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(II) A 0.060 -kg tennis ball, moving with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\), has a head-on collision with a \(0.090-\mathrm{kg}\) ball initially moving in the same direction at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision.

Short Answer

Expert verified
The tennis ball ends with a velocity of 3.00 m/s, and the other ball has a velocity of 4.50 m/s, both in the same direction.

Step by step solution

01

Understand the Concept of Elastic Collision

In a perfectly elastic collision, both momentum and kinetic energy are conserved. This means neither is lost or dissipated, which helps in setting up the equations needed to solve the problem.
02

Apply Conservation of Momentum

The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, it's expressed as: \\[ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \] \where \(m_1 = 0.060\) kg, \(v_1 = 4.50\) m/s, \(m_2 = 0.090\) kg, \(v_2 = 3.00\) m/s, \(v_1'\) and \(v_2'\) are the final velocities we need to find.
03

Apply Conservation of Kinetic Energy

For an elastic collision, kinetic energy is conserved as: \\[ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1(v_1')^2 + \frac{1}{2}m_2(v_2')^2 \] \Substitute the known values into this equation, using \(m_1 = 0.060\) kg, \(v_1 = 4.50\) m/s, \(m_2 = 0.090\) kg, and \(v_2 = 3.00\) m/s.
04

Solve the Equations Simultaneously

Use the two equations derived from the conservation of momentum and kinetic energy to create a system of equations. Solve these simultaneously to find \(v_1'\) and \(v_2'\), the final velocities of the two balls. You will typically need to solve one equation for one variable and substitute back into the other equation.
05

Calculate the Final Velocities

From solving the simultaneous equations, you find: \\[ v_1' = 3.00 \text{ m/s} \] \\[ v_2' = 4.50 \text{ m/s} \] \This means the tennis ball slows down to 3.00 m/s, while the 0.090 kg ball speeds up to 4.50 m/s.

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Key Concepts

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

Conservation of Momentum
Momentum is a fundamental concept in physics that remains conserved in an isolated system. It's all about the mass of an object multiplied by its velocity, giving it the quantity known as momentum. In any elastic collision, like the one with the tennis balls in our problem, the momentum before the collision will be equal to the momentum after the collision.

Here's the basic equation for momentum conservation:
  • Initial momentum = Final momentum
  • \( m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \)
In this equation, \( m_1 \) and \( m_2 \) are the masses of the balls, \( v_1 \) and \( v_2 \) are the initial velocities, while \( v_1' \) and \( v_2' \) are the final velocities.

To solve the problem, you need to substitute these values and solve for the unknowns, which are the final velocities after collision. Remember, both balls are in motion, but their combined momentum remains constant.
Conservation of Kinetic Energy
Kinetic energy, just like momentum, is conserved in a perfectly elastic collision. Kinetic energy is the energy an object possesses due to its motion and can be calculated with the formula \( KE = \frac{1}{2} mv^2 \). In an elastic collision, the total kinetic energy before and after is the same. This means no energy is "lost"; it is merely transferred between the colliding objects.

The equation for kinetic energy conservation in the given problem is:
  • Initial kinetic energy = Final kinetic energy
  • \[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2 \]
Here, you'll also input the masses and velocities, solving for the velocities after the collision. It’s crucial to understand that even though kinetic energy is not intuitively "visible," it is fully governed by principles of energy conservation, making it a powerful tool in physics problem-solving.
Physics Problem Solving
Tackling physics problems efficiently often requires breaking down complex ideas into simpler concepts, just like we do with the conservation laws. Remember a few key strategies when dealing with problems involving collisions:

  • First, carefully read and understand the problem. Identify what's known and what's unknown.
  • Apply conservation laws first, such as momentum and energy, to set up the necessary equations.
  • Look for symmetries and repetitions, which can simplify complex situations.
  • Visualize the problem where possible, using diagrams or drawings.
  • After applying the physical concepts, solve the mathematical equations by substitution or simultaneous solving if necessary.
  • Finally, check your answers physically and mathematically to ensure they make sense.
By following structured problem-solving techniques, you can make daunting physics problems more manageable and even enjoyable. Building these skills will serve you well in both academics and real-world scenarios.

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Most popular questions from this chapter

(II) A child in a boat throws a 5.70 -kg package out horizontally with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\), Fig. 9-37. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is \(24.0 \mathrm{~kg}\) and that of the boat is \(35.0 \mathrm{~kg}\).

(II) A 144-g baseball moving \(28.0 \mathrm{~m} / \mathrm{s}\) strikes a stationary 5.25-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at \(1.10 \mathrm{~m} / \mathrm{s} .\) (a) What is the baseball's speed after the collision? (b) Find the total kinetic energy before and after the collision.

The space shuttle launches an \(850-\mathrm{kg}\) satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.0 s to give it a velocity of \(0.30 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction relative to the shuttle. The mass of the shuttle is \(92,000 \mathrm{~kg} .\) ( \(a\) ) Determine the component of velocity \(v_{\mathrm{f}}\) of the shuttle in the minus \(z\) -direction resulting from the ejection. (b) Find the average force that the shuttle exerts on the satellite during the ejection.

(II) A 6.0 -kg object moving in the \(+x\) direction at \(5.5 \mathrm{~m} / \mathrm{s}\) collides head-on with an \(8.0-\mathrm{kg}\) object moving in the \(-x\) direction at \(4.0 \mathrm{~m} / \mathrm{s} .\) Find the final velocity of each mass if: ( \(a\) ) the objects stick together; \((b)\) the collision is elastic; \((c)\) the \(6.0-\mathrm{kg}\) object is at rest after the collision; \((d)\) the \(8.0-\mathrm{kg}\) object is at rest after the collision; \((e)\) the \(6.0-\mathrm{kg}\) object has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the \(-x\) direction after the collision. Are the results in \((c),(d),\) and \((e)\) "reasonable"? Explain.

(II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball? (b) What fraction of the original kinetic energy \((\Delta K / K)\) gets transferred to the second ball?

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