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.

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 0.15-kg baseball is pitched with a speed of \(35 \mathrm{m} / \mathrm{s}\) \((78 \mathrm{mph}) .\) When the ball hits the catcher's glove, the glove moves back by \(5.0 \mathrm{cm}(2 \text { in. })\) as it stops the ball. (a) What was the change in momentum of the baseball? (b) What impulse was applied to the baseball? (c) Assuming a constant acceleration of the ball, what was the average force applied by the catcher's glove?
A boy of mass \(60.0 \mathrm{kg}\) is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was \(8.0 \mathrm{m}\) above the net. The firefighters lower their arms as he lands in the net so that he is brought to a complete stop in a time of \(0.40 \mathrm{s}\). (a) What is his change in momentum during the 0.40 -s interval? (b) What is the impulse on the net due to the boy during the interval? [Hint: Do not ignore gravity.] (c) What is the average force on the net due to the boy during the interval?
A sled of mass \(5.0 \mathrm{kg}\) is coasting along on a frictionless ice- covered lake at a constant speed of \(1.0 \mathrm{m} / \mathrm{s} .\) A \(1.0-\mathrm{kg}\) book is dropped vertically onto the sled. At what speed does the sled move once the book is on it?
An intergalactic spaceship is traveling through space far from any planets or stars, where no human has gone before. The ship carries a crew of 30 people (of total mass \(\left.2.0 \times 10^{3} \mathrm{kg}\right) .\) If the speed of the spaceship is \(1.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) and its mass (excluding the crew) is \(4.8 \times 10^{4} \mathrm{kg},\) what is the magnitude of the total momentum of the ship and the crew?
A tennis ball of mass \(0.060 \mathrm{kg}\) is served. It strikes the ground with a velocity of \(54 \mathrm{m} / \mathrm{s}(120 \mathrm{mi} / \mathrm{h})\) at an angle of \(22^{\circ}\) below the horizontal. Just after the bounce it is moving at \(53 \mathrm{m} / \mathrm{s}\) at an angle of \(18^{\circ}\) above the horizontal. If the interaction with the ground lasts \(0.065 \mathrm{s}\), what average force did the ground exert on the ball?
See all solutions

Recommended explanations on Physics Textbooks

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