The only laws of classical mechanics are Newton’s laws. All other “laws” or general principles are deduced from Newton’s laws. The physicist is especially interested in statements about the behavior of systems which do not depend on the detailed nature of the forces which are involved. The best-known example of such a statement is the Principle of Conservation of Momentum:
To understand this statement we must, of course, first define “momentum.” If a particle of mass has velocity , the momentum of the particle (usually denoted by the vector ) is defined by
The momentum of a system of particles is defined as the sum of the momenta of the individual particles It is true that for any system
where is the total external force acting on the system.
To derive (2) we added up the force equations for all the particles in the system; the internal forces canceled out in pairs as a consequence of Newton’s third law.
If we have which is the Principle of Conservation of Momentum.
A body of mass 1 kg and a body of mass 2 kg collide on a smooth horizontal surface. Before the collision the 1 kg body has velocity 3 m/s directed due north and the 2 kg body has velocity 5 m/s directed northeast (i.e., east of due north). The two bodies stick together, forming a body of mass 3 kg. Find the magnitude and direction of the velocity of the 3 kg body.
We orient our axes so that the -axis points east, the -axis points north, and the -axis is perpendicular to the surface. We define our system as the two bodies. Because the surface is smooth there is no external force in the - or -direction (note that there are internal forces in the system since the two bodies exert forces on each other during the time when they are colliding). Gravity exerts a force in the -direction on each body, but the normal force exerted by the surface is equal in magnitude and opposite in direction to the gravitational force. Thus there is no net external force on the system and we can apply the Principle of Conservation of Momentum which states
Note that (3) is a vector equation, equivalent to the three equations
A common error is to think that (3) implies where is the magnitude of the velocity vector . This does not follow from (3) and is not, in general, true.
In the present example (4c) is uninteresting, merely stating that . If we call the unknown final velocity vector , with components and , then (4a) and (4b) imply Thus we find m/s, m/s. The speed of the 3 kg body is m/s. The velocity vector is directed north of east .
Consider the same two bodies as in the previous example, colliding with the same initial velocities. However, they do not stick together. After the collision the 2 kg body has velocity 4 m/s directed east of due north. Find the velocity of the 1 kg body.
Calling the unknown velocity , we find from (4a) and (4b) and thus m/s, m/s.
Note that when the two bodies stick together (this is called a completely inelastic collision) the Principle of Conservation of Momentum uniquely determines the final velocity. When the two bodies do not stick together, the Principle of Conservation of Momentum does not uniquely determine the two final velocity vectors; if one of the final velocity vectors is given, as in the present example, then the other is determined by Conservation of Momentum. More generally, a vector in the - plane is specified by two numbers (e.g., the two components of the vector, or the length of the vector and the angle it makes with the -axis); thus, four numbers are required to specify the two final velocity vectors. Conservation of - and -momenta imposes two conditions ((4a) and (4b)) which must be satisfied by these four numbers. Accordingly, the final state will be determined if any two of these numbers (e.g., the directions of the two final velocities) are specified. The multiplicity of possible final states corresponds to the fact that the bodies have shapes (and the contact may take place at various points on their surfaces) and various degrees of “squishiness” (e.g., two steel balls versus two old tennis balls).
We now consider a skyrocket fired vertically from the ground. At the instant when it is rising with velocity 100 m/s it explodes into three fragments of equal mass. Just after the explosion one fragment has velocity 50 m/s directed vertically downward and another fragment has velocity 75 m/s directed horizontally. Find the velocity vector of the third fragment just after the explosion.
Digression (fairly important)
Most students will immediately solve this problem by equating the momentum of the skyrocket just before the explosion to the sum of the momenta of the fragments just after the explosion. This procedure is correct but requires some discussion since the system is not free of external forces; gravity acts on the skyrocket and also on the fragments. How do we justify neglecting the effect of gravity? If we integrate both sides of (2) with respect to time from to , where and are arbitrary, we obtain
Let us choose as a time just before the explosion and just after the explosion. In this problem where is the total mass of the system and is a unit vector directed vertically upward. Then the left side of (5) becomes . An “ideal” explosion is one in which the skyrocket flies apart in an infinitesimal time; that is, . In this case the left side of (5) vanishes or is negligible, and thus the momentum just after the explosion is equal to the momentum just before the explosion.
Consider the problem just stated of an exploding skyrocket.
If we define as a unit vector parallel to the velocity of the horizontally moving fragment, momentum conservation requires where is the velocity of the third fragment. Thus we find .
Invent a problem in which you know the height at which the explosion occurred and you also know the locations (relative to the point directly under the explosion) of the points where the fragments landed and the times (relative to the time of the explosion) when the fragments landed. From this information you can calculate the velocity of the skyrocket just before the explosion. The solution will involve Conservation of Momentum as well as kinematic equations.
Two children, one with a mass of 30 kg and the other with a mass of 45 kg, are on a frozen pond (the ice is assumed perfectly smooth). Initially they are both at rest, 30 meters apart, each holding an end of a 30 meter weightless rope. Both children then start pulling on the rope until they collide. Where will the collision occur? (The method of solution should make it clear that the location of the collision point does not depend on the details of how they pull on the rope.)
We define our system to consist of the two children plus the rope. Since there is no external force on the system, we have where are the masses and velocities of the two children. Since and are initially zero, the value of the constant is zero and therefore where and are the positions of the children relative to a fixed origin. If the initial positions of the children are and and their position when they collide is , then
It is convenient (but not necessary) to take our origin at the initial position of child #1 so that . Then which states that if the initial distance between the children is , the collision occurs on the line between the initial positions at a distance from the initial position of #1. In the present example, the collision occurs 18 meters from the initial position of the less massive child.
The discussion of the preceding example illustrates the usefulness of the concept of center of mass. In general, if a system consists of particles enumerated by an index , located at positions , the location of the center of mass is defined by the equation
In words, the position vector of the center of mass is a weighted average of the position vectors of the individual particles, each particle being weighted by the ratio of its mass to the total mass.
If are the Cartesian coordinates of the center of mass, then (6) is equivalent to the three equations
If we rewrite (6) as where and differentiate both sides with respect to time, we obtain where . Differentiating both sides with respect to time again we obtain
Combining this result with , we obtain the very important result
which states that the motion of the center of mass of a system is identical with the motion of a particle of mass (where is the total mass of the system) subjected to a force (where is the total external force on the system). Therefore if you throw a chair into the air with an arbitrary amount of spin, the center of mass (generally abbreviated as CM) of the chair will move (we neglect air friction here) in a parabola.
In Example 4 the external force is zero and therefore and constant. Since initially it follows that at all times and is constant. Thus the center of mass never moves and the collision must take place at the center of mass of the two initial positions.
Frequently one is interested in the motion of a solid body of finite (i.e., not infinitesimal) size. The location of the center of mass is often obvious from symmetry considerations (e.g., the CM of a uniform rod is at its midpoint), but in other cases some calculation is necessary. Typically, we conceptually subdivide the body into infinitesimal pieces and perform the sums in (7a), (7b), and (7c) by means of integral calculus. As an example, let us calculate the location of the CM of a solid hemisphere of uniform density. For convenience we take our - and -axes in the flat face with the origin at the center of that face. From simple considerations of symmetry we see that the CM is on the -axis; that is, . To calculate we must convert the sums in (7c) into integrals. This can be done fairly simply in either of two ways. One way is to subdivide the body into thin slices by means of planes perpendicular to the -axis. A plane of constant intersects the hemisphere in a circle of radius where is the radius of the hemisphere. Thus the volume of the slice contained between the plane at height and the plane at height is and the mass of this slice is where is the mass density (mass per unit volume). Converting the sums in (7c) into integrals, we find