122. Next suppose a particle of mass m and moving with velocity v to impinge directly on a fixed plane, the coefficient of elasticity being e. In this case the particle will rebound from the plane with velocity ev. The impulse exerted by the plane on the particle must therefore destroy the velocity v with which it is moving, and generate a velocity ev in the opposite direction. Hence the whole change of the particle's velocity is numerically equal to v +ev, and the whole change in its momentum to mv (1+e). Hence the measure of the impulsive pressure between the particle and plane is mv (1+e).

If the plane be perfectly elastic we have e equal to 1, and the impulse is measured by 2mv.

123. Next suppose the plane on which the particle impinges to be moving with a velocity V in the same direction as the particle before impact. Then the velocity of the particle relative to the plane is v- V before impact, and after impact it is e (v - V), but in the opposite direction. If we adopt the usual convention with respect to sign, we may denote the velocity of the particle relative to the plane after impact bye (V). The whole change of the velocity of the particle is (v − V) (1 + e), and the change of its momentum is m(v-V) (1+ e), without regard to sign. This last expression is the measure of the whole impulsive pressure between the particle and the plane.

Since the velocity of the particle before impact was v, and the change of velocity produced by the impact is (v-V) (1 + e) in a direction opposite to that of v, it follows that the velocity of the particle after impact is v (v - V)(1 + e), and is in the same direction as before, or in the opposite direction, according as the sign of this expression is positive or negative.

If e be zero or the plane inelastic, the velocity after impact is V, as of course it should be. If e be unity or the elasticity perfect, the velocity after impact is 2V - v, and this is in the same direction as before, or in the opposite direction, according as V is greater or less than

V

The above investigations are true for spherical balls as well as particles, provided that their centres of gravity coincide with their centres of figure, and that they have no motion of rotation, unless their surfaces be perfectly smooth. If the spheres be rough and they have a motion of rotation, or if their centres of gravity do not coincide with their centres of figure, the problem becomes much more complicated, and requires the principles of Rigid Dynamics, to which subject all problems on the motion of spheres or of any rigid bodies of finite dimensions properly belong.

124. Suppose a particle of mass m moving with velocity v along the line QP to impinge obliquely at P, upon the smooth plane AB, the coefficient of elasticity between the particle and plane being e. It is required to find the motion of the particle immediately after impact, and the impulsive pressure on the plane. Let PT be the direction

of motion after impact, PN the normal at P to the plane. Let the angle QPN be denoted by a, and the angle TPN by 0. Then the velocity of the particle before impact may be resolved into two components, viz.-v sin a along the plane, and v cosa perpendicular to the plane. Since the plane is smooth it is only the latter component which is altered by the impact, and this is replaced by a velocity ev cos a in the opposite direction. The velocity after impact is therefore the resultant of the two velocities, v sin a along

the plane and ev cos a perpendicular to the plane, and is therefore numerically represented by

v sina + e2 cos2 a ;

and if PT be the direction of motion after impact,

If the elasticity be perfect, or e equal to 1, we have equal to a, or the angle of reflexion equal to the angle of incidence.

It remains to determine the impulsive pressure between the particle and the plane. The component of the velocity of the particle perpendicular to the plane before impact is v cos a, and after impact it is ev cos a in the opposite direction. The whole change of velocity produced in the particle by the impulse is therefore v cos a (1+e), and the change of momentum is measured by my cos a (1 + e), which is therefore the measure of the impulse.

125. If a particle impinge obliquely on a rough plane whose coefficient of friction is u, then besides the impulsive pressure there will be an impulsive friction called into play, proportional to the pressure, and such that its effect is to diminish the velocity of the particle parallel to the plane.

Referring to the figure of the preceding Article, since the velocity of the particle perpendicular to the plane is reversed in direction and diminished in the ratio of 1 to e by the impact, it follows as before that the measure of the impulsive pressure is my cos a (1+e). Now the impulsive friction called into play is μ times this, that is

μ. mv cos a (1+e),

and diminishes the particle's velocity parallel to the plane. Hence the velocity, parallel to the plane, of the particle after impact is v sin x − μv cos a (1+e). Therefore, if, as in the

G. D.

10

preceding example, e denote the angle the direction of motion after reflexion makes with the normal, we have

which gives the direction of motion after impact, and the velocity can be found as in the previous case.

If the surface upon which the particle impinges be curved, the effect of the impact is the same as if the surface were replaced by its tangent plane at the point at which the particle strikes it. Thus in the preceding Article, if PN be the normal at P to the surface AB, the whole of the reasoning will be equally true whether the surface AB be plane or curved.

126. Hitherto we have considered the obstacle against which the moving particle impinges to be either fixed, or made to move in such a manner that its velocity is unaffected by the collision. We proceed now to the consideration of the collision of two particles, or two spheres, each free to move. It will be seen that in order that two spheres may impinge directly upon one another, the line joining their centres at the time of impact must be the line of motion of either relative to the other.

Suppose two particles, or spheres, whose masses are respectively M and m, to be moving in the same direction with velocities V and v respectively, of which V is the greater, and to impinge directly upon each other; it is required to find the motion of each after impact, the coefficient of elasticity being e.

Let V1, v, be their respective velocities after impact. Then whatever be the impulsive action between them at the moment of impact, the impulse on the first must be equal and opposite to that exerted on the second. Hence the momentum generated in the first must be numerically equal but opposite in direction to that generated in the

second. Hence, whatever momentum reckoned in the positive direction may be lost by the first, the same amount must be gained by the second, and vice versa; consequently the Algebraical sum of the momenta of the two balls must be the same after impact as before. Therefore

MV1+mv1 = MV+mv..

1

1

.(I).

Again, the relative velocity after impact is to that before impact ase to 1, and the velocity of the first relative to the second before impact is V-v, and after impact it is V-v1. Therefore

These two equations, (I) and (II), determine V, and v ̧.

1

Substituting in (I) we get

(M+m) v1 = MV + mv + eM (V − v) ;

MV+mv+ eM (V-v)

.(III).

(IV).

If the balls be inelastic e is zero, and from equation (II) we have V, equal to v1, or they proceed with a common velocity, in other words they do not separate. Substituting in equation (I) we have, in this particular case,

This of course follows immediately from the general expressions given above for V, and v1, by putting e equal to 0

in them.