1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Impulse of force correct here?

  1. Feb 28, 2016 #1
    • Originally posted in non-homework section, so missing the HW template
    Im sure Imma mess this up, probably because Im really tired....but

    I have a bullet that weights .015kg's and is moving at 285m/s, so 609.18J's as with KE.

    I hit a 70kg block, penetrates .3048 meters thus exerting 1998.62N's of force, which is a lot of force.
    .3048/285m/s so 0.00106 seconds, now we can that this with impulse of force for...

    Favg•∆t = m • ∆v -> (1998.62N's)•(0.00106) = 2.1374N•s impulse of force, now we use this to calculate the change in velocity ∆v= 2.1374N•s/ .015kg = ∆v:142.5m/s Vi:285m/s Vf:427.5 m/s
    .015kg • 142.5 = 2.1374 N•s

    If I take my impulse of 2.1374N•s over 70kg's...the velocity is .03053m/s and we divide by 0.00106 seconds we get 28.55 m/s^2 acceleration....

    We can rework this obviously, 1998.62N's over 70kg's = 28.55m/s

    d=(0m/s)(.00106s)+1/2(28.55m/s^2)(0.00106s^2)
    d=0.000016328 meters

    Object of 70kg Vf=(0m/s)(s)+(28.55m/s)(0.00106s)
    Object of 70kg Vf=0.03053m/s

    momentum p=70kg • 0.03053m/s = 2.13kg-m/s
    KE=0.032622 Joules.....

    The force applied by bullet is constant....

    So, what type of collision is this? why did the Vf of the bullet become 427 when it is stopped....and Im doing this correct RIGHT?

    This is a concept check, not homework.
     
    Last edited: Feb 28, 2016
  2. jcsd
  3. Feb 28, 2016 #2

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Is the velocity constant during penetration?
     
  4. Feb 28, 2016 #3
    I should have mentioned that, the force applied by the bullet is constant thus same with velocity...
     
  5. Feb 28, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, constant force does not lead to constant velocity. What does it lead to?
     
  6. Feb 28, 2016 #5
    well, meant to say we just have constant force..not constant velocity, my bad.
     
  7. Feb 28, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    So the way you calculated the time is wrong, as A.T. points out.
     
  8. Feb 28, 2016 #7
    How so? t=.3048m/285m/s that is .00106 Seconds...correct? If it is wrong, care to correct it?
     
  9. Feb 28, 2016 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The relationship speed=distance/time is only true for finding average speed. The average speed over the .3048 seconds is not 285m/s, that's the max speed.
     
  10. Feb 28, 2016 #9
    Okay, Okay so how do I calculate the time applied speed for impulse?
     
  11. Feb 28, 2016 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Usually, with these 'bullet enters block' questions, the first step is to find the speed of the combination just after impact.
     
  12. Feb 28, 2016 #11
    you also said it was .3048 seconds, it was .3048 meters.....
     
  13. Feb 28, 2016 #12
    Second, we don't know the speed after....we are calculating that.
     
  14. Feb 28, 2016 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Sure, but the comment stands.
    It would have helped to have mentioned that in the problem statement.
    The way to find it is by a conservation law. Which one?
     
  15. Feb 28, 2016 #14
    look, look....We are looking at a collision, the collision is inelastic....this is the conservation of energy and conservation of momentum.

    requires two equations im guessing?....
     
  16. Feb 28, 2016 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, to find the velocity of bullet+block at the point where the bullet reaches maximum penetration, given the two masses and the bullet's initial velocity, only requires one equation.
     
  17. Feb 28, 2016 #16
    Okay, your asking...never mind, I was thinking elastic collision....okay...but we are digressing...We know the bullet penetrates the block by .3048 meters, thus giving us an impact force....
     
  18. Feb 28, 2016 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, don't get stuck into forces again. Think conservation laws.
     
  19. Feb 28, 2016 #18
    okay, okay.....We need conservation of momentum....its an inelastic collision.....all I need to know is the impact time...or time the force of impact lasts...we can find impulse...and find displacement after..
     
  20. Feb 28, 2016 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You wrote in post #12 that you are trying to calculate the speed after impact. To do that you do not need to find the force or the time, just apply conservation of momentum.
    If that is not what you are trying to find, please state clearly what the objective is.
     
  21. Feb 28, 2016 #20
    Im asking if my calculations are correct but since my time was incorrect...how do I calculate impulse time...to multiple my force too, would I just use Mass times ∆v to get impulse?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Impulse of force correct here?
  1. Impulsive force (Replies: 10)

  2. Force and Impulse (Replies: 5)

  3. Impulse and force? (Replies: 2)

  4. Force and Impulse (Replies: 13)

Loading...