Question

A ball is dropped from rest 3.10 m directly above the vertex of a concave mirror having a radius of 1.80 m and lying in a horizontal plane. (a) Describe the motion of the ball's image in the mirror by filling in the table below. Object Distance, p Image Distance, 3.10 m m 0.900 m m Om m (b) At what two times do the ball and its image coincide? (List the lesser time first.) s and NAU

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Answer 1

T 1.80 m Given Height from vertex from where ball is dropped h=3.10 m. radius of concave mirror a focal length of mirror f = 1 / 2o.gm a) And according to mirror formula P-t 의 * 1 / 2 = 1 1 1 distance & P is object distance. q=ff i) So for P 3.10 m Pf P where q= 3.1 xong 1268m 3.1-0.9 q = 1:208m ii) for 0.90.9 P=0.90 m q= > 0.9-0.9 q=0 (not infinite ) ii) for for prom Ox0,9 2= Om 0-09 object distance Image distance q 3.10 m $1.268 m 0.goom 0 m om

6) Image and object will coincide at two distances & p=0 from vertex when from vertex as for para2f 2 Rxf P-f 26 of Zapf tak & for pro Oxf q= orf so the initial velocity of ball u=0ml (rest) acceleration of ball a=g = 9.8 m/s² and as PSG from vertex distance & from initial position és Sah-r According to second Bi Kinematic equation S, = ut, + 1/4 at,? a) hra oxt,+ 12x9.88t² = 3.1-1.8 = 4.9 t? -0.265 sec? 113 21.9 t=0.51 sec and P2 = 0 from vertex distance & from initial position is S₂ = h an S₂ = utzt hat? ha oxt, + 5* 9.8xt? 2 3.1 =4.gxte tz 3.1 = 0.6326 وهما = t = = 0.79 sec at 10:57 + 0.79 S ball & its image will coincide