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Conservation of Momentum in an Explosion Problem

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    An object at rest explodes into three pieces. Two of them are equal in mass and break off at an angle of 85.3° to each other, both with a speed of 33.2 m/s. The last piece has 4 times as much mass as each of the other pieces. What is the magnitude of the velocity of the last piece?

    2. Relevant equations
    pf = pi
    m1 = m2
    m3 = 4*m1
    v1 = v2 = 33.2 m/s

    3. The attempt at a solution
    I think that i need to consider the momentum that is conserved as components in the x and y direction. but im not too sure how to draw out my picture to get my angles.
    y: pf=pi
    0 = m1cos(4.7)(33.2) + (4m1)sin[tex]\theta[/tex]

    x: pf=pi
    0 = m1(33.2) + v3(m3)(cos[tex]\theta[/tex]) + m1sin(4.7)(33.2)

    im not sure if im approaching this correctly...?
  2. jcsd
  3. Oct 19, 2009 #2
  4. Oct 19, 2009 #3


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    Yes, the general approach is to make use of the fact that momentum is conserved (before vs. after the collision). In order for momentum to be conserved, the third mass (4m) will have to have a certain x-component and a certain y-component (you can apply conservation of momentum separately for each direction). The x- and y-components of this third piece together give you the information about both the magnitude and direction of its velocity.
  5. Oct 19, 2009 #4


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    Also, a nice shortcut is that by symmetry, the x (or y, depending on how you set up your coordinates) components of the two equal masses cancel each other out, and you don't even have to calculate them.

    EDIT: I don't understand where you get the angle 4.7 from???
  6. Oct 19, 2009 #5
    so i just said that one mass will go directly east and then the other mass will go 80.3 deg from it north.. so to make a right angle triangle 90-80.3 = 4.7

    I also dont understand what you mean that they will cancel out and you dont need to calculate them?
  7. Oct 19, 2009 #6


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    Why is one angle [itex]\theta[/itex] and the other 4.7?

    Assuming this is in two dimensions we can set up a coordinate with the positive x-axis in the direction that the first piece flies and the positive y-axis 90 degrees north of that. Then the momentum vector of the first piece is [itex]32.2m\vec{i}[/itex]. the piece that goes 85.3 degrees "north" of that will have momentum [itex]33.2m cos(85.3)\vec{i}+ 33.2m sin(85.3)\vec{j}[/itex] or, same thing, [itex]33.2m sin(4.7)\vec{i}+ 33.2m cos(4.7)\vec{j}[/itex]. Add those to get the total momentum vector of the two pieces. The momentum vector of the third piece will be in the opposite direction with with the same magnitude. You can calculate the speed from the magnitude of the momentum vector.
  8. Oct 19, 2009 #7


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    You might find it easier to do this problem if you set up your coordinate system as follows:

    Positive x-direction is to the right. You can call this 'east' if you like.

    Positive y-direction is upward. You can call this 'north' if you like.

    Mass 1 travels at an angle of (85.3)/2 deg = 42.65 deg in the counterclockwise direction from the positive x-axis (i.e. 42.65 deg north of east)

    Mass 2 travels at an angle of (85.3)/2 = 42.65 degrees in the clockwise direction from the positive x-axis (i.e. 42.65 degrees south of east).

    By setting up the the problem in this way, you have exploited the *symmetry* of the situation. Since the y-components of masses 1 and 2 are equal and opposite (one is going north or in the +y direction and the other is going south or in the -y direction), they cancel, and you don't even need to calculate them).

    You can also deduce, from this setup, that the direction of mass 3 must be entirely in the negative x (west) direction (NO y component), due to conservation of momentum. Calculating this component is then much easier: it is just equal in magnitude to the sum of the x-components of the momenta of the masses 1 and 2.
  9. Oct 19, 2009 #8
    OH! i hadnt understood that. so is it always true that in an explosion problem the momentum of the piece(s) flying in positive direction will equal the momentum of the piece(s) flying in negative direction?
  10. Oct 19, 2009 #9
    Yeah! The principle of conservation of momentum (some law, i forgot the name) says the momentum of a system before an event is equal to the momentum after the event, as far as no external forces are involved.
  11. Oct 19, 2009 #10
    ok i was thinking about this all wrong.. i had assumed that since it was at rest it would be sitting on a table or the floor so all the pieces would fly out (semi-circle-type) and none would fly out in the negative y-direction.
    THANK YOU for all your help!!!
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