Elastic Collision and speed of a ball

  • Thread starter Atlos
  • Start date
  • #1
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Homework Statement



A 120 g ball moving to the right at 4.5 m/s catches up and collides with a 420 g ball that is moving to the right at 1.2 m/s.

If the collision is perfectly elastic, what is the speed of the 420 g ball after the collision?

Homework Equations



[tex]m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f} [/tex]

The Attempt at a Solution



I manipulated the formula a little bit to solve for V2_f and came up with this:
[tex](m_1v_{1i}+}m_2v_{2i} - m_1v_{1f}) / m_2[/tex]

then I plugged in my numbers:
(.12*4.5 + .42*1.2 - .12*.63) / .42

simplified to this:
(.54 + .504 - .0756) / .42

and got 2.3057. Mastering physics said that is wrong however and I'm not sure what I'm doing wrong. Is the conservation of momentum the correct equation for this problem? I found the velocity of the 120 gram ball fine, which was .63 m/s, but am stuck on this one. Any help would be appreciated.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Hi Atlos, welcome to PF.
How did you get v1f = 0.63 m/s?
 
  • #3
11
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I came to find v1f by using the following equation:

[tex][(m_1-m_2) / (m_1+m_2) ]v_{1i} + [2m_2 / (m_1+m_2]v_{2i}[/tex]
 
  • #4
62
0
A useful fact for straight line elastic collision is that

v_2b - v_1b = - (v_2a - v_1a)

v_1a = velocity of m1 before collision
v_1b = velocity of m1 after collision etc.

Relative velocity before the collision is the same as after, just opposite in sign.
 

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