Two-Dimensional Motion

Galileo was the founder of modern physics not merely because of the ideas he introduced and studied, but mostly because of the way he thought about them. In particular, he is the first famous practioner of the gedanken or thought experiment. You imagine a situation in which your experience (what physicists call physical intuition) tells you what should happen under certain conditions. If in the course of imagining what will happen you find a contradiction between what you expect and what your understanding of the principles of physics would allow, then you know that your understanding is limited or that the principle needs to be questioned. The advantage of gendanken experiments is that you can imagine situations which would be difficult or practically impossible to achieve in a laboratory setting. Nonetheless physicists use gedanken experiments even today to test their notions of how objects which have never been directly seen, black holes for example, must behave according to physics principles. If the behavior of the system due to a certain principle of physics leads to a contradiction with another principle, the deeper principle is usually held fixed in the mind while physicists explore the "weakness" in the the contradicting principle. In this way researchers can choose which problems are likely to yield significant insights.

Galileo's thought experiment is shown in the picture below.

In the picture one observer on a boat moving with speed v with respect to the shore sees a ball fall from the top of the mainsail mast. The observer on shore also watches the ball fall. Do the two observers see the same thing for the motion? They cannot. The observer on the boat moves horizontally with the same speed as the horizontal motion of the ball (since both were attached to the boat and no other forces in the horizontal direction are acting once the ball is released). Therefore the observer on the boat sees a trajectory or flight path which is straight down. The observer on the shore, however, sees the trajectory shown below

The shape of the trajectory is parabolic. How do we know? We have to have some idea of what the shape should look like. From this thought experiment Galileo concluded that the trajectory results from the independence of motion along the horizontal and vertical directions. The simplest idea is to say that the vertical motion is unaffected by the horizontal motion of the boat. Directions which satisfy the constraint of being completely independent of each other are said to be orthogonal. Orthogonal directions are always at right angles to each other in Cartesian coordinate schemes.

Making use of this notion, the difference in the two trajectories seen by the observers results only because they see the ball with different relative horizontal velocities. The observer on the boat sees a relative horizontal velocity of zero for the whole time the ball falls since gravity only accelerates things in the vertical direction. The shore observer sees this vertical motion AND the relative horizontal velocity of the boat and ball. Both observers see the same vertical motion.

We can use Maple to tell us if this idea gives us the right behavior. Assuming no change in the form of the equations of motion, the shore observer should see the initial velocities and positions for x (horizontal) and y (vertical) used in the appropriate equations for motion as follows:
x motion y motion
x = x0 + v0xt y = y0 + v0yt - ½ gt2

To see an interactive view of two objects falling with the same acceleration, but having different horizontal motions, consider the program below. You can adjust the horizontal speed to the right of each ball (ball A is red and ball B is blue) so that, as it falls, it includes a constant horizontal speed and a vertical acceleration. In particular, note that no matter what horizontal speeds you give, the vertical motion of the two objects is always the same. The two objects are always at the same vertical height at the same time no matter what the horizontal part of the motion is doing.

Maple can do this trivially. We just specify the equations and the initial values, then ask Maple to plot it. Better yet, let's animate the plot and see the trajectory drawn out for us. Click here for the Maple code.


Our Friends, the Vectors

As with any advance in physical intuition, we need a concise way to describe our thoughts about motion in more than one dimension. The mathematical notion of a vector fits perfectly. The properties of vectors are that they possess just 2 pieces of information: direction and magnitude. The rules for how vectors are formed and used follows exactly the purpose for which we use them in multi-dimensional motion. The properties are set to be mathematically consistent with the notion that components of vectors (those pieces of a vector which lie along orthogonal directions) have the same mathematical rules applied to them, independently. These rules are trigonometric. In just two dimensions we can describe all the rules for two orthogonal directions and then assert that these rules apply in just the same way to any number of dimensions. We summarize these rules here:

  1. The vector direction is given by an angle theta with respect to an axis in a coordinate system (usually the x axis but any axis is OK). The magnitude is represented by the length of the vector in that coordinate system. So for a vector A,

    Here the components satisfy the following relationships:

    tanq = Ay/Ax
    sinq = Ay/A
    cosq = Ax/A
    A = sqrt[Ax2 + Ay2]

    The magnitude A and the direction given by theta completely describe the vector A. An alternative and equally valid way of describing a vector is to use the components directly by defining the unit vectors. A unit vector has a magnitude of 1 in whatever units the coordinate system is defined in. The direction of a unit vector can be set to be parallel to one of the coordinate axes. Typically, the x and y and z coordinate axes have associated unit vectors i, j, and k. Therefore, the vector A can be described by

    Axi + Ayj

  2. To add two vectors, we add the components as shown below.

    The sum of A and B is C. The general rule for vector addition is

    C = A + B
    Cxi + Cyj = (Ax + Bx)i + (Ay + By)j

    You can see this rule applied graphically with the following Java applet.


    Two-Dimensional Motion Concepts Applied

    It's not too hard to believe that objects under the influence of gravity fall at the same rate independent of whether they are moving horizontally or not. Galileo's thought experiment about the cannonball on the boat mast seems reasonable after all. The parabolic shape is also not unfamiliar to anyone who's tried to visualize the shape of a falling object with an initially horizontal velocity. But is the answer so easily visualized if the object starts off with both a horizontal and a vertical velocity? According to Galileo's prescription, we should assume no difference in how we approach the problem. Horizontal and vertical motion are completely independent. One way to see that this still works is to work out quantitatively what happens for the hunter-monkey demonstration.

    To remind you: a hunter fires a dartgun with a harmless sedative at a monkey hanging from a vine a distance h vertically above the dartgun and a distance d horizontally away from the dartgun. The hunter aims directly at the monkey and fires, but just as the hunter fires, the monkey, using its incredible spider-monkey sense, realizes what's up and drops from the vine. Does the monkey avoid the dart?

  3. Solution: Our approach is standard. We draw a picture, then analyze the words in the problem to come up with the values for the variables we define. First, we need a coordinate system. Let's define up and to the right as positive directions with the origin being the end of the dartgun.

    It's almost always useful to find the components of the acceleration and initial velocity. The acceleration is easy since only gravity is acting to change the motion, so ax = 0 and ay = - g. For the initial velocity, we have

    v0x = v0 cosq
    v0y = v0 sinq

    The equations for motion for the monkey and the dart are:

    dart monkey
    x: xdart = v0xt xmonkey = d
    y: ydart = v0yt - ½ gt2 ymonkey = h - ½ gt2

    Assuming that the dart does hit the monkey, the condition for a collision of dart and monkey is for xdart = xmonkey and ydart = ymonkey for some time t, i.e. we want to see if it's possible to find

    v0 cosq*t = d
    v0sinq*t - ½ gt2 = h - ½ gt2 ==>
    v0sinq*t = h

    If we divide the second equation by the first, then

    (v0sinq*t)/ (v0cosq*t) = h/d ==>
    tanq = h/d

    Therefore, for some time t, we will always find a solution which yields a collision, provided tanq = h/d. However, if we go back to the wording in the original problem we find that the definition of "hunter aims directly at the monkey" is precisely that tanq = h/d so we are guaranteed a "hit" if the monkey doesn't reach the ground before the dart gets to it. Note that with the exception of hitting the ground, the "collision" of dart and monkey is inevitable no matter what value of v0 we use. That is to say, the condition for collision does not depend on the magnitude of v0, just its direction.

    Suppose we wanted to know the time of the collision between dart and monkey; what would we need to do? The consideration of this is in the Maple file along with the solutions to the problems posed below.

  4. We've seen the demonstration of the hunter-monkey experiment so you know that the solution given above does in fact work. You don't need to pick any particular initial speed to be sure that the monkey gets hit, you just need to aim right at the monkey.


    Let's do something a little more practical as long as Maple is available to do the algebra for us. Consider the following problems:

    1. A baseball player hits a ball so that it's initial velocity is 40 m/s at an angle of 63° w.r.t. the horizontal. The ball leaves the bat at a height 1 meter above the ground and heads for a wall 3 meters high, located 100 meters from the batter.
      1. Does the batter get a homerun or is the ball likely to be caught for an easy out?
      2. What's the smallest speed with which the batter can hit the ball at 63° w.r.t. the horizontal and still make it over the wall?

      Check the Maple file for the solutions.

    2. An olympic skeet shooter wishes to shoot a clay target that flies through the air. The target leaves from ground level 10 meters from the shooter with a velocity of 60 m/s directed 45° to the horizontal. The shooter fires a bullet with a velocity of 400 m/s directed at angle theta w.r.t. the horizontal. The bullet leaves the gun 2 meters above ground level. If the shooter fires 1 second after the clay target is fired, at what angle theta would he/she have to aim and how soon after firing would the bullet hit the target?

      Again, check the Maple file for solutions.



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