A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

A 6 g bullet is fired horizontally into a 5 kg block of wood suspended by a long cord. The bullet sticks in the block. Compute the velocity of the bullet if the impact causes the block to swing 10 cm above its initial level.

  • This Question is Closed
  1. sohailiftikhar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let the mass of bullet is m1=6/1000kg and mass of block m2=5kg

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    116806.67 cm/s^2 ?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @sohailiftikhar

  4. sohailiftikhar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the initial velocity for both bullit and block is zero you have to find their final velocity which you can take as v because it is same for both as the got combined

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can't i use the formula \[ (m1 + m2)\sqrt{2gy} \over (m1) \]

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ?

  7. sohailiftikhar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol you got two madel nice answer

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Use the Principle of Conservation of Energy Before impact, The bullet moved with suppose velocity v, therefore it's kinetic energy was \[E_{1}=\frac{1}{2}mv^2\] Where m is the mass of the bullet After impact the system of block+bullet is raised to a height of 10cm Therefore increase in the potential energy would be \[E_{2}=(m+M)gh\] Here M is the mass of the block and h is the height to which it rises (10cm) Since m is quite small compared to M, we can approximate the potential energy as \[E_{2} \approx Mgh\] By the principle of conservation of energy, the kinetic energy before should be equal to the increase in the potential energy of the block+bullet system assuming no loss in energy due to sound heat etc. We are not considering initial potential energies of the block and bullet as they would just cancel out anyway \[E_{1}=E_{2}\]

  9. sohailiftikhar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok you can use the law of conservation of momentum here

  10. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OK, so just how can you use momentum here?? we know from energy conservation assumption that \(\frac{1}{2}m_1v_1^2 = (m_1 + m_2)gh\) \(v_1 = \sqrt{ \frac{2(m_1 + m_2)gh}{m_1} }\) [which solves the OP's question ] "if" this is a closed system [and momentum is conserved just between m1 and m2] we can also say that \(m_1 v_1 = (m_1 + m_2)v_2\) \(v_2 = \frac{m_1 }{m_1 + m_2} \sqrt{ \frac{2(m_1 + m_2)gh}{m_1}} = \sqrt{ \frac{2m_1gh}{m_1+m_2}}\) so the total energy of the system, again conserved, gives us \(\frac{1}{2}(m_1+m_2)v_2^2 = \frac{1}{2}(m_1 + m_2) \frac{2m_1gh}{m_1+m_2} = m_1gh\) which is plainly wrong IOW m1 and m2 is not a closed system and you cannot use momentum conservation, that is why the solution requires energy conservation to see this we can say \(\frac{1}{2}(m_1+m_2)v_2^2 = \frac{1}{2}m_1v_1^2\) \(v_2 = \sqrt{ \frac{m_1}{m_1+m_2} } \ v_1\) \(= \sqrt{ \frac{m_1}{m_1+m_2} } \sqrt{ \frac{2(m_1 + m_2)gh}{m_1} } \\ = \sqrt{2gh}\) which is the answer we should be getting

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So am I right?:P

  12. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Conservation of momentum does not apply in here as gravity (external force) is acting on the system.

  13. thomas5267
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess it would apply if you include Earth in the system as well?

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.