ACCELERATED MOTION Let’s begin by summarizing some of our earlier definitions. If position is represented by a vector x, then displacement is change in position (in m, cm, km etc, with direction), i.e. x x f x i . (Average or instantaneous) velocity is the displacement divided by the elapsed interval of time (in m/s, cm/s, km/h etc, with direction), i.e. x x f xi v t t f ti . (Average or instantaneous) acceleration is the change in velocity divided by the elapsed interval of time (in m/s2, cm/s2, km/hr2, etc, with direction), i.e. v v f vi a t t f t i . For simplicity, let’s suppose we can assume that acceleration is uniform, or constant. Then, since the acceleration doesn’t change, ―average acceleration‖ and ―instantaneous acceleration‖ are identical and we can just refer to ―acceleration‖. Rearranging the last equation will give at v v f vi , or (adding vi to both sides) v f v i at .  2 Equation  shows us what the final velocity will be if we know the initial velocity, the acceleration and the elapsed time, and it can also be used to find one of the other quantities if we know the remaining three. (We’ll see examples of this shortly.) Now, if the acceleration is constant the average velocity can actually be written in two ways: x vi v f v and v 2 . t It follows that vi v f v vi at 2v at x vt 2 t i t i t , 2 2 or at . 1 x vi t 2 2  Essentially, equations  and  are all you need to solve any problem about bodies moving with constant acceleration! Example 1 A car starts from rest and accelerates uniformly to the east at 2 m/s2. Find its velocity and its displacement after 8 seconds. Solution Since it starts from rest, vi = 0 m/s. So, from  and  v f vi at 0 28 16 m/s (east) and 3 x vi t at 08 28 64 m (east). 1 2 1 2 2 2 Example 2 A ball starts from rest and accelerates down a 6m ramp, reaching the bottom in 4 seconds. What is its velocity at the bottom? Solution First, using  to find the acceleration, x vi t at , so 6 04 a 4 2 8a , 1 2 1 2 2 3 i.e. a 4 m/s2 (down the ramp). Then, from , the velocity at the bottom of the ramp is v f vi at 0 4 3 m/s (down the ramp). 3 4 Example 3 The maximum braking acceleration of a car is –3 m/s2 (opposite to the direction of motion). What is the braking displacement from 18 m/s (north) to rest and from 36 m/s to rest? Solution First, the car starts at vi = 18 m/s (north) and ends at vf = 0 m/s. So, using , we can find how long it takes to stop, i.e. v f vi at so 0 18 3t and t 6 s . 4 From , it follows that the displacement during braking is x vi t at 186 36 108 54 54 1 2 1 2 2 2 meters (north). Repeating when vi = 36 m/s (north), v f vi at so 0 36 3t and t 12 s . Therefore, x vi t at 3612 312 432 216 216 1 2 1 2 2 2 meters (north). Note that doubling your starting speed results in a quadrupling of your stopping distance! FREE FALL NEAR THE EARTH’S SURFACE It turns out that the acceleration of gravity near the Earth’s surface is virtually constant. Specifically, measurements of gravitational acceleration taken at thousands of locations around the globe would produce results which vary only from about –9.78 m/s2 (down) to –9.83 m/s2 (down). For this reason, we take the average of these and say that the magnitude of the acceleration of gravity here on Earth is g = 9.81 m/s2. (In fact, to keep calculations simple, we often just use g = 9.8 m/s2.) This means that equations  and  above can also be used to describe objects in ―free fall‖ near the Earth’s surface, as long as we replace a by –g = –9.8 m/s2 (9.8 m/s2 down). 5 Example 4 A stone is dropped from rest at a height of 60.0 m. How long does it take to hit the ground and what is its velocity at impact? Solution Since the stone falls 60.0 m, its displacement is –60.0 m. Using equation , x vi t at , so 60.0 0t 1 9.8t 2 4.9t 2 . 1 2 2 2 Hence t 2 60.0 12.24 and t 12 .24 3.5 s. 4.9 Now, from , the stone’s velocity when it hits the ground is v f vi at 0 9.83.5 34 .3 m/s (34.3 m/s down) . Any problem dealing with free fall or uniformly accelerated motion can be solved using equations  and , so make sure you have these equations on your information sheet for the next test. 6 NEWTON’S LAWS Aristotle had believed (mistakenly) that a force was required not only to start something moving but also to keep it moving. Much later, Galileo realized that, in fact, all bodies have a tendency to keep doing what they are doing, i.e. if at rest, they will tend to remain at rest, and if moving, they will tend to keep moving. This tendency is today referred to as inertia, and it is the basis for Newton’s First Law of Motion. (In the law as stated below, the term net external force simply means the net effect of all forces acting from outside a body.) Newton’s First Law Every body remains at rest or moves with constant velocity except when compelled to change its motion as a result of a net external force acting upon it. (Note that ―constant velocity‖ means both constant speed and constant direction, i.e. the motion is steady and in a straight line.) Think of a book at rest on a desk. It is easy to identify two external forces which act vertically on the book: gravity pulling down on it and the desk pushing up on it. But, because the book remains at rest, we can conclude there is no net external force, i.e. these two external forces are equal and opposite, and hence cancel one another. This serves to illustrate that a body can remain at rest when no net external force is acting. Now suppose the desk was frictionless and infinitely long! Even a slight push (say to the right) for just a brief 7 moment would get the book moving, but, with no friction to slow it down and stop it, it would continue moving to the right forever (i.e. long after the slight push had ended), even though the same two external vertical forces are still acting. This illustrates that a body can continue to move uniformly with no net outside force on it. If we return to the book at rest on the desk, and suddenly take the desk away, there is no longer an upward push to oppose the downward pull of gravity. That is, there will now be a net external (downward) force on the book due to gravity, which explains why the book will neither remain at rest nor move uniformly. Instead, it will accelerate downward. Finally, imagine that you are riding in a car travelling east at 30 km/hr when the car suddenly hits a wall and stops. Since you are not part of the car, your body’s own inertia will make it tend to keep moving forward even after the initial impact. That is, your body will keep travelling east at 30 km/hr until an external force acts to stop it. If you’re lucky, that external force will be exerted by your seatbelt and perhaps your airbag. If you’re unlucky, it will be exerted by the dashboard and the windshield! Momentum Momentum is one of two additional physical quantities associated with motion (the other being kinetic energy, 8 which will be discussed later). Specifically, there are actually two kinds of momentum: linear momentum (which is associated with linear motion) and angular momentum (which is associated with rotational or spinning motion). In this discussion, we consider only linear momentum, which is denoted p and defined to be the product of mass and velocity. That is, p mv (with units kg-m/s) . (Note that, because velocity is a vector and has a direction, then linear momentum is also a vector, and has the same direction as the velocity.) Basically, this formula tells us that both a very small mass with a very large velocity and a very large mass with a very small velocity can possess a lot of momentum. Ask yourself if you’d want to stand against a wall in front of either an approaching slow moving transport truck or an approaching speeding bullet and you’ll get the idea! (Refer to Table 5.2 in the textbook for some examples of magnitudes of linear momentum.) Force We have already introduced the idea of force as something which feels like a ―push‖ or a ―pull‖. The term force is referred to specifically in Newton’s First Law as the ―agent‖ responsible for causing a body to change its state of rest or motion with constant velocity. 9 Since it is obvious that the effect of a push or pull can depend on its direction, it is clear that force is a vector. We sometimes divide everyday forces into two categories: ―contact‖ forces which result from bodies (or particles within bodies) ―touching‖ one another; and ―action-at-a-distance‖ forces, which can occur between bodies even when are physically separated. Suppose we return to a book sitting on a desk. The downward gravitational pull of the Earth on the book is clearly ―action-at-a-distance‖ while the supporting force of the desk (upward) against the book is considered a ―contact‖ force. (Other so-called ―contact‖ forces include tension in a string and friction which can either keep a body from starting to slide across a supporting surface or slow an already-sliding body until it finally stops.) In fact, however, all forces are explained in terms of the four fundamental forces identified earlier—gravity, the electromagnetic force, the weak nuclear force and the strong nuclear force, all of which are really ―action-at-a- distance‖! As we’ve mentioned before, for example, the upward push of a desk on a book is actually due to the electrostatic repulsion between the electrons in the atoms on the bottom of the book and the electrons in the atoms on the top of the desk. And, because these electrons never actually touch each other, then, when considered 10 microscopically, the book and the desk are not really in ―contact‖ at all! In that sense, ―contact‖ forces are a myth. If a net external force can cause a body to change its state of rest or motion with constant velocity, it stands to reason that such a force will change the body’s momentum. This idea inspired Newton’s Second Law of Motion. Newton’s Second Law (First Version – universally true) The net external force acting on a body is equal to the rate at which its linear momentum changes, with the direction of the momentum change being in the direction of the force. Quantitatively, this law can be written p mv F t t , where we have to remember that F is not interpreted as a single force on any body, but rather as the net external force, i.e. the net effect of all outside forces on a body. The unit associated with force can be seen from the equation to be the unit of momentum divided by time. This is kg-m/s divided by s, or kg-m/s2, but because force is so important in physical science, this combination has been named the Newton (N), so that 1 N = 1 kg-m/s2 . 11 In most (though not all!) situations, when a net external force acts on a body to change its momentum, the mass stays constant while the velocity changes. In such cases, we can replace the momentum change (mv) by mv (to reflect the fact that only the velocity actually changes). That is, mv mv F ma , t t (since acceleration has already been defined by v v f vi a t t f t i .) This results in a second, and certainly better known (though limited), version of Newton’s Second Law. Newton’s Second Law (Second Version – true provided mass is constant) When a net external force acts upon a body of constant mass, the body will experience an acceleration which is proportional to the force (and in the same direction) and inversely proportional to the mass, i.e. F a F ma  . m  or Before we consider examples involving this more familiar version of Newton’s Second Law, one situation where this version would not apply is the case of a rocket. As you have certainly witnessed if you’ve watched a space shuttle launch, rockets propel 12 themselves forward by expelling large amounts of fuel out the back. This means that the mass of a rocket in fact decreases (usually very rapidly), and certainly can’t be assumed to be constant. Thus, in applying Newton’s Second Law to a rocket, one would have to use the First p mv Version, F t t . (Unfortunately, the First Version is considerably more difficult to use, mathematically, and we won’t be able to discuss it further in this course.) Example 1 A net external force of 50 Newtons (right) is applied to a 2 kg mass at rest. Determine its velocity after 4 seconds. Solution Using equation , F 50 a 25 m/s2 (right) . m 2 We can now find the velocity after 4 seconds from equation , i.e. v f vi at 0 25 4 100 m/s (right) . Example 2 How much force is required to stop a 5,000 kg truck if it is brought to rest from 30 m/s (east) in 10 s? Solution From , if v f v i at , then 0 30 a10 . 13 So, 30 10a 30 , or a 3 m/s2 (3 m/s2 west) . 10 It follows from  that the required force is F ma 5000 3 15000 N (15,000 N west) .