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anonymous
 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 forceconstant 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.
anonymous
 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 forceconstant 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use v^2  u^2 / 2as formula ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, while waiting for yahoo, let me give a clue:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You cannot use v^2=blahblah 2as formula because the acceleration here is not constant. Use conservation of energy instead: Kinetic energy+potential energy of the system is conserved.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you get what i mean or just saying okay :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For potential energy, use spring's potential. What is the kinetic energy + potential energy of the system after collition?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0imron07 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
 3 years ago
Best ResponseYou've already chosen the best response.0okay now i understand :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what should i do for (a) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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
 3 years ago
Best ResponseYou've already chosen the best response.0what is displacement here ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first part E = 5000 J ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No. 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
 3 years ago
Best ResponseYou've already chosen the best response.0okay .. i'll try ! Thanks for the help :)
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