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anonymous

  • 5 years ago

1. Two clay balls collide head-on. The first ball has a mass of 0.500 kg and its velocity is initially 4.00 ms-1 to the east. The mass of the second ball is 0.250 kg, and it has an initial velocity of 3.00 ms-1 to the west. If the two balls stick together after colliding, find their final velocity after the interaction. Compare kinetic energies before and after the interaction. What type of collision is this and WHY?

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  1. JamesJ
    • 5 years ago
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    So first use Conversation of Momentum to find their new, combined velocity. Then the question is whether the collision is elastic or inelastic ...

  2. JamesJ
    • 5 years ago
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    If KE is conserved, then the collision was elastic. If it was not conserved, then the collision was inelastic.

  3. anonymous
    • 5 years ago
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    what is the formula for conservation of Momentum?

  4. anonymous
    • 5 years ago
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    also can you please do this question so i know how to do the other. thanks!

  5. JamesJ
    • 5 years ago
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    This is very important idea in physics. Let m1 and m2 be the masses of two objects with initial velocities v1 and v2. After the collision they have velocity v1' and v2'. Then \[ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \]

  6. JamesJ
    • 5 years ago
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    You might find these two lectures helpful, even if a bit advanced. The demonstrations are compelling: http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-15/ http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-16/

  7. anonymous
    • 5 years ago
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    (.500*4.00)+(.250*3.00)=

  8. anonymous
    • 5 years ago
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    I dont have M1V1' or M2V2'

  9. JamesJ
    • 5 years ago
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    after colliding, their velocities are the same, \[ v = v_1' = v_2' \] You also know \( m_1 \) and \( m_2 \). Hence \[ m_1v_1' + m_2v_2' = (m_1 + m_2)v \] Now set that equal to the momentum you calculated before the collision and solve for v. I strongly recommend you watch at least the beginning of the first lecture I posted. he works an example __exactly like this__.

  10. anonymous
    • 5 years ago
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    (.750)V

  11. anonymous
    • 5 years ago
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    V= 3.66

  12. anonymous
    • 5 years ago
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    is that correct?

  13. anonymous
    • 5 years ago
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    ???

  14. anonymous
    • 5 years ago
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    Since the two balls stick together, they will have the same post-collisions velocity. From conservation of momentum\[m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'\]You can readily solve for \(v'\) To determine the type of collision we need to compare kinetic energies. From conservation of energy\[{1 \over 2} \left ( m_1 v_1^2 + m_2 v_2^2 \right) = {1 \over 2}(m_1+m_2) v'^2\]where \(v'\) is found from conservation of momentum. Since conservation of momentum is always satisfied, regardless of collision type. We know that the first equation will give us the correct \(v'\) value. Plug this into the second equation. If COE is satisfied, the collision is elastic, otherwise it is inelastic.

  15. anonymous
    • 5 years ago
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    I would also suggest watching the lectures that JamesJ linked to. These will help you better grasp the concepts involved in the study of collisions.

  16. anonymous
    • 5 years ago
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    I found that 2.28125=5.02335 so theres KE left over so it would be elastic

  17. anonymous
    • 5 years ago
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    You should double check your calculations. Please remember that momentum is a vector quantity, it has magnitude and DIRECTION, while kinetic energy is a scalar.

  18. anonymous
    • 5 years ago
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    my bad I recalculated it and got 5.02335=5.125

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