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Chapter 6: Problem 1

If a \(60-\mathrm{kg}\) woman is riding in a car traveling at \(90 \mathrm{~km} / \mathrm{h}\), what is her linear momentum relative to (a) the ground and (b) the car?

Short Answer

Expert verified
(a) 1500 kg m/s relative to the ground, (b) 0 kg m/s relative to the car.

Step by step solution

01

Understanding Linear Momentum

Linear momentum is defined as the product of an object's mass and its velocity. It is given by the formula: \[ p = m imes v \]where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity.
02

Calculate the Momentum Relative to the Ground

First, use the woman's mass (60 kg) and the velocity of the car (90 km/h) to calculate her momentum relative to the ground. Convert the velocity from km/h to m/s by using the conversion factor (1 km/h = \( \frac{1}{3.6} \) m/s): \[ 90 \text{ km/h} = \frac{90}{3.6} \text{ m/s} = 25 \text{ m/s} \]Now apply the momentum formula: \[ p = 60 \times 25 = 1500 \text{ kg m/s} \]So, her linear momentum relative to the ground is 1500 kg m/s.
03

Determine the Momentum Relative to the Car

When calculating linear momentum relative to the car, consider the fact that if both the woman and the car are moving together, her relative velocity to the car is 0 m/s. Thus: \[ p = m \times v = 60 \times 0 = 0 \text{ kg m/s} \]Thus, her linear momentum relative to the car is 0 kg m/s.

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Key Concepts

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

Relative Motion
Relative motion is an important concept in physics, especially when analyzing the movement of objects from different perspectives. It describes how the velocity of one object appears to change when viewed relative to another moving object. In simpler terms, it represents how fast an object seems to be moving when observed from another moving object rather than from a stationary point.
  • When dealing with relative motion, choose a reference point or object to compare movements.
  • In the scenario of the woman in the car, we find the car itself and the ground as the two reference points.
  • Compared to the ground, both the woman and the car are moving at 90 km/h. Therefore, her relative velocity to the ground is 90 km/h.
  • With the car as a reference, however, her velocity becomes 0 km/h, because she is moving in unison with the car.
Understanding relative motion is essential to correctly apply physics equations and concepts, notably in exercises involving relative speeds or positions.
Momentum Calculation
Calculating momentum is straightforward once you grasp the fundamental formula of linear momentum, which is the product of mass and velocity: \[ p = m \times v \] where \( p \) represents momentum, \( m \) is mass, and \( v \) stands for velocity. In our example, the woman's mass is given as 60 kg and the velocity of the car is 90 km/h.

  • Convert velocity to meters per second by dividing by 3.6: \[ \frac{90}{3.6} = 25 \text{ m/s} \]
  • Insert these values into the momentum equation: \[ p = 60 \times 25 = 1500 \text{ kg m/s} \]
  • Consequently, her momentum relative to the ground is 1500 kg m/s.
When shifting the frame of reference to the car, her velocity becomes 0 m/s as she is stationary relative to it, making her momentum \( 0 \text{ kg m/s} \).
Grasping momentum calculations is key to solving physics problems effectively, as they often form the basis for more complex computations.
Physics Problems
Physics problems, like the one about the woman's momentum, often aim to reinforce understanding of key principles like relative motion and momentum. They present real-world scenarios to practice the translation of physical concepts into mathematical expressions. Here's how to tackle similar problems:
  • Identify Given Information: Clearly note the mass, velocity, and reference frames provided in the problem.
  • Apply Relevant Equations: Utilize specific formulas, such as those for momentum, keeping units consistent.
  • Think About Reference Frames: Consider different views (e.g., from the ground or from a moving object) to assess relative motion and momentum.
  • Verify Results: Double-check calculations and examine if the results make physical sense.
By systematically applying these steps, students enhance their problem-solving skills, making physics more intuitive and less daunting.

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Most popular questions from this chapter

A \(60-\mathrm{kg}\) astronaut floating at rest in space outside a space capsule throws his \(0.50-\mathrm{kg}\) hammer such that it moves with a speed of \(10 \mathrm{~m} / \mathrm{s}\) relative to the capsule. What happens to the astronaut?

Consider two string-suspended balls, both with a mass of \(0.15 \mathrm{~kg}\). (Similar to the arrangement in Fig. \(6.15,\) but with only two balls.) One ball is pulled back in line with the other so it has a vertical height of 10 \(\mathrm{cm},\) and is then released. (a) What is the speed of the ball just before hitting the stationary one? (b) If the collision is completely inelastic, to what height do the balls swing?

A \(1200-\mathrm{kg}\) car moving to the right with a speed of \(25 \mathrm{~m} / \mathrm{s}\) collides with a \(1500-\mathrm{kg}\) truck and locks bumpers with the truck. Calculate the velocity of the combination after the collision if the truck is initially (a) at rest, (b) moving to the right with a speed of \(20 \mathrm{~m} / \mathrm{s},\) and (c) moving to the left with a speed of \(20 \mathrm{~m} / \mathrm{s}\).

Two cups are placed on a uniform board that is balanced on a cylinder ( \(\mathbf{v}\) Fig. 6.40 ). The board has a mass of \(2.00 \mathrm{~kg}\) and is \(2.00 \mathrm{~m}\) long. The mass of \(\operatorname{cup} 1\) is \(200 \mathrm{~g}\) and it is placed \(1.05 \mathrm{~m}\) to the left of the balance point. The mass of cup 2 is \(400 \mathrm{~g}\). Where should cup 2 be placed for balance (relative to the right end of the board)?

A volleyball is traveling toward you. (a) Which action will require a greater force on the volleyball, your catching the ball or your hitting the ball back? Why? (b) A 0.45 -kg volleyball travels with a horizontal velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) over the net. You jump up and hit the ball back with a horizontal velocity of \(7.0 \mathrm{~m} / \mathrm{s}\). If the contact time is \(0.040 \mathrm{~s}\), what was the average force on the ball?
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