Relative Velocities and Classical Relativity
The relative velocities in this topic are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than 3,000 km/s. Most things we encounter in daily life move slower than this speed.
Relative Velocities and Classical Relativity
When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities are called relative velocities. For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground.
Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity, which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.
Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later tutorials.
The relative velocities in this topic are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than \(\text{3,000 km/s}\). Most things we encounter in daily life move slower than this speed.
Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward?
The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See the figure below.)
To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in the figure below.
Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer.
Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.)
Example: Calculating Relative Velocity: An Airline Passenger Drops a Coin
An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?
The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall straight down. (b) An observer on the ground sees the coin move almost horizontally.
Strategy
Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.
Solution for (a)
Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:
\({{v}_{y}}^{2}={{v}_{0y}}^{2}-2g\left(y-{y}_{0}\right)\text{.}\)
Substituting known values into the equation, we get
\({{v}_{y}}^{2}={0}^{2}-2\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(-1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{m}-0\text{ m}\right)=\text{29}\text{.}4\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}{\text{/s}}^{2}\)
yielding
\({v}_{y}=-5\text{.}\text{42 m/s.}\)
We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.
Solution for (b)
Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is \({v}_{y}=-5.42\phantom{\rule{0.25em}{0ex}}\text{m/s}\), the same as found in part (a).
In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and \({v}_{x}=\text{260 m/s}\). The x- and y-components of velocity can be combined to find the magnitude of the final velocity:
\(v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\text{.}\)
Thus,
\(v=\sqrt{\left(\text{260 m/s}\right)^{2}+\left(-5.42\text{ m/s}\right)^{2}}\)
yielding
\(v=\text{260}\text{.}\text{06 m/s.}\)
The direction is given by:
\(\theta ={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)={\text{tan}}^{-1}\left(-5\text{.}\text{42}/\text{260}\right)\)
so that
\(\theta ={\text{tan}}^{-1}\left(-0\text{.}\text{0208}\right)=-1\text{.}\text{19º}\text{.}\)
Discussion
In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground.
This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity.
Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not \(\left(\text{260 – 5}\text{.}\text{42}\right)\text{ m/s}\); rather, it is \(\text{260}\text{.}\text{06 m/s}\). The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane.
The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See the figure above.)
In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.
Making Connections: Relativity and Einstein
Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is stored as increased mass, and more surprises await.
This lesson is part of:
Two-Dimensional Kinematics