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Chapter 6: Problem 2

The linear momentum of a runner in a 100 -m dash is \(7.5 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). If the runner's speed is \(10 \mathrm{~m} / \mathrm{s}\), what is his mass?

Short Answer

Expert verified
The runner's mass is 75 kg.

Step by step solution

01

Understand the Problem

We are given the linear momentum of a runner and their speed. We need to determine the runner's mass.
02

Formula Identification

The formula for linear momentum is given by \( p = m imes v \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity (speed).
03

Rearrange the Formula

To find the mass, we rearrange the formula to solve for \( m \): \( m = \frac{p}{v} \).
04

Substitute the Values

Substitute the given values into the rearranged formula: \( m = \frac{7.5 \times 10^2 \mathrm{~kg} \cdot \mathrm{m/s}}{10 \mathrm{~m/s}} \).
05

Perform the Calculation

Calculate the mass by dividing the momentum by the speed: \( m = 75 \mathrm{~kg} \).

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Key Concepts

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

Momentum Formula
In physics, momentum is a fundamental concept that describes the quantity of motion an object has. It is a vector quantity, which means it has both magnitude and direction:
  • Momentum is denoted by the symbol \( p \).
  • The formula for linear momentum is \( p = m \times v \).
  • Here, \( m \) stands for mass and \( v \) is velocity.
The product of an object's mass and its velocity, momentum, provides insight into how difficult it is to stop the object.
The higher the momentum, the harder it is to bring the object to rest.
In our exercise, the runner's momentum is given as \( 7.5 \times 10^2 \, \text{kg} \cdot \text{m/s} \).
This tells us how much force, or energy, would be required to stop the runner in motion or change the speed.
Mass Calculation
Calculating mass from momentum requires rearranging the momentum formula. Once you have the formula, \( p = m \times v \), you can solve for mass:
  • Rearrange the formula to \( m = \frac{p}{v} \).
  • This rearrangement helps us express mass in terms of momentum and velocity.
In the exercise, we substitute the given values into this formula.
The linear momentum value is \( 7.5 \times 10^2 \, \text{kg} \cdot \text{m/s} \) and the velocity is \( 10 \, \text{m/s} \).
Substituting, \( m = \frac{7.5 \times 10^2}{10} = 75 \, \text{kg} \).
This calculation confirms that the mass of the runner is 75 kg.
Understanding this step is crucial as it teaches how physical properties are interconnected through mathematical expressions.
Velocity in Physics
Velocity is a critical physics concept frequently used in motion-related calculations. It informs us about the speed and direction of an object's movement:
  • Velocity is denoted by the symbol \( v \).
  • It is different from speed as it includes direction.
  • Measured in \( \text{meters per second (m/s)} \).
In the given exercise, the runner's velocity is \( 10 \, \text{m/s} \).
This tells us not only how fast the runner is moving but also gives context to the motion, which is a key factor when calculating momentum.
Velocity helps differentiate motion concepts from static elements, highlighting its importance in force calculations, kinetic energy, and more.

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Most popular questions from this chapter

An astronaut (mass of \(100 \mathrm{~kg},\) with equipment) is headed back to her space station at a speed of \(0.750 \mathrm{~m} / \mathrm{s}\) but at the wrong angle. To correct her direction, she fires rockets from her backpack at right angles to her motion for a brief time. These directional rockets exert a constant force of \(100.0 \mathrm{~N}\) for only \(0.200 \mathrm{~s}\). [Neglect the small loss of mass due to burning fuel and assume the impulse is at right angles to her initial momentum.] (a) What is the magnitude of the impulse delivered to the astronaut? (b) What is her new direction (relative to the initial direction)? (c) What is her new speed?

An explosion of a \(10.0-\mathrm{kg}\) bomb releases only two separate pieces. The bomb was initially at rest and a 4.00-kg piece travels westward at \(100 \mathrm{~m} / \mathrm{s}\) immediately after the explosion. (a) What are the speed and direction of the other piece immediately after the explosion? (b) How much kinetic energy was released in this explosion?

A 1600 -kg (empty) truck rolls with a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) under a loading bin, and a mass of \(3500 \mathrm{~kg}\) is deposited into the truck. What is the truck's speed immediately after loading?

(a) The center of mass of a system consisting of two 0.10-kg particles is located at the origin. If one of the particles is at \((0,0.45 \mathrm{~m}),\) where is the other? \((\mathrm{b})\) If the masses are moved so their center of mass is located at \((0.25 \mathrm{~m}\), \(0.15 \mathrm{~m}\) ), can you tell where the particles are located?

A major league catcher catches a fastball moving at \(95.0 \mathrm{mi} / \mathrm{h}\) and his hand and glove recoil \(10.0 \mathrm{~cm}\) in bringing the ball to rest. If it took 0.00470 s to bring the ball (with a mass of \(250 \mathrm{~g}\) ) to rest in the glove, (a) what are the magnitude and direction of the change in momentum of the ball? (b) Find the average force the ball exerts on the hand and glove.
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