P must obviously be applied perpendicularly to the rod, to be a minimum. xiii. A bucket and a counterpoise, connected by a string passing over a pulley, just balance one another, and an elastic ball is dropped into the centre of the bucket from a distance h above it; find the time that elapses before the ball ceases to rebound, and prove that the whole descent of 4mh where m, e 2M+m (1 − e)2' the bucket during this interval is M are the masses of the ball and the bucket, and e is the coefficient of restitution. Letv be the velocity of the ball just before the first impact. The relative velocity after the first impact is ev, and the relative acceleration is g, since the acceleration of the bucket is zero. Therefore the time during which the ball rebounds is 3 Let V, V, V,... be the velocities of the bucket during the intervals between the first, second, third, impacts. xiv. A particle is projected from the foot of an inclined plane and returns to the point of projection after several rebounds one of which is perpendicular to the inclined plane: if it takes more leaps in coming down than in going up, 2 √(1-e)-2 (1-e), where a is the prove that cota cot 0 = inclination of the plane, (1 − e) er the angle between the direction of projection and the plane, and e the coefficient of restitution. What is the condition that it may be possible to project the particle so that one of its impacts may be perpendicular to the plane? Let n be the number of leaps going up and nr the number coming down the plane, and v the velocity of projection. Considering the motion perpendicular to the plane, the 2v sin 2ev sin times occupied by the leaps are and g cosa g cosa considering the motion parallel to the plane, since the nth impact is perpendicular to the plane, the time occupied by v cose 9 sina' the first n leaps is equal to the time of going up; therefore and the time of coming down is the negative sign being taken to the radical because e<1; therefore The condition that it may be possible to project the particle so that one of its impacts may be perpendicular to the plane is that the equation (1) should give an integral value of n for an integral value of r. 15. If in BA, CA two sides of a triangle ABC two points D, E be taken respectively, such that BA: AC:: EA: AD and G the middle point of DE be joined to A, and if BH, CK be constructed in the same way as AG, shew that AG, BH, CK intersect in a fixed point 0. Prove also that if from O perpendiculars be drawn to the sides of the triangle the sum of their squares is less than the sum of the squares of the perpendiculars from any other point. If GQ, GR (fig. 32) be drawn perpendicular to CA, AB, GQ: GR:: AD: AE:: CA: AB, and therefore at O OL: OM: ON:: BC: CA: AB, if OL, OM, ON be the perpendiculars on the sides of the triangle ABC. (DE is parallel to the tangent at A to the circumscribing circle and may be called an anti-parallel to BC, and O may be called the centre of anti-parallels.) If x, y, z be the perpendiculars on the sides of the triangle ABC and a, b, c the sides, (a3 +b2 + c2) (x2 + y2 + z2) = (ax+by+cz)2 + (bz − cy)2 + (cx — az)2 + (ay — bx)”, a minimum when xyz::a:b: c, since ax+by+cz is constant, being double the area of the triangle ABC. xvi. A body is describing an ellipse about a centre of force in the focus, and when its radius vector is half the latus rectum it receives a blow which causes it to move towards the other focus with a momentum equal to that of the blow; find the position of the axis of the new orbit 1- e2 where e is the and shew that its eccentricity is eccentricity of the original orbit. 2e Let LQ, LR (fig. 33) represent the velocities of the body before and after the blow, then QR represents the velocity generated by the blow, and therefore QR LR, and QR is parallel to LS. The blow is therefore towards the centre of force and the latus-rectum is unaltered in magnitude. Hence SL being a radius vector of the new orbit equal to half the latus-rectum coincides with the latus-rectum in position. LS' being the tangent to the new orbit, S' is the foot of the directrix, and the eccentricity is SL 1- e2 = SS' 2e 17. A cone floats in liquid which fills a fixed conical shell: both the cone and the shell have their axes vertical and vertices downwards: the vertical angles of the cone and shell are equal and the axis of the shell is twice that of the cone. If the cone be pressed down until its vertex very nearly reaches the vertex of the shell, so that some. of the liquid overflows, and then released, it is found that the cone rises until it is just wholly out of the liquid and then begins to fall. Prove that the densities of the cone and the liquid are in the ratio 45 - 21 (7) : 4 V(7), the free surface of the liquid being supposed to remain horizontal throughout the motion. In the two positions in which the velocity of the cone is zero, the heights of the centre of gravity of the cone and liquid are equal. h Let ρ be the density of the liquid, o of the cone, the height of the cone; in the second position the height of the surface of the liquid is (7) h; therefore 2.2h. tan2a.p-3.h. tan'a (po) 3 Th3 xviii. A conical cup of uniform thickness floats in water, with its axis inclined to the vertical at an angle and the fraction m of the axis below the level of the surface of the water; prove that cos 0= vertical angle of the cone. 8 sin' a where 2a is the Let G (fig. 34) be the centre of inertia of the conical cup, H of the water displaced; then GH is vertical. Let AD=h, then AE=mh, AG=3h. 、 FN=3h sin 0 = {FR = 3 (FP+ FQ) 19. B stands in front of a plane vertical mirror, find the position that A must take in order to see B's profile directly and B's full face by reflection at the mirror; determine in what cases no such position exists. If B without changing his station turn on a vertical axis, prove that the locus of A will be a rectangular hyperbola whose vertices are B and the image of B in the mirror. If B' be the image of B (fig. 35), then if C be the point of the mirror towards which B is looking, the angle ABC |