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Newton's Third Law Problem

  • Thread starter hya_been
  • Start date
  • #1
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So I have a homework question that says:
A 100 kg hockey player a 112kg hockey player collide with each other each travelling with a force of 50N.

The text book answer says that the acceleration of the 112kg is 0.89m/s^2
and the acceleration of the 100kg player is -1.0 m/s^2.

I'm confused because according to Newton's third Law there will be equal and opposite reaction forces so wouldn't they not move? Is the text book wrong or am I?
 
Last edited:

Answers and Replies

  • #2
2,209
1
Remember by the second law:

[tex] F = m a [/tex]

They both impart a force of 50N on each other, but the acceleration is also scaled by their masses. 50N on a 100kg person will push them farther than 50N on a 112kg person.


Imagine pushing a shopping cart with 50N. An empty shopping cart will go really far, whereas a full shopping cart will go a much shorter distance. In both instances the force is the same, but:

[tex] a = \frac{F}{m} [/tex]
 
  • #3
3
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calculating acceleration

So you do 100/100 and 100/112 even though the total of the forces equals 0 because its 50 + -50??
 
  • #4
2,209
1
The net force of the entire system is 0, but the force on each player is of magnitude 100N. You can find the acceleration of each player using

[tex] a = \frac{F}{m} [/tex]. Knowing F = 100N and m = 100kg for person 1,

[tex] a = \frac{100N}{100kg} = 1.0m/s^2 [/tex]

F = 100N and m = 112kg for person 2:

[tex] a = \frac{100N}{112kg} = 0.89m/s^2 [/tex]
 
  • #5
3
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Thank You!!

Thank-you you are a saviour!
 

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