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8.8: Collisions in Multiple Dimensions

  • Page ID
    17960
  • Learning Objectives

    • Express momentum as a two-dimensional vector
    • Write equations for momentum conservation in component form
    • Calculate momentum in two dimensions, as a vector quantity

    It is far more common for collisions to occur in two dimensions; that is, the angle between the initial velocity vectors is neither zero nor 180°. Let’s see what complications arise from this.

    The first idea we need is that momentum is a vector; like all vectors, it can be expressed as a sum of perpendicular components (usually, though not always, an x-component and a y-component, and a z-component if necessary). Thus, when we write down the statement of conservation of momentum for a problem, our momentum vectors can be, and usually will be, expressed in component form.

    The second idea we need comes from the fact that momentum is related to force:

    $$\vec{F} = \frac{d \vec{p}}{dt} \ldotp$$

    Expressing both the force and the momentum in component form,

    $$F_{x} = \frac{dp_{x}}{dt}, F_{y} = \frac{dp_{y}}{dt}, F_{z} = \frac{dp_{z}}{dt} \ldotp$$

    Remember, these equations are simply Newton’s second law, in vector form and in component form. We know that Newton’s second law is true in each direction, independently of the others. It follows therefore (via Newton’s third law) that conservation of momentum is also true in each direction independently.

    These two ideas motivate the solution to two-dimensional problems: We write down the expression for conservation of momentum twice: once in the x-direction and once in the y-direction.

    $$p_{f,x} = p_{1,i,x} + p_{2,i,x} \label{9.18}$$

    $$p_{f,y} = p_{1,i,y} + p_{2,i,y}$$

    This procedure is shown graphically in Figure 9.22.

    Figure a, titled break initial momentum into x and y components shows vector p 1 i as a solid arrow pointing to the right and down. Its components are shown as dashed arrows: p 1 i y points down from the tail of p 1 i and p 1 i x points to the right from the head of p 1 i y to the head of p 1 i. Vector p 2 i is shown as a solid arrow with its tail at the head of vector p 1 i, and is shorter than p 1 i. Vector p 2 i points to the right and up. Its components are shown as dashed arrows: p 2 i x points to the right from the tail of p 2 i and p 2 i y points up from the head of p 2 i x to the head of p 2 i. Vector p f points from the tail of p 1 i to the head of p 2 i, pointing to the right and slightly down. Figure b titled add x and y components to obtain x and y components of final momentum shows the vector sums of the components. P 1 i y is a downward arrow. P 2 i y is a shorter upward arrow, aligned with its tail at the head of P 1 i y. P f y is a short downward arrow that starts at the tail of P 1 i y and ends at the head of P 2 i y. P 1 i x is a rightward arrow. P 2 i x is a shorter rightward arrow, aligned with its tail at the head of P 1 i x. P f x is a long rightward arrow that starts at the tail of P 1 i x and ends at the head of P 2 i x. Figure c, titled add x and y components of final momentum shows the right triangle formed by sides p f x and p f y and hypotenuse p f. Arrows from figure b indicate that p f x and p f y are the same in figures b and c.
    Figure a, titled break initial momentum into x and y components shows vector p 1 i as a solid arrow pointing to the right and down. Its components are shown as dashed arrows: p 1 i y points down from the tail of p 1 i and p 1 i x points to the right from the head of p 1 i y to the head of p 1 i. Vector p 2 i is shown as a solid arrow with its tail at the head of vector p 1 i, and is shorter than p 1 i. Vector p 2 i points to the right and up. Its components are shown as dashed arrows: p 2 i x points to the right from the tail of p 2 i and p 2 i y points up from the head of p 2 i x to the head of p 2 i. Vector p f points from the tail of p 1 i to the head of p 2 i, pointing to the right and slightly down. Figure b titled add x and y components to obtain x and y components of final momentum shows the vector sums of the components. P 1 i y is a downward arrow. P 2 i y is a shorter upward arrow, aligned with its tail at the head of P 1 i y. P f y is a short downward arrow that starts at the tail of P 1 i y and ends at the head of P 2 i y. P 1 i x is a rightward arrow. P 2 i x is a shorter rightward arrow, aligned with its tail at the head of P 1 i x. P f x is a long rightward arrow that starts at the tail of P 1 i x and ends at the head of P 2 i x. Figure c, titled add x and y components of final momentum shows the right triangle formed by sides p f x and p f y and hypotenuse p f. Arrows from figure b indicate that p f x and p f y are the same in figures b and c." class="internal default" style="width: 306px; height: 314px;" width="306px" height="314px" src="/@api/deki/files/4845/clipboard_eec08452508018b5532b1f921f7f56855" />

    Figure \(\PageIndex{1}\): (a) For two-dimensional momentum problems, break the initial momentum vectors into their x- and y-components. (b) Add the x- and y-components together separately. This gives you the x- and y-components of the final momentum, which are shown as red dashed vectors. (c) Adding these components together gives the final momentum.

    We solve each of these two component equations independently to obtain the x- and y-components of the desired velocity vector:

    $$v_{f,x} = \frac{m_{1} v_{1,i,x} + m_{2} v_{2,i,x}}{m}$$

    $$v_{f,y} = \frac{m_{1} v_{1,i,y} + m_{2} v_{2,i,y}}{m}$$

    (Here, m represents the total mass of the system.) Finally, combine these components using the Pythagorean theorem,

    $$v_{f} = |\vec{v}_{f}| = \sqrt{v_{f,x}^{2} + v_{f,y}^{2}} \ldotp$$

    Problem-Solving Strategy: Conservation of Momentum in Two Dimensions

    The method for solving a two-dimensional (or even three-dimensional) conservation of momentum problem is generally the same as the method for solving a one-dimensional problem, except that you have to conserve momentum in both (or all three) dimensions simultaneously:

    1. Identify a closed system.
    2. Write down the equation that represents conservation of momentum in the x-direction, and solve it for the desired quantity. If you are calculating a vector quantity (velocity, usually), this will give you the x-component of the vector.
    3. Write down the equation that represents conservation of momentum in the y-direction, and solve. This will give you the y-component of your vector quantity.
    4. Assuming you are calculating a vector quantity, use the Pythagorean theorem to calculate its magnitude, using the results of steps 3 and 4.

    Example 9.14

    Traffic Collision

    A small car of mass 1200 kg traveling east at 60 km/hr collides at an intersection with a truck of mass 3000 kg that is traveling due north at 40 km/hr (Figure 9.23). The two vehicles are locked together. What is the velocity of the combined wreckage?

    An x y coordinate system is shown. A large truck mass m T = 3000 kilograms is moving north toward the origin with velocity v T. A small car mass m c = 1200 kilograms is moving east toward the origin with velocity v c, which is less than v T.
    Figure \(\PageIndex{2}\): A large truck moving north is about to collide with a small car moving east. The final momentum vector has both x- and y-components.

    Strategy

    First off, we need a closed system. The natural system to choose is the (car + truck), but this system is not closed; friction from the road acts on both vehicles. We avoid this problem by restricting the question to finding the velocity at the instant just after the collision, so that friction has not yet had any effect on the system. With that restriction, momentum is conserved for this system.

    Since there are two directions involved, we do conservation of momentum twice: once in the x-direction and once in the y-direction.

    Solution

    Before the collision the total momentum is

    $$\vec{p} = m_{c} \vec{v}_{c} + m_{T} \vec{v}_{T} \ldotp$$

    After the collision, the wreckage has momentum

    $$\vec{p} = (m_{c} + m_{T}) \vec{v}_{w} \ldotp$$

    Since the system is closed, momentum must be conserved, so we have

    $$m_{c} \vec{v}_{c} + m_{T} \vec{v}_{T} = (m_{c} + m_{T}) \vec{v}_{w} \ldotp$$

    We have to be careful; the two initial momenta are not parallel. We must add vectorially (Figure 9.24).

    Arrow p c points horizontally to the right. Arrow p t points vertically upward. The head of p t meets the tail of p c. P t is longer than p t. A dashed line is shown from the tail of p t to the head of p c. The angle between the dashed line and p t, at the tail of p t, is labeled as theta.
    Figure \(\PageIndex{3}\): Graphical addition of momentum vectors. Notice that, although the car’s velocity is larger than the truck’s, its momentum is smaller.

    If we define the +x-direction to point east and the +y-direction to point north, as in the figure, then (conveniently),

    $$\vec{p}_{c} = p_{c}\; \hat{i} = m_{c} v_{c}\; \hat{i}$$

    $$\vec{p}_{T} = p_{T}\; \hat{j} = m_{T} v_{T}\; \hat{j} \ldotp$$

    Therefore, in the x-direction:

    $$m_{c} v_{c} = (m_{c} + m_{T}) v_{w,x}$$

    $$v_{w,x} = \left(\dfrac{m_{c}}{m_{c} + m_{T}}\right) v_{c}$$

    and in the y-direction:

    $$m_{T} v_{T} = (m_{c} + m_{T}) v_{w,y}$$

    $$v_{w,y} = \left(\dfrac{m_{T}}{m_{c} + m_{T}}\right) v_{T} \ldotp$$

    Applying the Pythagorean theorem gives

    $$\begin{split} |\vec{v}_{w}| & = \sqrt{\Big[ \left(\dfrac{m_{c}}{m_{c} + m_{T}}\right) v_{c} \Big]^{2} + \Big[ \left(\dfrac{m_{T}}{m_{c} + m_{T}} \right) v_{T} \Big]^{2}} \\ & = \sqrt{\Big[ \left(\dfrac{1200\; kg}{4200\; kg}\right) (16.67\; m/s) \Big]^{2} + \Big[ \left(\dfrac{3000\; kg}{4200\; kg}\right) (11.1\; m/s) \Big]^{2}} \\ & = \sqrt{(4.76\; m/s)^{2} + (7.93\; m/s)^{2}} \\ & = 9.25\; m/s \approx 33.3\; km/hr \ldotp \end{split}$$

    As for its direction, using the angle shown in the figure,

    $$\theta = \tan^{-1} \left(\dfrac{v_{w,x}}{v_{w,y}}\right) = \tan^{-1} \left(\dfrac{7.93\; m/s}{4.76\; m/s}\right) = 59^{o} \ldotp$$

    This angle is east of north, or 31° counterclockwise from the +x-direction.

    Significance

    As a practical matter, accident investigators usually work in the “opposite direction”; they measure the distance of skid marks on the road (which gives the stopping distance) and use the work-energy theorem along with conservation of momentum to determine the speeds and directions of the cars prior to the collision. We saw that analysis in an earlier section.

    Exercise 9.9

    Suppose the initial velocities were not at right angles to each other. How would this change both the physical result and the mathematical analysis of the collision?

    Example 9.15

    Exploding Scuba Tank

    A common scuba tank is an aluminum cylinder that weighs 31.7 pounds empty (Figure 9.25). When full of compressed air, the internal pressure is between 2500 and 3000 psi (pounds per square inch). Suppose such a tank, which had been sitting motionless, suddenly explodes into three pieces. The first piece, weighing 10 pounds, shoots off horizontally at 235 miles per hour; the second piece (7 pounds) shoots off at 172 miles per hour, also in the horizontal plane, but at a 19° angle to the first piece. What is the mass and initial velocity of the third piece? (Do all work, and express your final answer, in SI units.)

    A drawing of a scuba tank exploding, and the resulting three pieces of different sizes.
    Figure \(\PageIndex{4}\): A scuba tank explodes into three pieces.

    Strategy

    To use conservation of momentum, we need a closed system. If we define the system to be the scuba tank, this is not a closed system, since gravity is an external force. However, the problem asks for the just the initial velocity of the third piece, so we can neglect the effect of gravity and consider the tank by itself as a closed system. Notice that, for this system, the initial momentum vector is zero.

    We choose a coordinate system where all the motion happens in the xy-plane. We then write down the equations for conservation of momentum in each direction, thus obtaining the x- and y-components of the momentum of the third piece, from which we obtain its magnitude (via the Pythagorean theorem) and its direction. Finally, dividing this momentum by the mass of the third piece gives us the velocity.

    Solution

    First, let’s get all the conversions to SI units out of the way:

    $$31.7\; lb \times \frac{1\; kg}{2.2\; lb} \rightarrow 14.4\; kg$$

    $$10\; lb \rightarrow 4.5\; kg$$

    $$235\; \frac{miles}{hour} \times \frac{1\; hour}{3600\; s} \times \frac{1609\; m}{mile} = 105\; m/s$$

    $$7\; lb \rightarrow 3.2\; kg$$

    $$172 \frac{mile}{hour} = 77\; m/s$$

    $$m_{3} = 14.4\; kg - (4.5\; kg + 3.2\; kg) = 6.7\; kg \ldotp$$

    Now apply conservation of momentum in each direction.

    The three pieces of the scuba tank are shown on an x y coordinate system. The medium size piece is on the positive x axis and has momentum p 1 in the plus x direction. The smallest piece is at an angle theta above the positive x axis and has momentum p 2. The largest piece is at an angle phi below the negative x axis and has momentum p 3.

    x-direction:

    $$\begin{split} p_{f,x} & = p_{0,x} \\ p_{1,x} + p_{2,x} + p_{3,x} & = 0 \\ m_{1} v_{1,x} + m_{2} v_{2,x} + p_{3,x} & = 0 \\ p_{3,x} & = -m_{1} v_{1,x} - m_{2} v_{2,x} \end{split}$$

    y-direction:

    $$\begin{split} p_{f,y} & = p_{0,y} \\ p_{1,y} + p_{2,y} + p_{3,y} & = 0 \\ m_{1} v_{1,y} + m_{2} v_{2,y} + p_{3,y} & = 0 \\ p_{3,y} & = -m_{1} v_{1,y} - m_{2} v_{2,y} \end{split}$$

    From our chosen coordinate system, we write the x-components as

    $$\begin{split} p_{3,x} & = - m_{1} v_{1} - m_{2} v_{2} \cos \theta \\ & = - (14.5\; kg)(105\; m/s) - (4.5\; kg)(77\; m/s) \cos (19^{o}) \\ & = -1850\; kg\; \cdotp m/s \ldotp \end{split}$$

    For the y-direction, we have

    $$\begin{split} p_{3,y} & = 0 - m_{2} v_{2} \sin \theta \\ & = - (4.5\; kg)(77\; m/s) \sin (19^{o}) \\ & = -113\; kg\; \cdotp m/s \ldotp \end{split}$$

    This gives the magnitude of p3:

    $$\begin{split} p_{3} & = \sqrt{p_{3,x}^{2} + p_{3,y}^{2}} \\ & = \sqrt{(-1850\; kg\; \cdotp m/s)^{2} + (-113\; kg\; \cdotp m/s)} \\ & = 1854\; kg\; \cdotp m/s \ldotp \end{split}$$

    The velocity of the third piece is therefore

    $$v_{3} = \frac{p_{3}}{m_{3}} = \frac{1854\; kg\; \cdotp m/s}{6.7\; kg} = 277\; m/s \ldotp$$

    The direction of its velocity vector is the same as the direction of its momentum vector:

    $$\phi = \tan^{-1} \left(\dfrac{p_{3,y}}{p_{3,x}}\right) = \tan^{-1} \left(\dfrac{113\; kg\; \cdotp m/s}{1850\; kg\; \cdotp m/s}\right) = 3.5^{o} \ldotp$$

    Because \(\phi\) is below the −x -axis, the actual angle is 183.5° from the +x-direction.

    Significance

    The enormous velocities here are typical; an exploding tank of any compressed gas can easily punch through the wall of a house and cause significant injury, or death. Fortunately, such explosions are extremely rare, on a percentage basis.

    Exercise 9.10

    Notice that the mass of the air in the tank was neglected in the analysis and solution. How would the solution method changed if the air was included? How large a difference do you think it would make in the final answer?

    Contributors

    • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).