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!

Bullet Lodges In Baseball (vertical)

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A 0.028 kg bullet is fired vertically at 185 m/s into a 0.15 kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/ bullet rise to a height of 37 m.

    (a) What was the speed of the baseball/bullet right after the collision?

    (b) What was the average force of air resistance while the baseball/bullet was rising? (Consider the positive direction to be upwards.)


    2. Relevant equations

    conservation of momentum.

    3. The attempt at a solution

    m1v1=m2v2 yeilding v2 (velocity of ball/bullet combo) of 29.1 m/s.
    This was correct.

    However, I can't figure out b.

    I tried using Force of Air Resistance = KE - PE, but that doesn't work.

    I also tried using Force Air Resitance =KE/h - m(total)-g

    That didn't work either (found it on yahoo answers).

    What am I missing here?
     
  2. jcsd
  3. Apr 9, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    I think you need to consider that KE = PE_final + W_air

    Where

    W_air = AvgForce_air * distance.
     
  4. Apr 9, 2009 #3
    Yes, but I still don't get the right answer even then:

    KE = PE_final + W_air

    Thus, W_air = KE - PE

    specifically:
    Wair = ((.5)(.178)(29.1^2)) - ((.178)(9.8)(37)) = 75.366-64.543 = 10.823

    Then,

    W_air = AvgForce_air * distance

    becomes:

    AvgForce_air = w_air / distance

    AvgForce_air = 10.823 / 37 = 0.293 (which isn't accepted)

    Thoughts?
     
  5. Apr 10, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    Look at it another way.

    Your initial velocity is 29.1.

    Figure acceleration against the upward motion.

    V2 = 2*a*x

    a = 29.12/(2*37) = 11.44 m/s2

    Well you know g was 9.8 so you apparently had an additional force that affected a 1.64 m/s2 downward drag.

    1.64*.178 = ... oops ... looks like .293N again.
     
  6. Apr 10, 2009 #5
    So im still confused?
     
  7. Apr 10, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    Or maybe .293 is correct?
     
  8. Apr 10, 2009 #7
    http://dl.getdropbox.com/u/119186/Picture%201.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Apr 10, 2009 #8
    Webassign sucks...this is for practice and it gives no feedback as to what we may be doing wrong...
     
  10. Apr 10, 2009 #9

    LowlyPion

    User Avatar
    Homework Helper

    [/URL]

    They want -.293 I think. The force is against the direction of motion which is up.
     
    Last edited by a moderator: May 4, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Bullet Lodges In Baseball (vertical)
  1. Baseball pieces (Replies: 29)

Loading...