Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 38

Prove Eq. $(7-13) \Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}=M \overrightarrow{\mathbf{a}}_{\mathrm{CM}} .\( [Hint: Start with \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}}=\lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t),\( where \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}$ is the net external force acting on a system and \(\overrightarrow{\mathbf{p}}\) is the total momentum of the system.]

Short Answer

Expert verified
Question: Prove that the net external force acting on a system is equal to the product of the total mass of the system and the acceleration of its center of mass, given the expression \(\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}}=\lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t)\). Answer: By defining the total momentum of the system, expressing the change in momentum, defining acceleration, substituting the change in velocity, using the limit, and simplifying the expression, we obtain \(\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}} = M \overrightarrow{\mathbf{a}}_{\mathrm{CM}}\), which proves the required statement.

Step by step solution

01

Start with the given formula

We are provided with the formula \(\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}}=\lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t)\), which expresses the net external force acting on a system in terms of the change in the total momentum of the system over an infinitesimally small time interval.
02

Express the change in momentum

Recall that the total momentum of a system is given by \(\overrightarrow{\mathbf{p}} = M\overrightarrow{\mathbf{v}}_{\mathrm{CM}}\), where \(M\) is the total mass of the system and \(\overrightarrow{\mathbf{v}}_{\mathrm{CM}}\) is the velocity of the center of mass. To find the change in momentum, we can write \(\Delta \overrightarrow{\mathbf{p}} = M\Delta \overrightarrow{\mathbf{v}}_{\mathrm{CM}}\).
03

Define acceleration

Acceleration is defined as the change in velocity over time. Therefore, we can express the change in the center of mass velocity as \(\Delta \overrightarrow{\mathbf{v}}_{\mathrm{CM}} = \overrightarrow{\mathbf{a}}_{\mathrm{CM}}\Delta t\).
04

Substitute the change in velocity

Now, substitute the expression for the change in velocity from step 3 into the change in momentum from step 2: \(\Delta \overrightarrow{\mathbf{p}} = M(\overrightarrow{\mathbf{a}}_{\mathrm{CM}}\Delta t)\).
05

Use the limit

Now use the limit to find the net external force: \(\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}} = \lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t) = \lim _{\Delta t \rightarrow 0}\frac{M(\overrightarrow{\mathbf{a}}_{\mathrm{CM}}\Delta t)}{\Delta t}\).
06

Simplify the expression

Cancel the \(\Delta t\) terms in the numerator and the denominator: \(\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}} = M \overrightarrow{\mathbf{a}}_{\mathrm{CM}}\). This expression shows that the net external force acting on a system is equal to the product of the total mass of the system and the acceleration of its center of mass, as required.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rifle has a mass of \(4.5 \mathrm{kg}\) and it fires a bullet of mass $10.0 \mathrm{g}\( at a muzzle speed of \)820 \mathrm{m} / \mathrm{s} .$ What is the recoil speed of the rifle as the bullet leaves the gun barrel?
Two African swallows fly toward one another, carrying coconuts. The first swallow is flying north horizontally with a speed of $20 \mathrm{m} / \mathrm{s} .$ The second swallow is flying at the same height as the first and in the opposite direction with a speed of \(15 \mathrm{m} / \mathrm{s}\). The mass of the first swallow is \(0.270 \mathrm{kg}\) and the mass of his coconut is \(0.80 \mathrm{kg} .\) The second swallow's mass is \(0.220 \mathrm{kg}\) and her coconut's mass is 0.70 kg. The swallows collide and lose their coconuts. Immediately after the collision, the \(0.80-\mathrm{kg}\) coconut travels \(10^{\circ}\) west of south with a speed of \(13 \mathrm{m} / \mathrm{s}\) and the 0.70 -kg coconut moves \(30^{\circ}\) east of north with a speed of $14 \mathrm{m} / \mathrm{s} .$ The two birds are tangled up with one another and stop flapping their wings as they travel off together. What is the velocity of the birds immediately after the collision?
In the railroad freight yard, an empty freight car of mass \(m\) rolls along a straight level track at \(1.0 \mathrm{m} / \mathrm{s}\) and collides with an initially stationary, fully loaded boxcar of mass 4.0m. The two cars couple together on collision. (a) What is the speed of the two cars after the collision? (b) Suppose instead that the two cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one was moving at \(1.0 \mathrm{m} / \mathrm{s} ?\)
Two pendulum bobs have equal masses and lengths \((5.1 \mathrm{m}) .\) Bob \(\mathrm{A}\) is initially held horizontally while bob \(\mathrm{B}\) hangs vertically at rest. Bob A is released and collides elastically with bob B. How fast is bob B moving immediately after the collision?
An object of mass \(3.0 \mathrm{kg}\) is projected into the air at a \(55^{\circ}\) angle. It hits the ground 3.4 s later. What is its change in momentum while it is in the air? Ignore air resistance.
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Atoms and Radioactivity

Read Explanation

Famous Physicists

Read Explanation

Classical Mechanics

Read Explanation

Quantum Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free