elastic collisions is fundamental in many areas of physics as they provide insights into the behavior of materials and mechanisms involved in impact and collision studies.

• Elastic collisions conserve both momentum and kinetic energy.
• Perfect elastic collisions are ideal, as energy loss is a reality in everyday collisions.

Momentum Equation

The momentum equation is a fundamental principle derived from the conservation of momentum. It states that the total momentum of a closed system remains constant provided no external forces act upon it. The equation can be mathematically expressed as:
\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]Here,

• \( m_1 \) and \( m_2 \) are the masses of the two colliding objects.
• \( v_{1i} \) and \( v_{2i} \) are the initial velocities of the objects before collision.
• \( v_{1f} \) and \( v_{2f} \) are the final velocities of the objects after collision.

In our exercise, the momentum before the collision must equal the momentum after the collision. By solving the equation with the given values, we can determine the unknown velocity of the lighter ball.

• This doctrine is pivotal for solving collision problems in physics.
• It underscores the invariant nature of momentum in isolated systems.

Physics Problem Solving

Physics problem solving involves breaking down a problem into manageable parts and using known physical laws to find the solution. A structured approach often involves understanding the problem, identifying the relevant equations, substituting known values, and solving for unknowns.
In this exercise:

• First, we identify the type of collision and know that the conservation principles are applicable.
• We apply the momentum equation, a pivotal tool in collision physics.
• Finally, solving algebraically for the unknown, here the velocity of the lighter ball.

Patience and practice are key. Often, visualizing or sketching the problem can help to ensure all aspects are considered. Confirming units and checking results for reasonability are also essential steps in verifying a solution.

Kinetic Energy

Kinetic energy refers to the energy that an object possesses due to its motion. It is one of the forms of energy conserved in an elastic collision. The basic formula for kinetic energy is:
\[ KE = \frac{1}{2} m v^2 \]Where:

• \( KE \) is the kinetic energy.
• \( m \) is the mass of the object.
• \( v \) is the velocity of the object.

In our scenario, no kinetic energy is lost in the collision between the two balls, though it may be redistributed between them.
This highlights a key feature of elastic collisions: while momentum guides the motion, kinetic energy offers a view into the transformations occurring within that motion. For real-world applications, understanding how energy remains and transforms allows us insight into material properties and energy-transfer efficiencies.

• Conservation of kinetic energy makes elastic collisions unique.
• Changes in kinetic energy distribution can indicate internal energy changes or force interactions during collision.