Problem 6 - College Physics

Question

\(\bullet\) 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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \( K = \frac{1}{2} m \left( \frac{p}{m} \right)^2 \).

Step 3: 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.

Step 4: 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.

Step 5: 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 \( p = \sqrt{2 \times 1.15 \times 15.0} \).

Step 6: Calculating the Momentum

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

Step 7: 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 \( K = \frac{0.220^2}{2 \times 3.50} \).

Step 8: Calculating the Kinetic Energy

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

Key Concepts

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.