11.5 Momentum in Two Dimensions

The law of conservation of momentum $\vec{P}_{f} = \vec{P}_{i}$ is not restricted to motion along a line. Many interesting examples of collisions and explosions involve motion in a plane, and for these both the magnitude and the direction of the total momentum vector are unchanged. The total momentum is the vector sum of the individual momenta, so the total momentum is conserved only if each component is conserved:

(p_{f x})_{1} + (p_{f x})_{2} + (p_{f x})_{3} + \dots = (p_{i x})_{1} + (p_{i x})_{2} + (p_{i x})_{3} + \dots

(p_{f y})_{1} + (p_{f y})_{2} + (p_{f y})_{3} + \dots = (p_{i y})_{1} + (p_{i y})_{2} + (p_{i y})_{3} + \dots

(11.36)

A time-lapse photograph shows the collision and subsequent movement of two billiard balls. — Collisions and explosions often involve motion in two dimensions.

Let’s look at some examples of momentum conservation in two dimensions.

EXAMPLE 11.8 A peregrine falcon strike

Peregrine falcons often grab their prey from above while both falcon and prey are in flight. A $0.80 kg$ falcon, flying at $18 m / s$ , swoops down at a $45 °$ angle from behind a $0.36 kg$ pigeon flying horizontally at $9.0 m / s$ . What are the speed and direction of the falcon (now holding the pigeon) immediately after impact?

MODEL The two birds are the system. This is a perfectly inelastic collision because after the collision the falcon and pigeon move at a common final velocity. The birds are not a perfectly isolated system because of external forces of the air, but during the brief collision the external impulse delivered by the air resistance will be negligible. Within this approximation, the total momentum of the $falcon + pigeon$ system is conserved during the collision.

VISUALIZE FIGURE 11.27 is a before-and-after pictorial representation. We’ve used angle $ϕ$ to label the post-collision direction.

The figure illustrates the before-and-after pictorial representations of a falcon catching a pigeon.

FIGURE 11.27 Full Alternative Text

SOLVE The initial velocity components of the falcon are $(v_{i x})_{F} = v_{F} cos θ$ and $(v_{i y})_{F} = - v_{F} sin θ$ . The pigeon’s initial velocity is entirely along the x-axis. After the collision, when the falcon and pigeon have the common velocity $\vec{v}_{f}$ , the components are $v_{f x} = v_{f} cos ϕ$ and $v_{f y} = - v_{f} sin ϕ$ . Conservation of momentum in two dimensions requires conservation of both the x- and y-components of momentum. This gives two conservation equations:

\begin{matrix} (m_{F} + m_{P}) v_{f x} & = & (m_{F} + m_{P}) v_{f} cos ϕ \\ = & m_{F} (v_{i x})_{F} + m_{P} (v_{i x})_{P} = m_{F} v_{F} cos θ + m_{P} v_{P} \\ (m_{F} + m_{P}) v_{f y} & = & - (m_{F} + m_{P}) v_{f} sin ϕ \\ = & m_{F} (v_{i y})_{F} + m_{P} (v_{i y})_{P} = - m_{F} v_{F} sin θ \end{matrix}

The unknowns are $v_{f}$ and $ϕ$ . Dividing both equations by the total mass gives

\begin{array}{l} v_{f} cos ϕ = \frac{m_{F} v_{F} cos θ + m_{P} v_{P}}{m_{F} + m_{P}} = 11.6 m / s \\ v_{f} sin ϕ = \frac{m_{F} v_{F} sin θ}{m_{F} + m_{P}} = 8.78 m / s \end{array}

We can eliminate $v_{f}$ by dividing the second equation by the first to give

\begin{array}{l} \frac{v_{f} sin ϕ}{v_{f} cos ϕ} = tan ϕ = \frac{8.78 m/s}{11.6 m/s} = 0.757 \\ ϕ = {tan}^{- 1} (0.757) = 37 ° \end{array}

Then $v_{f} = (11.6 m/s)/cos(37 °) = 15 m/s$ . Immediately after impact, the falcon, with its meal, is traveling at $15 m / s$ at an angle $37 °$ below the horizontal.

REVIEW It makes sense that the falcon would slow down after grabbing the slower-moving pigeon. And Figure 11.27 tells us that the total momentum is at an angle between $0 °$ (the pigeon’s momentum) and $45 °$ (the falcon’s momentum). Thus our answer seems reasonable.

EXAMPLE 11.9 A three-piece explosion

A 10.0 g projectile is traveling east at $2.0 m / s$ when it suddenly explodes into three pieces. A 3.0 g fragment is shot due west at $10 m / s$ while another 3.0 g fragment travels $40 °$ north of east at $12 m / s$ . What are the speed and direction of the third fragment?

MODEL Define the system to be the projectile, one piece before the explosion and three fragments afterward. Although many complex forces are involved in the explosion, they are all internal to the system. There are no external forces, so this is an isolated system and its total momentum is conserved.

VISUALIZE FIGURE 11.28 shows a before-and-after pictorial representation. We’ll use uppercase M and V to distinguish the initial object from the three pieces into which it explodes.

The figure illustrates the before-and-after pictorial representations of a three-piece explosion of a projectile of mass 10 grams.

FIGURE 11.28 Full Alternative Text

SOLVE The system is the initial object and the subsequent three pieces. Conservation of momentum requires

\begin{array}{l} m_{1} {(v_{f x})}_{1} + m_{2} {(v_{f x})}_{2} + m_{3} {(v_{f x})}_{3} = M V_{i x} \\ m_{1} {(v_{f y})}_{1} + m_{2} {(v_{f y})}_{2} + m_{3} {(v_{f y})}_{3} = M V_{i y} \end{array}

Conservation of mass implies that

m_{3} = M - m_{1} - m_{2} = 4.0 g

Neither the original object nor $m_{2}$ has any momentum along the y-axis. We can use Figure 11.28 to write out the x- and y-components of $\vec{v}_{1}$ and $\vec{v}_{3}$ , leading to

\begin{array}{l} m_{1} v_{1} cos 40 ° - m_{2} v_{2} + m_{3} v_{3} cos θ = M V \\ m_{1} v_{1} sin 40 ° - m_{3} v_{3} sin θ = 0 \end{array}

where we used $(v_{f x})_{2} = - v_{2}$ because $m_{2}$ is moving in the negative x-direction. Inserting known values in these equations gives us

\begin{array}{l} - 2.42 + 4 v_{3} cos θ = 20 \\ 23.14 - 4 v_{3} sin θ = 0 \end{array}

We can leave the masses in grams in this situation because the conversion factor to kilograms appears on both sides of the equation and thus cancels out. To solve, first use the second equation to write $v_{3} = 5.79 / sin θ$ . Substitute this result into the first equation, noting that $cos θ / sin θ = 1 / tan θ$ , to get

- 2.42 + 4 (\frac{5.79}{sin θ}) cos θ = - 2.42 + \frac{23.14}{tan θ} = 20

Now solve for $θ :$

\begin{array}{l} tan θ = \frac{23.14}{20 + 2.42} = 1.03 \\ θ = {tan}^{- 1} (1.03) = 45.8 ° \end{array}

Finally, use this result in the earlier expression for $v_{3}$ to find

v_{3} = \frac{5.79}{sin 45.8 °} = 8.1 m / s

The third fragment, with a mass of 4.0 g, is shot $46 °$ south of east at a speed of $8.1 m / s$ .

11.5 Momentum in Two Dimensions

EXAMPLE 11.8 A peregrine falcon strike

FIGURE 11.27 Pictorial representation of a falcon catching a pigeon.

EXAMPLE 11.9 A three-piece explosion

FIGURE 11.28 Before-and-after pictorial representation of the three-piece explosion.