Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A tennis ball of mass 0.06 kgtravelling at a velocity of(9,-2,13) m/sis about to collide with an identical tennis ball whose velocity is(4,5,-10) m/s. a) What is the total momentum of the system of the two-tennis ball? b) What is the velocity of the center of mass of the two tennis balls?

Short Answer

Expert verified

a) the total momentum of the system of the two tennis ball is(0.78,0.18,0.18)kg.m/s

and b) the velocity of the center of mass of the two tennis ball is(6.5,1.5,1.5)m/s .

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The mass of the first tennis ball is 0.06kg.
  • The velocity of the first tennis ball is (9,-2,13)m/s.
  • The mass of the second tennis ball is the same as the first tennis ball.
  • The velocity of the second tennis ball is (4,5,-10)m/s.
02

Significance of the law of conservation of momentum and center of mass for the two rocks

The law of the conservation of momentum states that the total momentum for a body before and after the collision becomes equal.

The law of the center of mass states that if a rigid object is pushed then it will continue to move as a point mass.

The equation of the momentum gives the total momentum of the system of the two tennis ball and the equation of velocity gives the velocity of the two tennis ball.

03

Determination of the total momentum of the system and the velocity of the center of the mass of the two tennis ball

a) From the law of conservation of momentum, the equation of the total momentum of the system of the tennis ball can be expressed as:

p=m1v1+m2v2

Here, pis the total momentum of the system, m1andm2are the mass of the first and the second tennis ball respectively and v1andv2are the velocity of the first and the second tennis ball respectively.

Substituting the values in the above equation, we get-

p=(0.06kg)×((9,-2,13)m/s)+(0.06kg)×((4,5,-10)m/s)p=(0.78,0.18,0.18)kg·m/s

Thus, the total momentum of the system of the two tennis ball is (0.78,0.18,0.18)kg.m/s.

b) From the law of center of mass, the equation of the velocity of the center of mass of the tennis ball is expressed as:

v=m1v1+m2v2m1+m2

Substituting the values in the above equation, we get-

v=(0.06kg)×((9,-2,13)m/s)+(0.06kg)×((4,5,-10)m/s)(0.06kg)+(0.06kg)v=(6.5,1.5,1.5)m/s

Thus, the velocity of the center of mass of the tennis ball is(6.5,1.5,1.5)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

The mass of Mars is 6.4×1023and its radius is 3.6×106m. What is the value of constant g on Mars?

At t = 532.0s after midnight, a spacecraft of mass 1400 kg is located at position (3x105, 7x105,-4x105 ) m, and at that time an asteroid whose mass is 7 x 10 15 kg is located at position (9 x 10 ^ 5, - 3 x 10 ^ 5, - 12 x 10 ^ 5) m. There are no other objects nearby. (a) Calculate the (vector) force acting on the spacecraft. (b) At t=532.0: the spacecraft's momentum was piand at the later time t = 538.0s its momentum was pf.Calculate the (vector) change of momentum pf-pi.

Use data from the inside back cover to calculate the gravitational and electric forces two electrons exert on each other when they are apart (about one atomic radius). Which interactions between two electrons is stronger, the gravitational attraction or the electric repulsion? If the two electrons are at rest, will they begin to move toward each other or away from each other? Note that since both the gravitational and the electric forces depend on the inverse square distance, this comparison holds true at all distances, not just at a distance of .

Astar exerts a gravitational force of magnitude 4×1025Non a planet. (a) What is the magnitude of the gravitational force that the planet exerts on the star? (b) If the mass of the planet were twice as large, what would be the magnitude of the gravitational force on the planet? (c)If the distance between the star and planet (with their original masses) were three times larger, what would be the magnitude of this force?

A star of mass 7×1030kg is located at <5×1012,2×1012,0>m.A planet of mass 3×1024kg and is located at <3×1012,4×1012,0>m and is moving with a velocity of <0.3×104,1.5×104,0>m/s (a) At a time 1×106s later what is the new velocity of the planet? (b) Where is the planet at this later time? (c) Explain briefly why the procedures you followed in parts (a) and (b) were able to produce usable results but wouldn’t work if the later time had been 1×109s instead of 1×106s after the initial time. Explain briefly how you could use a computer to get around this difficulty.

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