Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
What is the law of conservation of momentum?
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
[tex]\rm m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2[/tex]
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
https://brainly.com/question/1113396
#SPJ1
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
