How Is Momentum Conserved in a Two-Body Collision?

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SUMMARY

The discussion focuses on the conservation of momentum in a two-body collision involving a 2.6 kg object moving at 7.9 m/s and a 3 kg object moving at 7.4 m/s at an angle of 56.1768°. The initial momentum equation was incorrectly simplified by only considering the cosine component of the second object's velocity. To accurately calculate the final speed of the combined objects post-collision, both the cosine and sine components of the second object's velocity must be included. The correct approach leads to a final speed of approximately 5.8745 m/s.

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kmikias
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Hi.I have some confusion on this question ...here is the question

1.A(n) 2.6 kg object moving with a speed of
7.9 m/s collides with a(n) 3 kg object mov-
ing with a velocity of 7.4 m/s in a direction
56.1768◦ from the initial direction of motion
of the 2.6 kg object.What is the speed of the two objects after
the collision if they remain stuck together?
Answer in units of m/s

here is what i did but my answer out to be wrong .
Mass * velocity + Mass * velocity =( Mass +Mass)velocity

2.6 * 7.9 + 3*7.4 *cos56.1768 = (2.6 +3)v
20.54 +12.357 = 5.6v
so velocity = 5.8745 m/s

thank you.
 
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You're only considering the component of the velocity in the direction of the first object. That's why you did cos(56.1768). Don't forget to factor in the other component, namely, sin(56.1768). You should get the right answer if you do.
 

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