1. Not finding help here? Sign up for a free 30min 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!

Please help me, this problem is killing me

  1. Jun 22, 2005 #1
    A 20 g bullet is fired horizontally into a .5 kg wooden block, resting on a horizontal surface ( kinetic friction coefficient is .2), THe bullet goes through the block and comes out the other side with a speed of 200m/s... if the block travels 1 m before coming to rest, what was the Vi of the bullet.... (ignore work of gravity on the bullet....ps, this is a momentum problem i think, i just cant get it, i would GREATLY appriciate any help on this.... i have the answer but i need to work to get credit..... this is what i have so far

    i tried using Wnet = change in KE

    Fx - .2(.52)(1m)(9.8) = x - 1.02 J but i have no idea what to do next, i think it is a perfectly inelastic collision during the one meter, but than the bullet come out!!?!! how the heck do you account for that? again any help is appriaciated,,, thanks for y our time
     
  2. jcsd
  3. Jun 22, 2005 #2
    i'm not sure about this... (not really sure about wat kinetic coefficient is) but what i tink u can do is since there is kinetic friction .2 .. let's take as if u'r block is movin so use [tex] KE = 1/2 m v^2[/tex] to find u'r 'velocity' of u'r block/.. After doin that, u can use [tex] M(1) U(1) + M(2) U(2) = M(1) V(1) + M(2) V(2) [/tex] to find u'r initial speed.. i hope so.. note: u'r V(2)= 0... oso dunnoe y..

    May not be rite.. though APOLOGIES! Juz try it.. hope u get u'r answer :biggrin:
     
    Last edited: Jun 22, 2005
  4. Jun 22, 2005 #3
    Friction force = coefficient of friction * normal force.
    thus friction force is 0.5*9.8*0.2=0.98 N
    this acts through 1 m, so the work done by friction is
    work done = F*s = 0.98 * 1 = 0.98 J.
    this work done by friction is absorbing the kinetic energy of the block initially,
    so the initial kinetic energy of the block is 0.98 J
    thus 0.98 = 0.5*m*v^2.
    so the velocity of the block just after impact is 1.980 m/s.

    now, apply momentum equation,
    momentum initial = momentum final
    = 0.5*1.980 + 0.020 * 200 = 4.99 kg m/s.

    initial momentum of the bullet is 4.99 kg m/s also, so
    the speed of the bullet is 4.99/0.02 = 249 m/s.

    the BIG assumption we adopt here is the external friction is negligible during the interaction process which happens very-very fast, otherwise momentum is not conserved.

    maybe that helps.

    Sniffer (poor physicist)
     
  5. Jun 22, 2005 #4
    thank a bunch, thats exactly the answer that was given, thanks again.....
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Please help me, this problem is killing me
Loading...