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I really need help with this problem!

  • Thread starter tumbler19
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  • #1
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I have not been able to figure out anything about this problem. I hope somebody will help me.

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (A): What is the mass of the other body? (B): What is the speed of the two-body mass center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?
 

Answers and Replies

  • #2
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I'm only in High School so dont depend on my word for it!

Here's what I'm thinking...
momentum=m*v
p1=p2
.25v(2kg)=.75v(m)

the other mass is: 2/3 kg

So, v=.25*4*2 = 2m/s

think this is correct?
 
  • #3
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In an elastic collision, both kinetic energy and momentum are conserved. If I denote the initial velocity of the 2kg mass by u and the final velocity of the unknown mass by v then,

[tex]\frac{1}{2}(4kg)(u^2) = \frac{1}{2}(4kg)(\frac{u}{4})^2 + \frac{1}{2}(m)(v^2)[/tex] (Kinetic Energy Conservation)

[tex](4kg)u = 4kg({\frac{u}{4}}) + mv[/tex] (Linear Momentum Conservation)

You have two equations and two unknowns (m and v). You can solve for them easily now.

UrbanXrisis, you have written only one equation--that for linear momentum conservation. Read the question carefully (note that both energy and linear momentum are conserved in an elastic collision).

Now for the second part, I can only tell you that the hint for solving it lies in this very post coupled with the fact that the net force on a system equals the time rate of change of linear momentum.

Hope that helps...

Cheers
Vivek
 
Last edited:
  • #4
1,789
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Are you supposed to get the answers in terms of u?
 

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