Newtonian Collision

We consider Newtonian -- as opposed to relativistic -- mechanics. By considering point-masses, we may ignore the angular-momentum. Hence, the only conservation laws applicable to the collision of a set of objects are (1) that of mass and (2) that of the linear-momentum vector, which is defined as the product of the mass of an object by its velocity. That is,

sum, over all of the objects, of (m-before(j) v-before(j)) = sum, over all of the objects, of (m-after(j) v-after(j)).

By transposition, we obtain

0 = sum, over all of the objects, of (-m-before(j) v-before(j)) + sum, over all of the objects, of (m-after(j) v-after(j)).

The conservation of mass may be written as

sum, over all of the objects, of (m-before(j)) = sum, over all of the objects, of (m-after(j)).

Again, by transposition, we obtain

0 = sum, over all of the objects, of (-m-before(j)) + sum, over all of the objects, of (m-after(j)).

Please observe that the mass of each object need not be conserved: There may be mass-transfer among the objects. Furthermore, since the objects may stick together or break apart, the amount of objects need not be conserved. Ordinarily, the kinetic-energy is not conserved. Usually, some kinetic-energy *must* be dissipated -- as heat --; as a consequence of the two conservation laws. However, occasionally, some internal potential-energy is converted to out-going kinetic-energy -- or incoming kinetic-energy becomes bound as internal potential-energy. Such inter-conversion of potential and kinetic energy is common in chemical (atomic) or physical (nuclear) reactions; but, may occur in mechanical devices, as well. Hence, the kinetic-energy balance may be either positive or negative.

The foregoing two conservation laws may be utilized to solve for the mass and velocity of any one of the incoming or outgoing objects, given these attributes of each of the other objects involved in a collision. We provide an Excel spreadsheet for this calculation. mailto::collision@rism.com Please ask for the collision spreadsheet. In this spreadsheet, the cells into which data is to be entered are bold -- one row for each object. In each row, the first column is for the mass of an object -- enter a negative value for an incoming or positive for an outgoing object. The second column is for the speed of the same object. The third column is for the angle (in degrees), measured counter-clockwise from the negative ordinate (y-axis) to the head of the velocity-vector. Enter data for all, except one, of the objects -- incoming or outgoing. There are nine rows provided for the entry of the data. Enter a zero for the mass of any excess rows. The spreadsheet will compute -- by means of the two conservation laws -- the data -- mass and velocity -- for the single missing object. It will be displayed in the top row. Then, as a check, that object is re-computed in the last row. In a partial extra row, it is demonstrated that the sum of the masses is zero and the sum of each Cartestian-component of the momenta likewise is zero. The sum of the kinetic-energy is computed and displayed. It is the difference of the outgoing less the incoming kinetic-energy. The spreadsheet displays all of the intermediate steps of each of the original computation and of the check computation. These intermediate steps include the Cartesian components of velocity and of momentum.

There is a second block with two angles -- thus, spherical-coordinates -- introduced to provide a three-dimensional calculation. The data to be entered into each row is the mass and the velocity -- radius (speed), always positive; longitude, west is positive, east is negative; and latitude, north is positive, south is negative. Thus, we have a right-handed coordinate system.

Geometrical interpretation

Since mass is a scalar, the law of conservation of mass is that the sum of the masses -- negative for incoming, positive for outgoing -- is zero. There is nothing more to say regarding this law. The law of conservation of linear-momentum states that the sum of these momenta is zero. Since the summation is commutative and associative, each of the sums is invariant under permutation of the order of the terms. Since linear-momentum is a vector, the law of conservation of linear-momentum is that these momenta constitute a directed-graph in the form of a polygon. Since this polygon is closed, if the polygon is planar; then the sum of its external angles must be an integral multiple of (2 pi). If there is only one unknown momentum; then the sum of all the known momenta is a fixed vector. The unknown momentum must be the negative of this vector. If there are two partly-known momenta; then the sum of all the fully-known momenta likewise is a fixed vector. Now, the two partly-known momenta, together with this vector, comprise a triangle -- of necessity a *plane* triangle. To be able to obtain a unique solution; we need:

something to specify the one plane out of the family through the vector.
appropriate angles or lengths of the partly-known momenta -- whatever is required. Please see the summary for the strategy to solve this triangle.