Problem 15 - Applied Physics

Question

A \(125-\mathrm{kg}\) pile driver falls from a height of \(10.0 \mathrm{~m}\) to hit a piling. (a) What is its speed as it hits the piling? (b) With what momentum does it hit the piling?

Step-by-Step Solution

Answer

The speed is 14.0 m/s, and the momentum is 1750 kg⋅m/s.

Step 1: Identify Known Values

We are given the mass of the pile driver as \( m = 125 \, \text{kg} \), the height from which it falls as \( h = 10.0 \, \text{m} \), and we need to find the speed and momentum of the pile driver as it hits the piling.

Step 2: Calculate Speed Using Energy Conservation

The potential energy (PE) at the top is converted into kinetic energy (KE) at the bottom. The potential energy is given by \( \text{PE} = mgh \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. So, \( \text{PE} = 125 \times 9.8 \times 10.0 = 12250 \, \text{J} \). The kinetic energy is \( \text{KE} = \frac{1}{2}mv^2 \), setting \( \text{PE} = \text{KE} \), we have: \[ 12250 = \frac{1}{2} \times 125 \times v^2 \]Solve for \( v \): \[ v^2 = \frac{2 \times 12250}{125} = 196 \] \[ v = \sqrt{196} = 14.0 \, \text{m/s} \]

Step 3: Calculate Momentum

Momentum \( p \) is given by the formula \( p = mv \). Plugging in the values, we have: \[ p = 125 \times 14.0 = 1750 \, \text{kg} \cdot \text{m/s} \]

Key Concepts

Energy conservation

Energy conservation is a fundamental concept in physics, stating that energy cannot be created or destroyed in an isolated system. It only changes forms. In the context of our problem, energy conservation can help us understand how gravitational potential energy transforms into kinetic energy.

• Gravitational Potential Energy (PE): When an object is at a certain height, it possesses potential energy due to gravity. This energy can be calculated using the formula \( \text{PE} = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
• Kinetic Energy (KE): As the object falls, the potential energy is converted into kinetic energy. Kinetic energy is the energy of motion and can be represented as \( \text{KE} = \frac{1}{2}mv^2 \), where \( v \) is velocity.
• Application: For our falling pile driver, initially at rest, all of its potential energy is converted into kinetic energy just before impact. Thus, setting \( \text{PE} = \text{KE} \) allows us to derive the speed at which it hits the ground. This is a clear demonstration of energy conservation at work.

Momentum calculation

Momentum is a measure of an object's motion and is calculated as the product of its mass and velocity. Understanding momentum is essential for analyzing collisions and other dynamic events.

• Formula: The momentum \( p \) of an object is given by the equation \( p = mv \), where \( m \) is mass and \( v \) is velocity.
• Connection to Velocity: In the exercise, once we have calculated the velocity of the pile driver using energy conservation, it becomes straightforward to determine its momentum. Simply multiply the mass by this velocity to find the momentum.
• Significance: Knowing the momentum of an object helps predict the result of interactions, such as collisions. Greater momentum indicates more forceful impacts " — essential in fields like engineering and automotive safety design.

Kinematic equations

Kinematic equations describe the motion of objects and are crucial to solving physics problems involving velocity, acceleration, time, and displacement.

• Basic Definitions: Key variables include:
  • \( v \): Final velocity
  • \( u \): Initial velocity
  • \( a \): Acceleration
  • \( t \): Time
  • \( s \): Displacement
• Equations of Motion: While our problem primarily focused on energy conservation to find velocity, kinematic equations can also be used for similar problems. They include:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
  These equations provide a powerful toolkit for motion analysis in other contexts where either energy or force details are less relevant.
• Relevance to the Problem: The provided solution does not use a kinematic equation directly, due to the straightforward conversion of potential to kinetic energy. However, understanding these equations offers deeper insight into various situations involving constant acceleration, such as freely falling objects.