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Chapter 5: Problem 6

The average speed of a nitrogen molecule in air is about \(6.70 \times 10^{2} \mathrm{m} / \mathrm{s}\) , and its mass is \(4.68 \times 10^{-26} \mathrm{kg}\) . (a) If it takes \(3.00 \times 10^{-13} \mathrm{s}\) for a nitrogen molecule to hit a wall and rebound with the same speed but moving in the opposite direction, what is the average acceleration of the molecule during this time interval? (b) What average force does the molecule exert on the wall?

Short Answer

Expert verified
The average acceleration of the molecule is \(4.46 \times 10^{15} \mathrm{m/s}^2 \), and the average force exerted on the wall is \(2.09 \times 10^{-10} \mathrm{N}\).

Step by step solution

01

Understanding Average Acceleration

Average acceleration is given by the change in velocity divided by the time interval over which the change occurs. The formula for average acceleration is: \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time interval.
02

Calculating the Change in Velocity

Since the molecule moves in the opposite direction with the same speed after hitting the wall, the change in velocity is twice the original speed. Calculate the change in velocity using the formula \( \Delta v = v_{final} - v_{initial} = -(6.70 \times 10^{2}) - (6.70 \times 10^{2}) = -2(6.70 \times 10^{2}) \mathrm{m/s} \). Note that we consider the opposite direction as the negative of the initial direction.
03

Calculating Average Acceleration

Use the calculated change in velocity and the given time interval to find the average acceleration: \( a = \frac{-2(6.70 \times 10^{2})}{3.00 \times 10^{-13}} \). Find the magnitude of this acceleration.
04

Understanding Average Force

Newton's second law states that force is the product of mass and acceleration: \( F = ma \). The average force can be calculated by multiplying the average acceleration from the previous step by the mass of the molecule.
05

Calculating the Average Force

Once the magnitude of the average acceleration has been determined, multiply it by the mass of the nitrogen molecule to find the average force: \( F = m \times a \), where \( m = 4.68 \times 10^{-26} \mathrm{kg} \) and \( a \) is the magnitude of average acceleration obtained in Step 3.

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Key Concepts

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

Average Force
When we talk about average force, we're looking at the overall push or pull that is exerted on an object during a certain period of time. It's not about the force at an instant, but rather an average over a time interval. Think of it as a way to smooth out the peaks and valleys of force to get an overall picture.

In our nitrogen molecule example, after hitting the wall, the molecule exerts a force on the wall as it rebounds. The average force is calculated based on the change in momentum of the molecule over the time it contacts the wall. Using Newton's second law, which we will explore next, we can connect acceleration and mass to find this force. It's essential here to use the mass of the nitrogen molecule and the average acceleration we calculated from the change in velocity and the time taken for this change.
Newton's Second Law
Newton's second law is one of the most fundamental principles in physics. It's elegantly simple yet powerful: The law states that the force on an object is equal to its mass times its acceleration (\( F = ma \)).

This means that if you know the mass of an object and how quickly its speed is changing鈥攊ts acceleration鈥攜ou can determine the force being applied. In the context of our nitrogen molecule exercise, once we've calculated the average acceleration of the molecule during its collision with the wall, we can simply multiply it by the molecule's mass to find the average force the molecule exerts on the wall. This highlights the direct proportionality between force and acceleration鈥攄ouble the acceleration, and you'll double the force. The same goes for mass.
Velocity Change
Velocity change is a measure of how much an object's speed and direction are altered over a certain period. In technical terms, it's the difference between the object's final velocity and its initial velocity. Velocity change is crucial in calculations involving motion because it's directly related to acceleration, which is a measure of how quickly velocity changes.

In our example with the nitrogen molecule, since it reverses direction, the velocity change is actually twice the original speed 鈥� once to stop and once to move in the opposite direction. By expressing this velocity change quantitatively, we're setting the stage to calculate average acceleration, which is essential for then determining the force involved. Understanding velocity change gives us insight into the dynamic behavior of objects and is a key component in many physical calculations, including those involving Newton's second law.

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Most popular questions from this chapter

A \(25.0-\mathrm{kg}\) block is initially at rest on a horizontal surface. A horizontal force of 75.0 \(\mathrm{N}\) is required to set the block in motion. After it is in motion, a horizontal force of 60.0 \(\mathrm{N}\) is required to keep the block moving with constant speed. Find the coefficients of static and kinetic friction from this information.

To model a spacecraft, a toy rocket engine is securely fastened to a large puck, which can glide with negligible friction over a horizontal surface, taken as the \(x y\) plane. The \(4.00\) -kg puck has a velocity of \(300 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\) at one instant. Eight seconds later, its velocity is to be \((800 \hat{\mathbf{i}}+10.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\). Assuming the rocket engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude.

A block weighing 75.0 \(\mathrm{N}\) rests on a plane inclined at \(25.0^{\circ}\) to the horizontal. A force \(F\) is applied to the object at \(40.0^{\circ}\) to the horizontal. pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.363 and 0.156 . (a) What is the minimum value of \(F\) that will prevent the block from slipping down the plane? (b) What is the minimum value of \(F\) that will start the block moving up the plane? (c) What value of \(F\) will move the block up the plane with constant velocity?

Three forces acting on an object are given by \(\mathbf{F}_{1}=(-2.00 \hat{\mathbf{i}}+2.00 \hat{\mathbf{j}}) \mathrm{N}, \mathbf{F}_{2}=(5.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}) \mathrm{N},\) and \(\mathbf{F}_{3}=(-45.0 \hat{\mathbf{i}}) \mathrm{N} .\) The object experiences an acceleration of magnitude \(3.75 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 \(\mathrm{s}\) ? (d) What are the velocity components of the object after 10.0 \(\mathrm{s}\) ?

A woman at an airport is towing her 20.0 -kg suitcase at constant speed by pulling on a strap at an angle \(\theta\) above the horizontal (Fig. P5.40). She pulls on the strap with a \(35.0-\mathrm{N}\) force, and the friction force on the suitcase is 20.0 \(\mathrm{N}\) . Draw a free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase?
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