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Momentum problem

  • Thread starter Sheneron
  • Start date
360
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1. Homework Statement

A ball of mas m1 = 2kg is moving with a speed of 5 m/s and collides with a ball of mass m2 = 2.5 kg (there are losses so the collision is inelastic). After the collision the incoming ball has deviated by 35 degrees from its origin. The struck ball has moved off at an angle of 50 degrees. Find both balls final velocities.


2. Homework Equations

Pxi = Pxf
Pyi = Pyf


3. The Attempt at a Solution

I think solved it, if someone could though, I would like to have it checked to make sure I did it right. Thanks in advance.

I found the Conservation of Momentum in the x:

[tex] m_{1}v_{1ix} = m_{1}v_{1fx} + m_{2}v_{2fx} [/tex]
[tex]m_{1}v_{1}cos(0) = m_{1}v_{1f}cos(35) + m_{2}v_{2f}cos(-50)[/tex]

[tex]m_{1}v_{1} = m_{1}v_{1f}cos(35) + m_{2}v_{2f}cos(-50)[/tex]

Then I found it in the y and solved for [tex]v_{2f} [/tex]

[tex]m_{1}v_{1iy} = m_{1}v_{1fy} + m_{2}v_{2fy}[/tex]
[tex]m_{1}v_{1i}sin(0) = m_{1}v_{1f}sin(35) + m_{2}V_{2f}sin(-50)[/tex]
[tex]0 = m_{1}v_{1f}sin(35) + m_{2}v_{2f}sin(-50)[/tex]

[tex]m_{1}v_{1f}sin(35) = -m_{2}v_{2f}sin(-50)[/tex]

[tex]v_{2f} = \frac{-m_{1}v_{1f}sin(35)}{m_{2}v{2f}sin(-50)} [/tex]

Then I plugged it into the x equation.

[tex]m_{1}v_{1} = m_{1}v_{1f}cos(35) + m_{2}\frac{-m_{1}v_{1f}sin(35)}{m_{2}sin(-50)}cos(-50)}[/tex]

Then solved for [tex] v_{1f} [/tex]

[tex] v_{1f}[\frac{-m_{1}sin(35)cos(-50)}{sin(-50)} + m_{1}cos(35)] = m_{1}v_{1i}[/tex]

[tex]v_{1f}[\frac{-2sin(35)cos(-50)}{sin(-50)} + 2cos(35)] = 10 [/tex]

For v_{1f} I got 3.84. Is this the right way to solve this problem?
 

Answers and Replies

Doc Al
Mentor
44,828
1,083
Looks good to me! :approve:
 
360
0
Thanks, you have helped me with a lot of problems, I really appreciate it.
 

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