332

NEW EXPERIMENTS

TRACT 34.

have on the truth of our theorem, or on the velocity of theball, as computed from it.

In order to this, let the notation em-ployed in Art. 21 be supposed here; andlet abc he a side-view of the pendulum,moved out of the vertical position a d, bythe perpendicular blow of the ball againstthe point d or c. Alsolet x = d c the space moved by the pointof impact c,

Z = c B the depth penetrated by the ball,y = velocity of the ball at b,ii = velocity of the point c of the pendulum, andR = the uniform resisting force of the wood.

Then is j the retarding force of the ball, which is con-stant. Again, as the motion of the pendulum arises fromthe resisting force R of the wood, Ri will be its momentum ;and as the sum of the forces in the pendulum was found tobe = p g o, the accelerating force of the point c will be

—, which force is constant also. But, in the action ofpgo

forces that are constant, the time t is equal to the velocity'divided by the force, and by 2 h or 2 x 16‘09 feet, and thespace is equal to the square of the velocity divided by theforce and by r 4 h; consequently

4 _ p gou ^ _ p g o uu

1 ~ WiiR* x ~ 4 him*

. — b v — h v v

~ 2 hit’ x + z —

or by correc. t = x (v - v), a- + z = ^ x (w 1 - v ! ).

The two values of the time t being equated, we obtainpgou = bii(y — v), or pgou + biiv — biiv.

And when v = u, or the action of the ball on the pendulumceases, this equation becomes pgou = biiu biiv, and henceM = p g 0 ‘+' l ' bii the greatest velocity of the point c, at theinstant when the ball has penetrated to the greatest depth,and ceases to urge the pendulum farther. So that thisvelocity is the same, whatever the resisting force of the