Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bullet Momentum Problem

  1. Nov 27, 2006 #1
    This problem is causing me quite a bit of grief and I could really use a guiding hand to tell me where I am going wrong. Here is the set up, then after that is what I've used to tackle it so far.

    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 6.00 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.

    Frictional Constant (wood-wood) = .02

    What I've done:


    M_Chair = 20.0 kg
    M_Bullet = .01 kg
    M_System = 20.01 kg
    friction_chair = (.02)(9.8 m/s^2)(20.01kg) = 39.2196 N
    V_bullet = 450 m/s

    Force of the bullet on the chair:

    Acceleration of bullet as it approaches 0th Velocity

    V_f^2 - V_i^2 =2*a*s

    [(-450^2)/(2*.06)] = a
    = -1,687,500 m/s^2

    (bullet force)

    F = m*a = .01 *-1,687,500 m/s^2
    = 16875 N

    (time that chair and bullet are moving)

    V_f = V_i + a*t

    (-450 m/s)/(-1,687,500 m/s^2) = t
    = .0002667 s


    F_bullet - Force_friction = ma
    = (16875 - 39.216 )/(20.01kg) = a
    = 841.3685157m/s

    a*t^2 = X (X = Distance that chair has moved)

    (841.369)(.0002667^2) = .0000598456 m

    .00598 CM


    I've tried this several times and I've reached this answer about 3 different ways. What is going wrong? Friction on the bullet inside the chair?
  2. jcsd
  3. Nov 27, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    The usual approach to this problem is to use conservation of momentum to analyze the collision followed by the decelartion from the frictional force after the collsion is over. Trying to analyze the movement of the chair during the collision is more difficult, and probably not the approach taken by the author of the problem.

    To incorporate the simultaneous motion of the bullet and the chair during the collision process you would probably want to look at the impulse on the chair-bullet system during the collision time. I doubt the difference between the momentum conservation approach and the more detailed approach would be enough to be ineresting, and I think there are still some assumptions you would have to make about how the chair accelerates before it decelerates.
  4. Nov 27, 2006 #3
    Isn't muzzle velocity the velocity of the bullet relative to the gun? Thats how our teacher described it.
  5. Nov 27, 2006 #4
    Muzzle Velocity is the velocity that the bullet leaves the barrel. As far as I can tell the barrel length is just useless information. I've tried several momentum approaches and none are working still. I'll post another attempt in one moment.
  6. Nov 27, 2006 #5
    here is another attempt, that has failed:

    Law of momentum

    m_bullet* velocity_bullet + m_chair *v_chair = M_System * V_System

    (.01 * 450 ) + 0 = (20.01) ( V_s)

    V_s = .2249 m/s



    V_f^2-V_i^2 = 2*a*(delta)x
    -450^2/(.06*2) = a
    a = -1687500m/s^2

    vf = vi +a*t

    -450/-1687500m/s^2 = .000266666666667

    V_system * time = distance

    .2249 * .000266666666667 = .00005997 meters


    Tried friction

    F_bullet - F_Floor =ma
    [(.01)(1687500) - (9.8)(.2)(20.01)]/(20.01) = a

    a = 841.36833

    x_f = x_i + v_i*t+1/2*a*t^2

    = .0000299153 m


    This is why I am beating my brain so hard. I don't get what I am doing incorrectly.
  7. Nov 27, 2006 #6
    Thank you everyone who attempted to help me out with this problem. I did some re thinking and I've solved it.

  8. Dec 11, 2007 #7
    how did you solve this?
  9. Dec 11, 2007 #8
    The mass of the chair in my case is 30kg, that's the only difference!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook