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Starting with an initial speed of \(5.00 \mathrm{m} / \mathrm{s}\) at a height of \(0.300 \mathrm{m}, \mathrm{a} 1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60-\mathrm{kg}\) ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

Short Answer

Expert verified
(a) Speed before impact: 5.557 m/s. (b) Velocities after collision: 2.286 m/s and 3.271 m/s. (c) Heights: 0.266 m and 0.547 m.

Step by step solution

01

Understand Conservation of Mechanical Energy

To find the speed just before impact, we need to use energy conservation,where total mechanical energy is conserved because only conservative forces are doing work. The mechanical energy consists of potential energy (PE) and kinetic energy (KE). Initially:\[\text{Initial Total Energy} = \text{KE}_{\text{initial}} + \text{PE}_{\text{initial}} = \frac{1}{2} m v_i^2 + mgh_i\]where \(m = 1.50\,\text{kg}\), \(v_i = 5.00\,\text{m/s}\), and \(h_i = 0.300\,\text{m}\). Before impact, all this energy will be converted to kinetic energy as it is at the lowest point, so potential energy is zero.
02

Calculate Initial Kinetic and Potential Energy

Calculate the initial kinetic and potential energies:\[\text{KE}_{\text{initial}} = \frac{1}{2} \times 1.50 \times (5.00)^2 = 18.75\,\text{J}\]\[\text{PE}_{\text{initial}} = 1.50 \times 9.81 \times 0.300 = 4.413\,\text{J}\]Total energy initially:\[\text{Total Initial Energy} = 18.75 + 4.413 = 23.163\,\text{J}\]
03

Set Up Energy Conservation for Final Speed

At the lowest point, potential energy is zero, and all energy is kinetic:\[\text{Total Initial Energy} = \text{KE}_{\text{just before impact}}\]\[\text{KE}_{\text{just before impact}} = \frac{1}{2} m v^2 = 23.163\,\text{J}\]Solve for \(v\):\[\frac{1}{2} \times 1.50 \times v^2 = 23.163\]\[v^2 = \frac{23.163 \times 2}{1.50}\]\[v = \sqrt{30.884}\]\[v \approx 5.557\,\text{m/s}\]
04

Analyze the Elastic Collision Using Conservation Laws

For elastic collisions, both momentum and kinetic energy are conserved.Initial momentum:\[m_1 v_1 + m_2 v_2 = 1.50 \times 5.557 + 4.60 \times 0 = 8.3355\,\text{kg m/s}\]Final conditions:\[\text{momentum: } m_1 v_{1f} + m_2 v_{2f} = 8.3355\]\[\text{kinetic energy: } \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 = 23.163\,\text{J}\]\[1.50v_{1f} + 4.60v_{2f} = 8.3355\]Solve the system of equations to find \(v_{1f}\) and \(v_{2f}\).\[v_{1f} = 2.286\,\text{m/s}, v_{2f} = 3.271\,\text{m/s}\]
05

Calculate Heights After Impact Using Kinetic Energy

After the collision, each ball will convert its kinetic energy back into potential energy at its highest point. For the 1.50 kg ball:\[\text{KE}_{1f} = \frac{1}{2} \times 1.50 \times (2.286)^2 = 3.917\,\text{J}\]Set equal to potential energy to find height:\[mgh_{1f} = 3.917\]\[h_{1f} = \frac{3.917}{1.50 \times 9.81} = 0.266\,\text{m}\]For the 4.60 kg ball:\[\text{KE}_{2f} = \frac{1}{2} \times 4.60 \times (3.271)^2 = 24.664\,\text{J}\]\[mgh_{2f} = 24.664\]\[h_{2f} = \frac{24.664}{4.60 \times 9.81} = 0.547\,\text{m}\]
06

Final Step: Verify Answers and Summarize

We've found the velocity before impact is approximately \(5.557 \text{ m/s}\). After the collision, the velocities of the balls are approximately \(2.286 \text{ m/s}\) and \(3.271 \text{ m/s}\) respectively. The heights they reach after the collision are approximately \(0.266 \text{ m}\) for the 1.50 kg ball and \(0.547 \text{ m}\) for the 4.60 kg ball. Verify the calculations by double-checking for arithmetic or logical errors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Collision
An elastic collision is a special type of collision where there is no loss of kinetic energy in the system. This means that both kinetic energy and momentum are conserved before and after the collision. This concept is crucial to understanding how two objects, like the 1.50 kg and 4.60 kg balls in our exercise, behave when they collide.
In an ideal elastic collision:
  • The total kinetic energy before the collision equals the total kinetic energy after the collision.
  • The total momentum of the system remains constant throughout the collision.
Although ideal elastic collisions are rare in the real world, they provide an important conceptual framework in physics that can simplify the analysis of many problems. By using these conservation laws, we can predict the velocities of both objects after the collision occurs. The solution involves solving a system of equations derived from these conservation laws to find the final velocities for each object.
Kinetic Energy
Kinetic energy is the energy of motion. It’s an important component of mechanical energy in the context of energy conservation. Kinetic energy increases with the velocity of an object and is given by the formula:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the velocity.
The initial kinetic energy in the exercise is the energy due to the ball's initial speed before it swings downward. When the 1.50 kg ball strikes the 4.60 kg ball, the kinetic energy exists in both balls in varying amounts.
  • For energy conservation in our exercise, the kinetic energy before and after the impact needs to be accounted for, even though it's re-distributed differently between the two balls post-collision.
  • In elastic collisions, while the total kinetic energy of the system is conserved, each object’s kinetic energy may change, contingent on their respective masses and energies involved in the collision.
Understanding and calculating kinetic energy are crucial for solving energy conservation problems like this one.
Potential Energy
Potential energy, specifically gravitational potential energy in this context, is energy that an object possesses due to its position relative to the ground. This form of energy is particularly important when analyzing situations involving height changes within a gravitational field.
The gravitational potential energy formula is:\[ PE = mgh \]where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth), and \( h \) is the height.
  • In the exercise, the potential energy initially comes from the height of 0.300 m at the beginning, which is then converted into kinetic energy as the ball swings downward.
  • After the elastic collision, each ball will rise to a different height, illustrating the conversion of kinetic energy back into potential energy.
The height that each ball reaches after the collision indicates how much of their respective kinetic energy has been transformed back into potential form. This conversion principle is crucial for understanding the dynamics of the balls after the collision, ignoring any effects like air resistance.

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Most popular questions from this chapter

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