AP PHYSICS 1 AND 2 › UNIT 1, KINEMATICS › TOPIC 1.5

Vectors and Motion in Two Dimensions

Everything so far has happened along a line, where a sign was enough to carry a direction. This topic is what changes when the motion leaves that line, and the answer is smaller than it looks: nothing new happens, twice at once.

What you should be able to do

1.5.A Resolve a vector into perpendicular components using a coordinate system you choose, and rebuild the vector from those components.

1.5.B Analyze motion in two dimensions by separating it into components, and treat projectile motion as the special case with zero horizontal acceleration.

1. A vector is two vectors you have not separated yet

A displacement of 50 m at 37° above the horizontal is one arrow. It is also, exactly and without approximation, a walk of 40 m east followed by a walk of 30 m north. Those are not two descriptions of nearly the same thing. They are the same thing, and you may swap between them whenever one is more convenient.

That is the whole method of this topic. A two-dimensional problem is hard because two things are happening at once. Split the vector into perpendicular pieces and each piece is a one-dimensional problem of the kind you have already solved four times.

Perpendicular is the point. Two components at right angles cannot influence each other. That is why the split works: what happens along x never leaks into y. Split a vector into two pieces that are not perpendicular and you have made the problem worse, not better.

2. Resolving onto axes that you choose

Drop a perpendicular from the tip of the arrow onto each axis. The side lying against the angle is the adjacent one, and the side opposite the angle is the other. For a vector of magnitude A at angle θ measured from the x axis:

A_x = A cos θA_y = A sin θ

The cosine belongs to the axis the angle is measured from. That is the only thing to remember, and it is worth remembering as a sentence rather than as a pair of formulas, because the moment somebody measures the angle from the vertical instead the two swap over.

Drag the tip of the arrow. The magnitude and the angle are read off the arrow itself; the two components are read off the axes. Then rotate the axes with the slider and watch what happens. The arrow never moves. The numbers describing it move a great deal, which is the point: components belong to a coordinate system, and the vector does not.

Rotate the axes

3. Rebuilding the vector from its components

Going back is Pythagoras for the size and a tangent for the direction:

A = √(A_x² + A_y²)tan θ = A_y / A_x

Two habits are worth forming here. The first is to state a direction against something: above the horizontal, north of east. An angle on its own is not an answer, because it does not say what it was measured from. The second is to check the magnitude against the larger component. The resultant is always longer than either piece and always shorter than their plain sum, so an answer outside that range is arithmetic, not physics.

Why the sum is too big. Adding 40 and 30 gives 70, but the vector is 50. The two walks are at right angles, so neither one helps the other along. Only when two vectors point the same way do their magnitudes simply add.

4. Two dimensions are two one-dimensional problems

Once a motion is split into components, each component obeys the equations from Topic 1.3 on its own, with its own acceleration, and neither knows the other exists. The two motions share only one thing: the clock. That shared time is what stitches them back together at the end.

One ball is launched, and at the same instant a second is simply dropped from the same height. Watch the horizontal guide line that joins them. Change the launch speed and angle as much as you like: the guide line stays horizontal until the launched ball reaches the ground, because the vertical motions are identical. The horizontal motion is doing nothing to the falling at all.

Launch speed Launch angle

5. Projectile motion, the special case

A projectile is the case where the split is at its most useful, because one of the two accelerations is zero:

a_x = 0a_y = −g ≈ −10 m/s²

Horizontally, nothing accelerates, so the distance is just speed multiplied by time. Vertically, it is free fall, exactly as in Topic 1.3. Nothing in this topic is new physics. It is the old physics, applied twice, to two directions that do not talk to each other.

Two consequences follow immediately, and both are worth stating out loud because both are routinely got wrong. The time a projectile spends in the air is decided by the vertical motion alone, so throwing it harder sideways does not keep it up any longer. And at the very top of its path the vertical velocity is zero while the horizontal velocity is unchanged, so the projectile is still moving, and its acceleration there is still g downward.

Check yourself

1. A displacement of 50 m points 37° above the horizontal. Find both components, then check that they rebuild the original.

Horizontal 50 cos 37° = 40 m, vertical 50 sin 37° = 30 m. Rebuilding: √(40² + 30²) = 50. If your components add to 50 rather than rebuilding to it, you have added them as though they pointed the same way.

2. The same vector is resolved again, this time onto axes rotated so that one axis lies along the vector. What are the components now, and what has changed about the vector?

The components become 50 and 0. Nothing whatever has changed about the vector: it has the same magnitude and points the same way in the room. Only the description changed, because you changed the axes. This is the single most useful idea on the page.

3. A ball rolls off a table at 3.0 m/s from a height of 1.25 m. How long is it in the air, and which given did you not need?

Vertically it starts from rest and falls 1.25 m, so 1.25 = 5t² gives t = 0.50 s. You did not need the 3.0 m/s. A horizontal speed cannot change a vertical fall, and noticing which given is unnecessary is usually the sign that you have separated the components properly.

4. A classmate says the acceleration of a projectile is zero at the top of its path, because it stops for an instant. Identify what is momentarily zero, and say what the acceleration actually is there.

The vertical component of the velocity is zero for an instant. The horizontal component is not, so the projectile has not stopped at all; it is moving horizontally at its full launch speed. The acceleration is g downward there, exactly as it is everywhere else on the path. If it were zero the projectile would carry on in a straight line.

5. Two balls are launched from the same point at the same speed, one at 30° and one at 60°. Which lands farther away, and which is in the air longer?

They land the same distance away, because complementary angles give equal ranges. The 60° ball is in the air longer, because its vertical launch component is larger and hang time is settled by the vertical motion alone. Range and time are different questions and they do not have to agree.

Unit 2, Force and Translational Dynamics. Unit 1 has described motion in full without once asking what causes it. Unit 2 starts there, and the component method you have just learned is the tool it uses on every inclined plane in the course.