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s.h_math s.h_math
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Posts: 37
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1 months ago
A 30.0 kg object moving to the right at a velocity of 1.00 m/s collides with a 20.p kg object moving to the left with a velocity of 5.00 m/s. If the 20.0 kg object continues to move to the left at a velocity of 1.25 m/s, what is the velocity of the 30.0 kg object.

Hi there this is my question, I always have trouble with putting the direction at the end and I'm unsure how to know the correct direction at the end. Thanks a lot!
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duddyduddy
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1 months ago
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Hi s.h_math

Using law of conservation of linear momentum

Note: Assume right is positive, left is negative. This is why is set some velocities as positive, and others as negative.

Momentum of the system before collision = Momentum of the system after collision

\[ (m_1 \cdot u_1) + (m_2 \cdot u_2) = (m_1 \cdot v_1) + (m_2 \cdot v_2) \]\[ (30.0\;kg \cdot 1\frac{m}{s}) + (20.0\;kg \cdot -5\frac{m}{s}) = (30.0\;kg \cdot v_1) + (20.0\;kg \cdot -1.25 \frac{m}{s}) \]

Now solve for \(v_1\)

It should equal to -1.5 m/s, or simply 1.5 m/s to the left.
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- Master of Science in Biology
- Bachelor of Science (Biology)
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wrote...
1 months ago
Could you show steps on how to solve for v1?
wrote...
Staff Member
1 months ago
Sure,

30 - 100 = 30v - 25

-70 = 30v - 25 ..... now move the -25 over.

-70 + 25 = 30v

-45 = 30v ..... now divide both sides by 30.

-45/30 = v

-1.5 = v

remember that we defined negative as going to the left, so the answer is 1.5 m/s to the left.
- Master of Science in Biology
- Bachelor of Science (Biology)
- Bachelor of Education
wrote...
1 months ago Edited: 1 months ago, s.h_math
Thanks a lot sir!
Post Merge: 1 months ago

Also the way you wrote out the initial equation first, really helped me out!
wrote...
Staff Member
1 months ago
Yes, the entire premise is based on that idea. I will mark my answer as best answer.
- Master of Science in Biology
- Bachelor of Science (Biology)
- Bachelor of Education
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