The Story of "Acts Upon"

After the concept of inertia was understood, what remained in understanding "motion" was to figure out just how an external body "acts upon" another one to change its motion. For philosophers, the beginning of the argument would be in defining "motion". We have assumed that motion is given by the velocity simply because displacement and the time over which it takes place are now easily measured and can be clearly defined. However, for Descartes, the clockwork mechanism by which he sought to understand the behavior of the universe consisted of assuming that the universe had a fixed amount of "motion" and a fixed amount of "matter". His view was that the easiest way to express the combination of these conserved quantities was as the product of mass times speed. Hence, the universe had a fixed quantity of mass times speed and, in a collision of two objects, the speeds of each object might change, but the quantity of mass times speed would stay the same.

Descartes' idea was clearly in contradiction to experimental data that could be gathered by scientist of almost any skill level. Two soft, sticky objects of about the same mass can be made to collide on a smooth surface in such a way that they both come to rest. Hence their speeds both go to zero while their total mass stays unchanged and nothing else in the universe seems to pick up the slack. Still, Descartes' idea of a conserved motion was enticing.

In his study of collisions, Newton performed many careful experiments on collision between all kinds of objects. He observed, as did others, that Descartes was almost correct. The quantity mass times velocity, is conserved as long as there are no external forces on the two colliding objects. In other words, the vector quantity p = mv is conserved no matter what kinds or shapes of objects collide. He also observed that the scalar quantity mv2 was also conserved as long as the objects colliding were hard and spherical. The importance of the last statement is that mv2 for a system is conserved for special conditions of collisions while mv for the system is always conserved. Newton therefore defined p as the true quantity of "motion" and formulated his second law of motion by stating that the "motive force" which changes the "motion" follows the relationship

Fnet ext = dp/dt

The "motive force" is the sum of all external forces on the particles in the system. This is equivalent to our previous statement of the Second Law if the mass of the system is not changing with time since dp/dt = d(mv)/dt = m dv/dt = ma. If we choose our system to be large enough, we can always find a situation in which there are no external forces acting and then p is conserved. Conserved quantities are monumental in that they not only indicate something fundamental about the operation of the universe, they are enormously simple to handle mathematically since they do not change no matter what happens during the observation.

We have to get comfortable with applying the ideas however. Consider the figure below which shows two objects colliding upon a frictionless, level surface.

We can define our system as including only particle 1, only particle 2, or as including both particles. If our system includes only particle 1, then particle 2 is external and "acts upon" particle 1 to change particle 1's motion. If we consider only particle 2 as a system, then particle 1 is external and acts to change 2's motion. If both particles are included in our system, then there are no external forces which affect motion in the horizontal direction and the momentum of the system is conserved.

We can make use of these ideas in a quantitative way for problems for which Newton's Law as we originally formulated it would not be as helpful. A set of such problems is depicted below.

  1. We have a man and boy standing together on frictionless ice. If they each give a sharp push, they slide apart. What is the relationship between their velocities?

    Solution: Newton's Third Law tells us that the forces exerted have to be the same magnitude no matter how hard the boy or man think they are pushing. Therefore, the acceleration of the man and boy will be related as follows: mbab = mmam ==> ab = mmam/ mb. So the boy's acceleration is twice that of the man's since his mass is half that of the man's. We do not know over what distance or time this push is occuring, but we know that the time over which its applied should be the same, so we expect vb = mmvm/ mb = 2 vm.

    Using conservation of energy, we could get this result immediately since the system that includes the boy and the man has no external forces other than gravity and the normal force of the ice acting on them. Since these forces act only in the vertical direction, they cannot change the horizontal motion and hence there are no external horizontal forces, so we need mbvb = -mmvm ==> vb = -2vm. Where the minus sign assures us that the direction of the velocities are opposite.

  2. A bullet with mass m1 is moving with velocity v1. It strikes a wooden block of mass m2 which is sitting motionless on a frictionless surface. If the bullet lodges inside the block after impact, find the velocity, Vf of the block and bullet just after the impact.

    Solution: Again, we can use Newton's Laws, in principle, but the action of the bullet interacting with the block takes place over such a short period of time that it is impractical to talk about acceleration of the block and bullet during the interaction. Momentum conservation allows a much easier solution. The system is taken to include both the bullet and the block. The initial momentum of the system is therefore p0 = m1v1 since only the bullet is moving initially. The final momentum has both the bullet and the block moving as one body, so pf = (m1 + m2) Vf. Since there are no external forces (any interaction between the bullet and block is internal to the system we defined), we must have p0 = pf and we can solve for the final velocity to get Vf = m1v1/ (m1 + m2)

  3. Below is a movie of a car and truck crashing together.

    Is the force the car exerts on the truck
    1. greater than the force the truck exerts on the car
    2. less than the force the truck exerts on the car
    3. equal to the force the truck exerts on the car
    4. impossible to determine without knowing the speeds of the car and the truck

    Of course you know the answer is supposed to be c., but is that intuitive? To be honest, probably not. However, Newton's Second and Third Laws and the concept of momentum conservation are consistent with each other and with our intuitive notion although it may not seem so at first glance. First, let's set up the problem with some numbers so we can try it out. The picture below shows the crash. Let's assume m1 = 1500 kg, m2 = 20,000 kg, u1 = 8m/s, and u2 = 5 m/s.

    We can see that the solution here for the final velocity can be derived from momentum conservation

    p0 = pf
    p1 + p2 = (m1 + m2)V
    (m1u1 - m2u2) = (m1 + m2)V ==>
    V = (m1u1 - m2u2)/ (m1 + m2)
    = [(1500 kg)( 8 m/s) - (20,000 kg)( 5 m/s)]/( 1500 kg + 20,000 kg)
    V = -4.09 m/s
    So we can find the forces acting on the driver in each car by noting that the change in velocity of the car or truck is the same as the change in velocity of the driver. Let's assume the crash takes place over a time period of about 0.2 s. The acceleration of the truck is a2 = vf - v0/tcrash = (V - u2)/ tcrash = (-4.09 m/s - -5 m/s)/0.2 s = +4.6 m/s2. The driver experiences an acceleration of (V - u1)/tcrash = (-4.09 m/s - 8 m/s)/0.2 s = -60.5 m/s2. If the truck driver and car driver are about the same mass, then the force on the car driver is about 13 times as great as the force on the truck driver! Note that the magnitude of the forces are equal since, Ftruck = (20,000 kg)(4.6 m/s2) = 92,000 Nt and Fcar = (1500 kg) (60.5 m/s2) = 91,000 Nt with the difference being round off error. Note also that since the accelerations of the car and truck are opposite signed, the directions of the forces on both are in opposite directions just as Newton's Third Law demands.

To see an interactive demonstration of 1D momentum conservation in action, turn to the interactive textbook by using this Java Applet.

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