Multiplying these last n equations together and rejecting factors common to both sides, we have

tan an= e" tan a.

When n∞, we see from this equation that tan a1 = 0, and therefore, by (1), that the terminal velocity of the ball is V cos a.

Again, r denoting the radius of the hoop, the time through the chord between the nth and (n+1)th impact, is equal to

hence, the whole time between the first and (n+1)th impact is less than

and therefore the ball arrives at its terminal velocity in a time less than

(3) A particle, of given elasticity e, is projected along a horizontal plane, from the middle point of one of the sides of an isosceles right-angled triangle, so as, after reflection at the hypotenuse and remaining side, to return to the same point; to prove that the cotangents of the angles of reflection are e+ 1 and e+2 respectively.

Let ABC, fig. (122), be the triangle, A being the right angle: let PQRP be the course of the particle: let

Again, by the geometry,

PQ sin (0 +0') = PR sin (4 + 4'),

whence, AP being equal to BP,

cot ' + 2 cot = cot & (1 + cot 0) + 1,

(2).

(e2 + 2e+1) cot 0

and therefore, by (2),

tan 0 = e + 1,

tan = e +2.

(4) A locomotive is travelling at the rate of 60 miles an hour: shew that, if it were suddenly stopped, the violence of the concussion would be nearly as great as if it fell from a height of 40 yards.

(5) A ball, of given elasticity, impinges against a circular arc: determine the point of impact, in order that it may rebound at right angles to its direction of incidence.

Let A be the point of the arc, the radius of which is parallel to the direction of incidence, and let P be the point of impact: then, denoting the radius and e the elasticity,

AP=r tan1 (e1).

(6) ABC is a horizontal triangle; D, E, F, the points where the circle inscribed in it meets the sides BC, CA, AB, respectively prove that, if a ball, of elasticity e, be projected from D, so as to strike AC in E and then rebound to F, AE= e. CE: if the ball return to D, prove that also AB=e.AC.

(7) A ball is projected from the middle point of one side of a billiard-table, so as to strike in succession one of the sides adjacent to it, the side opposite to it, and a ball placed in the centre of the table; shew that, if a and b be the lengths of the sides of the table, and e the elasticity of the ball, the inclination of the direction of projection to the side a of the table, from which it is projected, must be

(8) The tangents of the angles of a triangle ABC are in geometrical progression, tan B being the mean proportional: a ball is projected in a direction parallel to the side CB, so as to strike the sides AB, BC, successively: shew that, if its course after the first impact be parallel to AC, its course after the second will be parallel to BA; and that, if e be the modulus of elasticity,

sec B = e1+e".

(9) A heavy particle, of given elasticity, impinges on a fixed rough plane in a given direction: determine the direction of the motion of the particle after impact, the impulsive friction being proportional to the impulsive pressure between the particle and plane.

If a, a', denote the angles of incidence and reflection, e the elasticity of the particle, and μ the coefficient of friction,

(10) Shew that it is possible to project a ball on a smooth billiard-table from a given point in an infinite number of directions, so as, after striking all the sides in order once or oftener, to hit another given point; but that this number is limited, if it have to return to the point from which it was projected.

CHAPTER V.

COLLISION OF SPHERES.

SECT. 1. Direct Collision.

(1) Two spheres A and B, moving with equal velocities in opposite directions, collide directly: supposing that, after collision, A is at rest, and that B moves back with the velocity which it had before collision, to compare the masses of A and B.

Let u denote the velocity of either sphere before collision, and let m, m', represent their respective masses: then, the algebraic sum of the momenta being the same after collision as before, we have

(2) A ball A strikes a ball B, which is at rest, directly, and, after collision, their velocities are equal and opposite: to find the mutual elasticity of the balls, supposing the mass of B to be times that of A.

Let m be the mass of A, u its velocity before, and v its velocity, in the opposite direction, after collision: the mass of B is accordingly Am, and its velocity v. Let e be the common elasticity of the balls.

Then, the algebraic sum of the momenta being the same before and after collision,

Again, by the equation of elasticity, u being the relative velocity before and 2v after collision,

COR. Since e is less than unity, we must have 2<λ − 1, or >3: thus we see that the problem is impossible unless B's mass is at least three times as great as that of A.

(3) A number of balls, of the same given elasticity, A1, ▲, А................ are placed in a straight line: A, is projected with a given velocity so as to impinge on A,; A, then impinges on A,, and so on; to find the relation between the masses of the balls, in order that each of them may be at rest after impinging on the next; and to find the velocity of the nth ball after its collision with the (n-1)th.

Let u1, u2, u,,... denote the velocities with which A ̧, А„, A,,... impinge respectively on A,, A., 4.,...., and suppose the masses of the balls to be represented by A, A„, A..........

Then, since the momentum of A before striking A, must be the same as that of A, after the collision,

Anun= An-1Un-1

Also, by the law of elasticity,

Un = eun-1••

(1).

.(2).

From (2) we see that u, u2, u,,... form a geometrical progression of which e is the common ratio, u, being therefore equal to e1-1u1.

Again, from (1) and (2), we see that

hence the masses of the balls are in geometrical progression, the