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2 Dimensional Collision Problem

  1. Nov 12, 2011 #1
    Hey guys! I have this problem for an online homework assignment and I just can't get it right. My teacher never covered it in class and our textbook only covered it when the two colliding objects were the same weight. I would really appreciate some help!

    1. The problem statement, all variables and given/known data

    A(n) 1.9 kg object moving at a speed of 4.9 m/s strikes a(n) 1.4 kg object initially at rest. Immediately after the collision, the 1.9 kg object has a velocity of 1.2 m/s directed 41° from its initial line of motion. What is the speed of the 1.4 kg object immediately after the collision? Answer in units of m/s.

    2. Relevant equations

    m1(vi1 - vf1) = m2(vf2 - vi2)
    vi1 + vf1 = vf2 + vi2
    These are the equations he wants us to use. He doesn't like us using equations he doesn't give us in class so any way to solve this problem using only them would be great. The problem doesn't say the collision is elastic though so I don't understand why they still apply.

    3. The attempt at a solution
    Using the first equation I plugged in all the known information and got 5.02 m/s while the second gave me 6.1 m/s. Obviously something isn't right here. I also tried solving the problem using conservation of KE. I found all the KE for both objects in the x and y direction before and after the collision. Using the conservation principle I was able to solve for the KE of the second object in both the x and y direction. Because I have the mass of the object I was able to get the v in the x and y directions and I used the Pythagorean Theorem to solve for the overall vf of the second object but that was wrong too. I'm not sure what to try next so please help me!

    Last edited: Nov 12, 2011
  2. jcsd
  3. Nov 13, 2011 #2


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    Homework Helper

    These are not the appropriate equations for the problem.
    It is a two-dimensional collision, both components of the momentum has to be conserved. Make a drawing and write the equations for the components, both parallel and perpendicular to the initial velocity of the 1.9 kg object. There are enough data, you do not need conservation of energy.

  4. Nov 13, 2011 #3
    Okay, but how do you write the equations for the components? And once I have them what do I do with them? That just gives me the components of the initial velocity of the 1.9 kg object, what about the final velocity and both velocities for the 1.4 kg object. What do I do for them? If I do this for all of the velocities do I then use conservation of energy or is there some other equation I don't have?
    Last edited: Nov 13, 2011
  5. Nov 13, 2011 #4


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    Do you know how to get the x and y components of a vector if you know the magnitude and the angle the vector encloses with the x axis?
    Do you know what momentum is? What does conservation of momentum mean?

    Show something you did to solve this problem. I will tell you if it is correct or not.

  6. Nov 13, 2011 #5
    I have the components, I just don't get what to do with them.
    vi1 = 4.9cos0 + 4.9sin0 = 4.9 m/s
    vi2 = 0cos0 + 0sin0 = 0 m/s
    vf1 = 1.2cos41 + 1.2sin41 = 1.7 m/s
    vf2 = xcosθ + xsinθ = ? (with x being the final velocity of the 1.4 kg object and θ the angle from the x axis)

    All we've covered on momentum is that P = mv. I've never heard of conservation of momentum but I would guess it's the same as any conservation. Does that mean that my total initial momentum should equal my total final momentum?
    P = mv = (1.9 kg)(4.9 m/s) = (1.4 kg)(1.7 m/s + xcosθ m/s + xsinθ m/s)
    6.65 m/s = 1.7 m/s + xcosθ m/s + xsinθ m/s
    4.95 m/s = x m/s(cosθ + sinθ)
    x m/s = (4.95 m/s) / (cosθ + sinθ)
    How do I find my final velocity if I don't know what my second θ is?
  7. Nov 14, 2011 #6


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    I see... I seriously suggest that you read your book or notes about vectors.

    Velocity is a vector. It has magnitude and direction. A vector can be represented in a Cartesian coordinate system by an arrow, inclined with an angle with respect to the x axis. Its components are projections on the axes, see figure. The red vector, [itex]\vec a[/itex] has components ax and ay, the components of the blue vector [itex]\vec b[/itex] are bx and by.
    When you add vectors, their x and y components add separately. The resultant vector [itex]\vec c=\vec a +\vec b[/itex] has the components cx=ax+bx and cy=ay+by.

    Never add x component to y component!!

    The momentum is vector, too, [itex]\vec p=m \vec v[/itex] Its components are
    px=mvx and py=mvy.

    The total momentum of two objects is the sum of the individual momenta. [itex]\vec p=\vec p_1+\vec p_2[/itex]. That means two equations for the components:

    px=p1x+p2x and py=p1y+p2y.

    Conservation of momentum during a collision means that the total momentum before the collision is the same as the total momentum after the collision.

    [itex]\vec p(initial)=\vec p(final)[/itex]

    That is valid to the x and y components separately.

    Now we start with this problem, writing out the individual components of the initial and final momenta for both objects.

    The mass of object 1 is m1=1.9 kg,, the mass of object 2 is m2=1.4 kg.

    We set up a coordinate system with x axis in the direction of the velocity of the first object.

    Initially the components of momentum of the first object are

    p1x(i)=m1v1x(i)=1.9*4.9 cos(0)=1.9*4.9
    p1y(i)=m1v1y(i)=1.9*4.9 sin(0)=0

    Te momentum of the second object is zero:

    After the collision, the components of momentum for the first object:

    p1x(f)=1.9*1.2 cos(41)
    p1y(f)=1.9*1.2 sin(41)

    Those for the second object (with speed v2(f) and direction of velocity θ)

    p2x(f)=1.4*v2(f) cos(θ)
    p2y(f)=1.4*v2(f) sin(θ)

    Apply conservation of momentum for both the x and y components separately.


    1.9*4.9 =1.9*1.2 cos(41)+1.4*v2(f) cos(θ)

    0=1.9*1.2 sin(41)+1.4*v2(f) sin(θ)

    these are two equations for two unknowns: the speed of object 2, v2(f), and the angle of its velocity, θ. You can solve.


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