Physics
posted by LingWarm .
i really dnt understand wen i revise it nw so cn u xplain the solutions 2 me. Thnks.[take g=10 ms^2] Two particles P & Q, of equal mass m, are attached to the ends of a light inextensible string. The string passes through a small smooth hole at the vertex of a cone fixed with its axis vertical. The cone has semivertical angle x, where x = tan^1(4 /3). The particle P hangs inside the cone & the particle Q is free to move on the smooth surface of the cone. (a) the particles are released from the rest & move in a fixed vertical plane, with the string taut & P moving vertically downwards. Find, in terms of m & g, the tension in the string. Find the time taken for P to fall a distance of 0.2 m from the rest, giving your answer correct to 3 significant figures. (b) the particles are brought to rest. Particle Q is projected so that it moves in a horizontal circle of radius 0.5m on the surface of the cone with constant angular speed w rad s^1 . Particle P remains at rest. Find, in either order, (i) the magnitude of the force exerted on Q by the cone, giving your answer in terms of m & g, (ii) the value of w, giving your answer correct to 3 significant figures. Ans : (a) (4 mg)/5, 0.447 s (b) i) 1/2 mg N (ii) 3.6 rad s^1

i help wn wrk is shwn

Cn u plz gv explaination 2 the solutions bit by bit in brief ? Bcz i actually cnt wrk out hw 2 use d formula 2 answer it wen i revise nw. Much obliged. A particle of mass 0.5 kg is free to move on a smooth horizontal plane. A force of magnitude 0.8 N acts on the particle in a direction due east.A second force of magnitude of P N in the direction N c degree E also acts on the particle. The resultant of the 2 forces has magnitude 2 N in the direction N 60degree E. Calculate P & c,giving your answer correct to two places of decimals & to 0.1 degree respectively. The particle moves from rest under the action of the two forces, which remain constant in magnitude & direction. Find the time taken for the particle to move a distance of 50 m. Find also the total work done by the forces in the same period. Answer: P=1.37 N,c 43.0degree ; 5 s, 100 J.