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A point object located at a distance of 20 cm from the pole of a concave mirror of focal length 30 cm with height 2 cm is moving with a velocity ( bar(V_(OG)) =4 hati - 5 hatj) m/s and velocity of the mirror is (bar(V_(mg)) = -6 hati + 10 hatj) m I s as shown. Find the velocity of image w.r.t ground. |
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Answer» Solution :From the velocity diagram of object w.r.t mirror is `(V_(om))_(II)= 10 m//s implies (V_(om)) = -15 m//s` `implies m= F/(f-u) =3 implies v=60cm` `(V_(IM))_(II) = - v^(2)/u^(2) (V_(om))_(II) implies (V_(IM))_(II)` = -`(60/-20)^(2) (10)= -90`m/s `(dm)/(dt) = [(u(V_(IM)_(II) - V(V_(om)_(II))))]/u^(2)` = `[( (-20)xx(-90)xx(-60)xx10^(2)xx(10))/(-20xx10^(-2))^(2)]` per SEC `(dm)/(dt) = 3XX10^(2)` per sec `(V_(IM)) = h_(0) (dm)/(dt) + m(V_(om))` `bar(V_(IM))=(V_(IM))_(II) hati + (V_(IM)) hatj implies bar(V_(IM)) = (-90 hati + -51 hatj)` `bar(V_(IG)) = (-90 hati + -51 hatj) + (-6 hati + 10 hatj) implies ` `:. bar(V_(IG)) = (-96 hati + -41 hatj)` m/s |
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