4-1   4-2 - 4-4   4-5   next        

Chapter Review

---

In chapter two you studied motion in one dimension. Everything you learned there applies to two-dimensional motion. In fact, two-dimensional motion is really just two completely independent cases of motion in one dimension. We call these two dimensions the horizontal (x) and vertical (y) directions. The running theme of this chapter is the following: horizontal and vertical motions are independent.

4-1 Motion in Two Dimensions

(A) Constant Velocity in Two Dimensions

Knowing that two-dimensional motion is really two separate cases of one-dimensional motion makes it easy to find the equations for describing constant velocity in two-dimensions. Simply take the one-dimensional equation, qx = vt, and apply it to both the x- and y-directions. The key difference is that (a) only the horizontal component of velocity v_0x applies to the horizontal motion, and (b) only the vertical component of velocity v_0y applies to the vertical motion. Therefore, the equations are

x = x₀ + v_0xt
y = y₀ + v_0yt.

Notice that since the velocity is constant v = v₀ at all times during the motion.

---

Example 4.1 Pocket Billiards: A billiard ball is struck so that it rolls across the pool table with a velocity of 25.0 cm.s at an angle of 52.0^o above the horizontal. The horizontal direction is the short side of a standard 122 cm x 244 cm table. If the ball was initially located at a point (x₀, y₀) = (11.0 cm, 17.0 cm) from the leftmost corner on the short side, (a) how much time will pass before it strikes a side of the table? (b) Will it go into a pocket?

Picture the Problem Our picture shows a rough sketch of the table and the ball. The lower left corner of the pool table is the origin of the coordinate system.

Strategy Treating the vertical and horizontal motions as independent, we need the x- and y-components of the velocity. Then we can solve this problem by finding out where the ball would be at certain times.

Solution

Determine the components of the velocity:

v0x = v0cosq = (25.0 cm/s)cos(52.0o) = 15.39 cm/s v0y = v0sinq = (25.0 cm/s)sin(52.0o) = 19.70 cm/s

Part (a)

1. Solve for the time it takes to reach the right edge along the x-direction:
2. Determine the vertical position of the ball:	y = yo + v0yt = 17.0 cm + (19.70 cm/s)(7.212 s) = 159 cm

Since the upper edge is 244 cm from the lower edge the ball has not reached the upper edge by the time it reaches the right edge. It strikes the right edge first.

Part (b)

1. Determine the ball's distance from the corner when it hits the right edge:

qy = 244 cm - 159 cm = 85 cm

The ball is 85 cm from the corner, which is too far to fall into the pocket.

Insight Clearly we were able to treat the horizontal and vertical motions as simultaneous, yet independent. Also notice that an additional digit was retained for the values of t, v_0x, and v_0y because they were used in intermediate steps.

---

Physlet Illustration: Two Dimensional Motion With Constant Velocity
	Interactive Help on      off
The blue ball moves with constant velocity across the floor.  The distance grid is in cm, and the times are shown in seconds. Can you determine the x- and y- components of the velocity of the ball? Start
Hints Choose a specific time interval in the animation, that is, specific starting and stopping times. Measure the ball's x- and y-positions at the starting time and again at the stopping time. Divide the changes in positions by the time it takes to undergo that change. How would you calculate the overall speed (the magnitude of the velocity)?

Practice Quiz

If you walk in a direction of 50.0o north of east at a pace of 2.30 m/s for one hour, how far north of your starting position will you be? 8.28 km 138 m 6.34 km 5.32 km

A bird flies with a westerly speed of 20 m/s and a northerly speed of 15 m/s. How far will it fly over a time period of 32 seconds? 800 m 640 m 480 m 160 m 1100 m

(B) Constant Acceleration in Two Dimensions

The same strategy used with constant velocity applies to motion with constant acceleration. Just apply the one-dimensional equations from chapter two to both directions using only x-components of velocity and acceleration for the horizontal motion and only y-components for the vertical motion. Thus, the four equations become eight:

Horizontal Motion	Vertical Motion
vx = v0x + axt	vy = v0y + ayt

These equations can be applied precisely as in chapter two. Here they represent two independent, yet simultaneous, cases of one-dimensional motion.

---

Example 4.2 Horizontal Rocket Launch: A toy rocket is launched horizontally from rest off the edge of a 0.850 m high table. While in flight the rocket manages to remain pointed horizontally and its thrust causes it to accelerate forward with 7.48 m/s² as gravity pulls it down. (a) How far from the edge of the table will it be when it hits the floor? (b) With what speed will it hit the floor?

Picture the Problem The picture shows the rocket leaving the edge of the table and the path it takes. The positive coordinate directions are also indicated.

Strategy The rocket has constant acceleration in both directions, so the full set of eight equations applies.

Solution

Part (a): We need to determine when it hits the floor and get its horizontal position. Let's choose (x₀, y₀) = (0,0). Also notice that the initial velocities are zero.

1. Choose a vertical equation that contains t with all other quantities known:
2. Solve this equation for the time to hit the floor:
3. Choose a horizontal equation involving x with all other quantities known:
4. Solve to get the horizontal position:

Part (b):

1. Get the horizontal velocity when it hits the floor:	vx = axt (7.48 m/s2)(0.4163 s) = 3.114 m/s
2. Get the vertical velocity when it hits the floor:	vy = ayt = (-9.81 m/s2)(0.4163 s) = -4.084 m/s
3. Determine the magnitude of v to get the speed:

Insight Be sure you understand the minus signs in part (a)-2.

---

4-1   4-2 - 4-4   4-5   next