Example iii. A small particle of mass m lbs. moving with u velos impinges directly on a large mass of M lbs. at rest; find their subsequent motion, the coefficient of restitution being e.

m

Now suppose M to be so much larger than m, that the ratio M

is too small to be observed, and may be considered to be nothing; then

This is the case when a ball of mass in lbs. comes into collision with the Earth of mass M lbs. Hence we say that

73. When an elastic ball impinges against any surface fixed to the Earth, its velocity after impact is - e times that before impact.

Example. A particle let fall from a height of 16 ft., impinges on a horizontal pavement, the coefficient of restitution being e; it again falls and rebounds; and so on; find when the particle will come to rest.

The particle has an acceleration 32 celos downwards.

Let its velocity on striking the ground be v velos, and on reaching the ground after each rebound, v1 velos, v2 velos, etc.; then we have = 2as = 2 × 32 × 16; or, v = 32;

V1 = ev; v2 = ev1 = e2v; V3 = ev2=e3v; and so on.

To ascertain the intervals, we observe that to produce 32 velos, 32 celos require 1 sec.; to produce e × 32 velos, 32 celos require e times

1 sec., and so on;

In the successive bounds ev velos, e2v velos,... are in turn first destroyed and then reproduced by the acceleration due to gravity.

Hence the sum of the intervals occupied by the bounds are

{ze + 2e2 + 2e3 + ..... .} seconds.

2e

Therefore the particle comes to rest in

Ite

seconds, from the instant at which it was let fall.

I-e

I-e

EXAMPLES. XXI.

The coefficient of restitution of an elastic ball is sometimes called its elasticity.

1. A particle let fall from a height of 16 feet, impinges on a horizontal pavement; find the interval occupied by the first rebound, the coefficient of restitution being .

2. A particle let fall from a height of 48 ft., impinges on a horizontal pavement; find the elasticity that in its first rebound it may rise to a height of 32 ft.

3. A particle let fall from a height of h ft., impinges on a horizontal pavement, the coefficient of restitution being e; find

(i) the height of the first rebound,

(ii)

the interval between the first two impacts,

(iii) the time in which the particle comes to rest.

4. An elastic ball falls from a height h ft. to a horizontal plane and then rebounds; falls again and rebounds; and so on; find the suni of the vertical distances passed over.

5. Supposing that in a direct impact between two particles of elasticity e, we consider the impulse between the particles to be divided into two parts, the first part being that which causes the two particles to be relatively at rest; shew from Art 71 that these two parts of the impulse are in the ratio of 1 to e.

6. The 110 ton gun projects its shot of 1800 lbs. with a muzzle velocity of 2100 velos; what will be the initial velocity of the recoil of the gun, supposing the shot projected horizontally, and the gun to be placed on a smooth horizontal plane?

7. Two masses m lbs. and m' lbs. of elasticity e impinge directly with u velos and u' velos respectively; prove that if v velos and v velos are their velocities after impact, then

* CHAPTER V.

CHANGE OF UNITS.

74. Our units have been, a foot, a second, a pound; from which are derived the units, a velo, a celo, a poundal, a pulse.

The velocity chosen for the unit velocity is always that of a point which passes uniformly over the unit distance in the unit interval.

The acceleration chosen for the unit acceleration is always that of a point whose velocity is increased uniformly by the unit velocity in the unit interval.

The force chosen for the unit force is that which when acting on the unit mass produces in it unit acceleration.

75. We give some examples involving a change of units.

Example i. When the unit acceleration is 32 feet per sec. per sec. and the unit velocity is 10 feet per sec., find the unit distance and the unit interval.

The unit acceleration is 10 velos per x seconds;

Thus the unit distance is ft. the unit interval secs.

Example ii. When the area of a field of 10 acres is the unit area, and the acceleration due to gravity the unit acceleration, find the unit interval.

A field of 10 acres contains 48400 sq. yds.,

that is, (220)2 sq. yds. ;

hence, the unit distance is 220 yds., or 660 ft.

Let the unit interval be x seconds,

The unit velocity is 660 ft. per x seconds;

Example iii. Given, that the unit of acceleration is a celos, the unit of velocity, u velos; find the units of time and distance.

1. The unit distance being a yard and the unit interval a minute, express the unit velocity in velos and the unit acceleration in celos.

2. The unit distance being a mile and the unit interval an hour, express the unit velocity in velos and the unit acceleration in celos.

3. The unit distance being a yds. and the unit interval 6 seconds, express the unit velocity in velos and the unit acceleration in celos.

4. The unit velocity being 60 miles an hour, and the unit distance I yd., find the unit interval.

5. The unit velocity being 100 yds. per minute and the unit interval 5 secs., find the unit distance.

6. The unit acceleration being the double of that due to gravity and the unit distance 1 mile, find the unit velocity.

7. The unit velocity being 40 ft. in II secs. and the unit acceleration 3 yds. per minute per minute, find the units of time and distance.

8. The unit mass being I cwt. and the unit force the weight of 7 lbs., the unit velocity 10 velos, find the unit distance.

9. The unit force being the weight of 1 lb., and the unit interval and unit distance being 1 sec. and 1 ft. respectively, find the unit mass.

10. The unit mass being 1 ton, the units of interval and of distance an hour and a mile, find the unit force.

11. Find the measure of the acceleration due to gravity when the unit distance is a metre, say 3°28 ft., and the unit interval a second.

12. If f is the measure of an acceleration when m secs. and n ft. are the unit interval and unit distance respectively, express the acceleration in celos.

13. If fand F be the measures of the same acceleration when the unit interval and unit time are t secs. s ft. and 7 secs. and σ ft., respectively, prove that

+2

F=

14. Taking the centimetre (3ʊ of a metre=0328...ft.) as the unit distance, the second as the unit interval and the gram (='0022... lbs.) as the unit mass, find the unit velocity in velos, the unit acceleration in celos, and the unit force in poundals.

N. B. This system of units is called the C.G.S. [Centimetre, Gramme, Second] system, and the unit force is called a dyne.

NOTE. We may remark that in Statics the unit force generally chosen is the weight of 1 lb.; in this case, since we ought to take as the unit-mass the mass in which unit force produces unit acceleration, our unit mass must be glbs.; the disadvantage of this is that g has different values for different latitudes.

L. D.

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