Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A 7.28-kilogram bowling ball traveling 8.50 meters per second east collides head-on with a 5.45 kilogram bowling ball traveling 10.0 meters per second west. Determine the magnitude of the total momentum of the two-ball system after the collision

Sagot :

Giovi
Considering Conservation of Momentum, the momentum [tex]p=mv[/tex] before and after must remain the same:
so:
before: [tex]7.28*8.50-5.45*10.0=7.38 kg \frac{m}{s} [/tex]
Cricetus

The magnitude of the total momentum will be "7.38 kg.ms⁻¹".

Given:

Mass,

  • [tex]m_1 = 7.28 \ kg[/tex]
  • [tex]m_2 = 5.45 \ mg[/tex]

Speed,

  • [tex]v_1 = 8.5 \ m/s[/tex]
  • [tex]v_2 = 10 \ m/s[/tex]

By using the law of conservation of linear momentum, we get

→ [tex]P_{initial} = P_{final}[/tex]

or,

→ [tex]P_{final } = m_1\vec{v_{1i}}+m_2\vec{v_{2i}}[/tex]

By substituting the values, we get

 [tex]|P_{final}|= 7.28\times 8.5-5.45\times 10[/tex]

             [tex]=61.88-54.5[/tex]

             [tex]= 7.38 \ kg.ms^{-1}[/tex]

Thus the above approach is correct.

Learn more about momentum here:

https://brainly.com/question/12091981

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.