Lecture 3

Motion in two Dimensions

If you master the equations of motion in one dimension, you are in very good shape, since you will be able to handle any problem in any number of dimensions (even though we will limit ourselves to the case of two dimensions only).

The rule is very simple : what happen in x is independent of what happens in y. We can solve any problem by writing down the projections of the equations of motion on the x and y axis respectively, and solve separately in x and in y.

Formally : if $\vec{r}$ is the displacement vector in the x-y plane, the equations of motion (for the case of constant acceleration a) are written as:

$\vec{r}(t) = \vec{r}_{0}+\vec{v}_{0}t+1/2~\vec{a}t^{2}$

$\vec{v}(t) = \vec{v}_{0}+\vec{a}t$

These are equations among vectors, that we would not know how to solve. But what we can do is to project the vectors on the two axes, to get:

x(t)=x₀+v_0xt+1/2 a_xt², y(t)=y₀+v_0yt+1/2 a_yt²

v_x(t) = v_0x+a_xt,  v_y(t) = v_0y+a_yt

and these we know how to handle.

The most common case,encountered in endless situations (kicking a football, dropping a package from a plane, throwing a stone off a cliff, hitting a golfball, doing the long jump, etc., etc.) is that of a body with a given initial velocity $\vec{v}_{0}$ (whose components are $v_{0}\cos\theta$and $v_{0}\sin\theta$), that is subjected to the acceleration of gravity. The most important thing to remember for these situations is that the horizontal motion will continue unperturbed with constant velocity $v_{0x}=v_{0}\cos\theta$. For the acceleration, one has obviously

a_x = 0, a_y=g

(but pay attention to the sign of g...) In other words the equations of motion are :

x(t)=x₀+v_0xt, y(t)=y₀+v_0yt+1/2 gt²

v_x(t) = v_0x,  v_y(t) = v_0y+gt

As an immediate application, we get the range of a projectile : suppose you hit a golf ball, giving it an initial velocity $\vec{v}$ (of magnitude v) and at an angle $\theta$, and we want to find out how far it will go. The reasoning is that the furthest x will be reached when y=0, i.e. when the ball hits the ground. We can then write :

$0 = v\sin\theta~t + 1/2~gt^{2}$, i.e. $0 = t(v\sin\theta + 1/2~gt)$

This expression is 0 for either t=0 (which is obviously true) or for

$v\sin\theta + 1/2~gt = 0$, i.e. $t = -(2v\sin\theta)/g$

This would give me the "time of flight" of the ball, but I want to know how far it goes. Substituting in the expression for x, we get:

$x = -(v\cos\theta~2v\sin\theta )/g = -v^{2}\sin (2\theta )/g$

(remember : $2\sin\theta\cos\theta = \sin (2\theta )$

Question : do we have a sign problem ????


General conclusion : the range is proportional to the square of the initial velocity and, for a given velocity , is the largest for $\theta = 45^{0}$.

Famous problem : the monkey and the hunter. A monkey drops a coconut and, at the same instant, a hunter shoots his gun,trying to hit the nut. Question : in which direction should the gun point ? You might think that the answer depends on the distance to the tree and the bullet initial velocity, but it turns out that the solution is independent of all that, and you wil hit the target provided you aim at it at t=0. (See also problem 2.47)

Proof :

As usual, hitting the target with the bullet means that target and bullet must have the same x and y at the same time. Let us choose as positive x the direction from the gun to the tree, and positive y pointing up. We set the gun at x=y=0, and the target (coconut) at x₀ and y₀. If the bullet is shot at an angle $\alpha$and velocity of magnitude v, the equations of motion for the two objects are (we write explicitely the minus sign to indicate that g is pointing downward) :

• bullet :
  $x = v\cos\alpha ~t$, $y=v\sin\alpha ~t -1/2~gt^{2}$
• coconut :
  x = x₀, y = y₀-1/2 gt²
  (the coconut obviously remains always at the same x=x₀, it only moves in y)

We now impose the condition of y equality :

$v\sin\alpha~t-1/2~gt^{2} = y_{0}-1/2~gt^{2}$
$v\sin\alpha~t = y_{0}$
$t = y_{0}/(v\sin\alpha )$

This gives us the time t at which the two bodies are at the same y, and it will have to be plugged into the expression for the equality of x, which is :

$x_{0} = v\cos\alpha~t$, i.e.
$x_{0}=v\cos\alpha~y_{0}/(v\sin\alpha )$
$y_{0}/x_{0} = \sin\alpha /\cos\alpha = \tan\alpha$

But $\alpha$ is the direction (until now undetermined) at which the gun was pointing. And y₀/x₀ is exactly the tangent of the angle formed by the line joining the gun with the coconut . The result we have obtained then tells us that the only condition we need to satisfy is that, at t=0, the gun points at the target.

Relativity of Velocities

Would you be able to play Table Tennis if the ball were moving at 500 miles/hour?

The answer is yes, provided also you and the table move at the same speed. And this is what actually happens if your game takes place while flying on a jet-plane : with respect to the ground, the ball is moving at a few hundred meters/sec (plus or minus 1 or 2 m/s as it goes back and forth). But of course you are not aware of it since you are moving at high speed too.

The lesson we learn is that velocity is not an absolute, but a relative quantity. You probably associate the concept of relativity with Einstein, but relativity was a familiar concept in Physics well before him. In fact, the concept of relativity of velocities goes under the name of Galileian Relativity (what Einstein did, was to show that not only space, but also time is relative).

In a large class of problems, we face a situation where there are several objects in motion with respect to each other, and we need to determine their relative velocities. The problem is simpler if we need to determine velocities with respect to something which appears to be at rest. Examples :

• the velocity of the ping-pong ball with respect to the ground, is the (vectorial) sum of the velocity of the ball w.r.t. the plane + the velocity of the plane w.r.t. the ground
• the velocity of a boat in a river is the (vectorial) sum of the velocity of the boat w.r.t. the water + the velocity of the water w.r.t. the shore
• the velocity of a point on a car tire w.r.t. the ground is etc. etc....

The problem is only a little bit trickier if we have to find the relative velocity between two objects that are both in motion, e.g. two cars moving towards each other. A good way to attack such a problem is to pretend that one of the two objects (cars) is at rest, and perform the vector sum of the other velocities. Specifically :

car A is moving with velocity $\vec{v}_{A}$ and car B with velocity $\vec{v}_{B}$ (both velocities referred to the ground). Question : what is the velocity of B w.r.t. A?

To find the answer we pretend that A is at rest : the answer we look for will be given by the (vector) sum of the velocity of the ground w.r.t. A + the velocity of B w.r.t. the ground.

The second term is known, what is the first one ? You should convince yourself that the velocity of the ground w.r.t. A is the vector opposite to the velocity of A w.r.t. the ground, i.e. $-\vec{v}_{A}$ .