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Chapter 8: Problem 6

\(\bullet\) Some useful relationships. The following relationships between the momentum and kinetic energy of an object can be very useful for calculations: If an object of mass \(m\) has momentum of magnitude \(p\) and kinetic energy \(K,\) show that (a) \(K=\left(p^{2} / 2 m\right),\) and \((b) p=\sqrt{2 m K}\) . (c) Find the momen- tum of a 1.15 kg ball that has 15.0 J of kinetic energy. (d) Find the kinetic energy of a 3.50 kg cat that has 0.220 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) of momentum.

Short Answer

Expert verified
(c) The momentum is approximately 5.873 kg m/s. (d) The kinetic energy is approximately 0.00691 J.

Step by step solution

01

Understanding Kinetic Energy Formula

The formula for kinetic energy is given by \( K = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity. We need to relate this to momentum.
02

Establishing the Relationship with Momentum

Momentum \( p \) is defined by \( p = m v \). Solving for velocity, we find \( v = \frac{p}{m} \). Substituting \( v \) in the kinetic energy equation gives \( K = \frac{1}{2} m \left( \frac{p}{m} \right)^2 \).
03

Simplifying the Kinetic Energy Formula

Simplifying the expression \( K = \frac{1}{2} m \left( \frac{p}{m} \right)^2 \), we get \( K = \frac{p^2}{2m} \). This proves part (a) of the problem.
04

Deriving the Relationship for Momentum

From \( K = \frac{p^2}{2m} \), rearrange to solve for \( p \): \( p^2 = 2mK \), thus \( p = \sqrt{2mK} \). This proves part (b) of the problem.
05

Solving for Momentum of a Ball

Given that the ball's mass \( m = 1.15 \) kg and its kinetic energy \( K = 15.0 \) J, use the formula \( p = \sqrt{2mK} \). Substitute to get \( p = \sqrt{2 \times 1.15 \times 15.0} \).
06

Calculating the Momentum

The calculation gives \( p = \sqrt{34.5} \). Compute \( p \approx 5.873 \) kg m/s.
07

Finding Kinetic Energy of a Cat

Given the cat's mass \( m = 3.50 \) kg and its momentum \( p = 0.220 \) kg m/s, use the formula \( K = \frac{p^2}{2m} \). Substitute to find \( K = \frac{0.220^2}{2 \times 3.50} \).
08

Calculating the Kinetic Energy

Compute the expression \( K = \frac{0.0484}{7.0} \). Thus, \( K \approx 0.0069143 \) J.

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Key Concepts

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

Kinetic Energy Formula
Kinetic energy is a fascinating concept that helps us understand how energy is expressed in moving objects. The kinetic energy (\( K \)) of an object with mass (\( m \)) and velocity (\( v \)) is determined by the formula:\[K = \frac{1}{2} mv^2\]This formula tells us that kinetic energy depends on two factors:
  • The mass of the object; heavier items have greater kinetic energy if they move at the same speed as lighter ones.
  • The velocity of the object, squared; meaning small increases in speed lead to significant increases in kinetic energy.
In physics problem-solving, it's essential to connect this formula to momentum for more advanced calculations, such as determining how kinetic energy varies with different physical parameters using the unique relationships between these fundamental concepts.
Momentum Calculation
Momentum represents the quantity of motion an object has and is given by the equation:\[p = mv\]where (\( p \)) is the momentum. It's a vector quantity, meaning it has both magnitude and direction, reflecting the nature of the object's movement.

A crucial aspect of momentum is its relationship with kinetic energy. To find velocity (\( v \)) using momentum, rearrange the formula:\[v = \frac{p}{m}\]By substituting this back into the kinetic energy equation, we establish a direct relationship between kinetic energy and momentum:\[K = \frac{p^2}{2m}\]This shows us how momentum and mass can directly influence the kinetic energy of a moving object and helps us solve complex physics problems involving motion dynamics.
Physics Problem-Solving
Physics problems often require a deep understanding of both kinetic energy and momentum concepts. Let's explore how these principles apply in specific scenarios:
  • Finding the momentum of a ball: For a ball with a mass of 1.15 kg and 15.0 J of kinetic energy, use the relationship:\[p = \sqrt{2mK} = \sqrt{2 \times 1.15 \times 15.0}\] The calculation results in a momentum of approximately 5.873 kg m/s.
  • Calculating the kinetic energy of a cat: For a cat with a mass of 3.50 kg and momentum of 0.220 kg m/s, we use:\[K = \frac{p^2}{2m} = \frac{0.220^2}{2 \times 3.50}\] This results in about 0.0069143 J of kinetic energy.
These calculations illustrate how dynamic these relationships are and the methodical steps required to accurately solve these physics problems.

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Most popular questions from this chapter

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the bat traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of 65.0 \(\mathrm{m} / \mathrm{s}\) . (a) What are the horizontal and vertical components of the impulse the bat imparts to the ball? (b) If the ball and bat are in contact for 1.75 \(\mathrm{ms}\) , find the horizontal and vertical components of the average force on the ball.

\(\bullet\) In outer space, where gravity is negligible, a \(75,000 \mathrm{kg}\) rocket (including \(50,000 \mathrm{kg}\) of fuel) expels this fuel at a steady rate of 135 \(\mathrm{kg} / \mathrm{s}\) with a speed of 1200 \(\mathrm{m} / \mathrm{s}\) relative to the rocket. (a) Find the thrust of the rocket. (b) What are the initial acceleration and the maximum acceleration of the rocket? (c) After the fuel runs out, what happens to this rocket's acceleration? Does it (i) remain the same as it was just as the fuel ran out, (ii) suddenly become zero, or (iii) gradually drop to zero? Explain your reasoning. (d) After the fuel runs out, what happens to the rocket's speed? Does it (i) remain the same as it was just as the fuel ran out, (ii) suddenly become zero, or (iii) gradually drop to zero? Explain your reasoning.

. Two identical objects traveling in opposite directions with the same speed \(V\) make a head-on collision. Find the speed of each object after the collision if (a) they stick together and (b) if the collision is perfectly elastic.

\(\cdot\) Three identical boxcars are coupled together and are moving at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) on a level track. They collide with another identical boxcar that is initially at rest and couple to it, so that the four cars roll on as a unit. Friction is small enough to be neglected. (a) What is the speed of the four cars? (b) What percentage of the kinetic energy of the boxcars is dissipated in the collision? What happened to this energy?

A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of 6000 \(\mathrm{kg}\) and ejects gas at a relative velocity of magnitude 2000 \(\mathrm{m} / \mathrm{s}\) , how much gas must it eject in the first second to have an initial acceleration of 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) .
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