(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 6 kg object moving with a speed of 9.9 m/s collides with a 19 kg object moving with a velocity of 7.2 m/s in a direction 24* from the initial direction of motion of the 6 kg object. What is speed of the two objects after the collision if they remain stuck together?

2. Relevant equations

P_{x}= m_{1}v_{1i}cos[tex]\theta[/tex]+m_{2}v_{2i}cos[tex]\theta[/tex] = m_{1}v_{1f}cos[tex]\theta[/tex] + m_{2}v_{2f}cos[tex]\theta[/tex]

P_{y}= m_{1}v_{1i}sin[tex]\theta[/tex]+m_{2}v_{2i}sin[tex]\theta[/tex] = m_{1}v_{1f}sin[tex]\theta[/tex] + m_{2}v_{2f}sin[tex]\theta[/tex]

3. The attempt at a solution

Edit I realized my original way was faulty in its logic.

So what I did was I began by finding the initial and final momenta components in both the x and y directions for the two particles. Being that they stick together their final velocity ought to be the same.

[tex]\Sigma[/tex]P_{xi}= m_{1}v_{1}cos[tex]\theta[/tex] + m_{2}v_{2}cos[tex]\theta[/tex]

=184.37302

[tex]\Sigma[/tex]P_{yi}= m_{1}v_{1}sin[tex]\theta[/tex] + m_{2}v_{2}sin[tex]\theta[/tex]

=55.641573

As I mentioned their final velocity should be equal since they are stuck together. Equating my initial momenta to my final momenta in the x direction I get the following:

V_{f}= [tex]\frac{(m1v1cos\theta + m2v2cos\theta}{(m1+m2)cos\theta}[/tex]

=8.072856 m/s

When I checked it using the momenta of the y direction I come to a different value for the final velocity which doesn't make sense so I know I made a mistake somewhere... Thanks in advance.

Joe

Edit: I tried a new way, but

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# Homework Help: Perfect inelastic collison

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