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Homework Help: World Famous Physicist Lawyer

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  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data

    You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged that your client, Mr. Smith, shot the victim, Mr. Wesson. The detective who investigated the scene of the crime found a second bullet, from a shot that missed Mr. Wesson, that had embedded itself into a chair. You arise to cross-examine the detective. You: In what type of chair did you find the bullet? Det: A wooden chair. You: How massive was this chair? Det: It had a mass of 20.0 kg. You: How did the chair respond to being struck with a bullet? Det: It slid across the floor. You: How far? Det: Three centimeters. The slide marks on the dusty floor are quite distinct. You: What kind of floor was it? Det: A wood floor, very nice oak planks. You: What was the mass of the bullet you retrieved from the chair? Det: Its mass was 10 g. You: And how far had it penetrated into the chair? Det: A distance of 4.00 cm. You: Have you tested the gun you found in Mr. Smith's possession? Det: I have. You: What is the muzzle velocity of bullets fired from that gun? Det: The muzzle velocity is 450 m/s. You: And the barrel length? Det: The gun has a barrel length of 62 cm.


    2. Relevant equations



    3. The attempt at a solution


    I found the force on chair by the bullet if the bullet is going 450 m/s

    0=(450)^2 +2a(.04m)
    a= -2531250 m/s^2

    F= (.01kg)(-2531250m/s^2)
    So force by bullet on chair is 25312.5 N
    This force acts this time:
    for 0=(450m/s) + (-2531250m/s^2)(t)
    t= 1.78 * 10^-4 s

    Force of friction on chair : (20.01kg)(9.80)(.20) = 39.2 N

    sum forces on x = Fb-kinetic friction = 25312.5N-39.2N= 25273.3 N
    acceleration of chair= (20.01kg)(a)= 25273.3N a=1263 m/s^2

    Impulse-momentum theorem: Impulse plus initial momentum equals final momentum
    (25273.3N)(1.78*10^-4s) + 0 = (20.01kg)v
    v=.225m/s this is the velocity of the chair when the bullet stops moving

    distance it has gone during the impulse:
    (.225m/s)^2= (0m/s)^2 + 2(1263 m/s^2)x
    x=2.00*10^-5 m

    after this: Force of friction = 39.2 N

    (20.01kg)(a)=(-39.2N) a=-1.95 m/s^2

    distance traveled during this time

    (0m/s)^2=(.225 m/s)^2 + 2(-1.95m/s^2)x
    x=.013 m =1.29 cm.

    Is this correct, or am I making this way too complicated? If it is not correct, could you please explain how you would have done it?
     
  2. jcsd
  3. Apr 19, 2008 #2
    why does the problem give you the barrel length of the gun?
     
  4. Apr 21, 2008 #3
    I think that is useless info. Any thoughts on this one yet?
     
  5. Apr 21, 2008 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    Why does the chair not move across the entire room?

    You need another piece of information: hint: it has to do with the fact that the floor is wood, as well as the chair feet?

    AM
     
  6. Apr 21, 2008 #5
    i took friction into account i think
     
  7. May 20, 2010 #6
    I think you are making this too complicated
     
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