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ashna

  • 3 years ago

A bullet of mass 20g strikes a block of mass 980g with a velocity V and is embedded with it . The block is in contact of a spring whose force-constant is 100N/M. After the collision the spring is compressed upto 10cm . find (a) the velocity of the block after collision (b) magnitude of the velocity v of the bullet (c) loss in KE due to the collision.

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  1. ashna
    • 3 years ago
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    use v^2 - u^2 / 2as formula ?

  2. ashna
    • 3 years ago
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    then ?

  3. imron07
    • 3 years ago
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    Okay, while waiting for yahoo, let me give a clue:

  4. ashna
    • 3 years ago
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    okay

  5. imron07
    • 3 years ago
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    You cannot use v^2=blah-blah 2as formula because the acceleration here is not constant. Use conservation of energy instead: Kinetic energy+potential energy of the system is conserved.

  6. ashna
    • 3 years ago
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    okay

  7. imron07
    • 3 years ago
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    Did you get what i mean or just saying okay :)

  8. imron07
    • 3 years ago
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    For potential energy, use spring's potential. What is the kinetic energy + potential energy of the system after collition?

  9. eashmore
    • 3 years ago
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    imron07 is right, we need to track energy. Let's study the flow of energy. Let's ignore potential energy, we will assume everything occurs in a horizontal plane. Let's identify the components of the system. We have a bullet, block, and spring. First we fire the bullet. It has kinetic energy of\[KE_b = {1 \over 2} m_b v^2\] Now, when the bullet impacts the block, we will assume all the energy is transferred. The kinetic energy of the bullet/block system is\[KE_B = {1 \over 2} (m_b + m_B) V^2\](Big B is the block). Now, all that energy is transferred to the spring. The energy stored in a spring is\[E_s = {1 \over 2} kx^2\]where x is the displacement and k is the spring constant.

  10. ashna
    • 3 years ago
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    okay now i understand :)

  11. ashna
    • 3 years ago
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    what should i do for (a) ?

  12. eashmore
    • 3 years ago
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    You need to work backwards from what I presented. We know how much energy goes into the spring from what is given in the problem statement.

  13. ashna
    • 3 years ago
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    what is displacement here ?

  14. ashna
    • 3 years ago
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    @eashmore ?

  15. eashmore
    • 3 years ago
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    10 cm

  16. ashna
    • 3 years ago
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    first part E = 5000 J ?

  17. eashmore
    • 3 years ago
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    No. Remember to convert grams to kg. I made a mistake. The collision of the bullet and block is inelastic and therefore requires that we study the collision using conservation of momentum. \[p_b = p'_b + p'_B\]where ' indicates after the collision. They move at the same velocity so we have\[m_b v_b = (m_b + m_B)V\] So, first find the energy stored in the spring. \[E_s = {1 \over 2} k x^2\] This energy will be equal to the kinetic energy of the bullet/block system AFTER collision of the bullet. \[E_s = KE_B \rightarrow KE_B = {1 \over 2} (m_b+m_B)V^2\]We can find V from this. Then we can use the momentum equation to find the velocity of the bullet. Then find the energy from \[E_l = KE_b - KE_B = {1 \over 2} m_bv^2 - {1 \over 2} (m_b + m_B)V^2\]

  18. ashna
    • 3 years ago
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    okay .. i'll try ! Thanks for the help :)

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