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Homework Help: Center of Mass Projectile Problem

  1. Nov 14, 2006 #1
    Hi, sorry stuck again!

    Here is the problem:

    A projectile of mass 19.4 kg is fired at an angle of 57.0 degrees above the horizontal and with a speed of 81.0 m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance.

    And here is what I did:

    I first found the center of mass when y = 0 since the center of mass follows the trajectory of the parabola I used the equation:

    y(x) = tan(theta)x - .5g(x/(V_i*cos(theta))^2

    y(x) = 0
    theta = 57 degrees
    g= 9.8m/s^2
    V_i = 81.0 m/s so...

    0 = tan(57)x - 4.9(x/(8.1*cos(57))^2
    0 = 1.54x - 0.00252x^2

    x = 0, 776.1m = center of mass

    Then it says that the first fragment drops vertically at the highest point and this is partly where I get confused since with my math I assumed that means straight down, correct me if I'm wrong. So I found t when V_y = 0 to find x at that point.

    V_y = 0 = V_yi + a_y*t

    0= 81.0*sin(57) - 9.8t
    t= 6.93s

    x when t = 6.93s which should give me the position fragment 1 landed:

    x_f1 = x_i + V_ix*t + 0.5a_x*t^2
    x_f1 = 0 + 81.0cos(57) + 0
    x_f1 = 305.72m

    Then I just had to find x of fragment 2:

    x_cm = (m_f1*x_f1) + (m_f2*x_f2) / (m_f1 + m_f2)

    so, I made the center of mass the 0 coordinate:

    776.1m = [(9.7kg*-470.4m) + (9.7kg*x_f2)] / (9.7kg*2)

    x_f2 = 2020m, which turned out to be wrong! :eek:

    I'm sorry this problem is so long :frown:
  2. jcsd
  3. Nov 14, 2006 #2


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    Why did [tex]x_{f1}[/tex] become -470.4m, when it clearly states it's 305.27m above?

    Also [tex]m_1 + m_2[/tex] is 19.4, not 9.7
  4. Nov 14, 2006 #3
    ya sorry I had it written 9.7kg*2, I did use 19.4.

    But for the x_f1 I put it to -470.4m because I put the center of mass that I found as the origin in a new axes for calculating x_f2's position.

    so 776.1m became 0, which means that 776.1m-305.72m became x_f1 and since it's to the left of the new origin it would be negative. Should I not have done that?
  5. Nov 14, 2006 #4
    If I left it at 305.27m then the math would be

    776.1m = [(9.7kg*305.27m) + 9.7kg*x_f2] / 19.4kg = 1246.93 m and that answer was not correct either :(
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