At a certain instant a particle is moving in the +xdirection with momentum+8kg·m/s. During the next 0.13sa constant force acts on the particle, with Fx=-7N and xFy=+5N. What is the magnitude of the momentum of the particle at the end of this v interval?

Short Answer

Expert verified

The magnitude of the particle's momentum at the end of this 0.13s interval is x8.26\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}.

Step by step solution

01

Identification of the given data

The given data can be listed below as

- The particle has the momentum of +8kg·m/s.

- The force acts on the particle 0.13\mathrm{~s}.

- The constant force which is acting on the particle is xFx=-7Nand xFy=+5N respectively.

02

Significance of the momentum principle for the particle

This principle states that if there is a collision between two objects, the total momentum before and after the collision will be equal as there is no external force.

The equation of the principle of momentum gives the magnitude of the momentum of the particle.

03

Determination of the magnitude of the momentum of the particle

From the momentum principle, the magnitude of the momentum of the particle can be expressed as:

Here xp2is the final momentum of the particle and Fnel=Fx2+Fy2+Fz2p1is the initial momentum of the particle.

Fnetis the net force acting on the body that can be expressed as:

Fnet=Fx2+Fy2+Fz2

For xFx=-7N,Fy=+5Nand Fz=0net force can be calculated as:

Fz=0

The time interval\Deltat=0.13\mathrm{~s}.

For Δt=0.13s,Fnel=8.6027N, and p1=+8kg·m/s; from the equation, (1) the Final momentum p2can be calculated as:

\begin{aligned}&p_{2}=8\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}+8.6027\mathrm{~N}\times0.13\mathrm{~s}\\&p_{2}=8\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}+(8.6027\times0.13)\cdot\left(1\mathrm{~N}\cdot1\mathrm{~s}\times\frac{1\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}^{2}}{1\mathrm{~N}}\right)\\&p_{2}=8\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}+1.118351\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\\&p_{2}=9.118351\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\end{aligned}

Thus, the magnitude of the momentum of the particle at the end of 0.13sthe time interval is 9.118351kg·m/s.

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 cart rolls with low friction on a track. A fan is mounted on the cart, and when the fan is turned on, there is a constant force acting on the cart. Three different experiments are performed:

(a) Fan off: The cart is originally at rest. You give it a brief push, and it coasts a long distance along the track in the +x direction, slowly coming to a stop.

(b) Fan forward: The fan is turned on, and you hold the cart stationary. You then take your hand away, and the cart moves forward, in the +x direction. After traveling a long distance along the track, you quickly stop and hold the cart.

(c) Fan backward: The fan is turned on facing the “wrong” way, and you hold the cart stationary. You give it a brief push, and the cart moves forward, in the +x direction, slowing down and then turning around, returning to the starting position, where you quickly stop and hold the cart. Figure 2.57 displays four graphs of px (numbered 1–4), the x component of momentum, vs. time. The graphs start when the cart is at rest, and end when the cart is again at rest. Match the experiment with the correct graph.

(a)On a piece of graph paper, draw the vector g=<4,7,0>m. Put the tail of the vector at the origin.

(b)Calculate the magnitude ofg.

(c)Calculateg^, the unit vector pointing in the direction ofg.

(d)On the graph, drawg. Put the tail of the vector at <1,0,0>so you can compareg^andg.

(e)Calculate the product of the magnitudegtimes the unit vectorg^,(g)(g^).

A thin diverging lens of focal length 25cm is placed 18cm to the right of a point source of blue light on the axis of the lens. Where is the image of the source? Is it a real or a virtual image? If you placed a sheet of paper at the location of the image, what would you see on the paper?

In a lab experiment you observe that a pendulum swings with a “period” (time for one round trip) of 2s. In an iterative calculation of the motion, which of the following would NOT be a reasonable choice for t, for either hand or computer iterative calculations? a) 1sb) 0.1sc) 0.05sd) 0.01s.

Here are questions about human diet. (a) A typical candy bar provides 280 calories (one “food” or “large” calorie is equal to ). How many candy bars would you have to eat to replace the chemical energy you expend doing 100 sit-ups? Explain your work, including any approximations or assumptions you make. (In a sit-up, you go from lying on your back to sitting up.) (b) How many days of a diet of 2000 large calories are equivalent to the gravitational energy difference for you between sea level and the top of Mount Everest, 8848 m above sea level? (However, the body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

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