anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
use v^2 - u^2 / 2as formula ?
anonymous
  • anonymous
then ?
anonymous
  • anonymous
Okay, while waiting for yahoo, let me give a clue:

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anonymous
  • anonymous
okay
anonymous
  • anonymous
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.
anonymous
  • anonymous
okay
anonymous
  • anonymous
Did you get what i mean or just saying okay :)
anonymous
  • anonymous
For potential energy, use spring's potential. What is the kinetic energy + potential energy of the system after collition?
anonymous
  • anonymous
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.
anonymous
  • anonymous
okay now i understand :)
anonymous
  • anonymous
what should i do for (a) ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
what is displacement here ?
anonymous
  • anonymous
@eashmore ?
anonymous
  • anonymous
10 cm
anonymous
  • anonymous
first part E = 5000 J ?
anonymous
  • anonymous
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 \[E_l = KE_b - KE_B = {1 \over 2} m_bv^2 - {1 \over 2} (m_b + m_B)V^2\]
anonymous
  • anonymous
okay .. i'll try ! Thanks for the help :)

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