InterviewSolution
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InterviewSolution
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The system shown in the figure can move on a smooth surface. They are initially compressed by `6cm` and then released, then choose the correct options. (a) The system performs, SHM with time period `(pi)/(10)s` (b) The block of mass `3kg` perform SHM with amplitude `4 cm` ( c) The block of mass `6kg` will have maximum momentum of `2.40kg - m//s` (d) The time periods of two blocks are in the ratio of `1:sqrt(2)` |
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Answer» a. `T = 2 pisqrt((mu)/(k))` Here, `mu = (m_(1) m_(2))/(mm_(1) + m_(2)) = (3 xx 3)/(9) = 2 kg` `T = 2 pi sqrt((2)/(800)) = 2 pi xx (1)/(20)` `T = (pi)/(10) sec` b. Here compression of spring `x_(0) = 6 cm` Let displacement of block (1) and (2) w.r.t. center of mass be `x_(1) and x_(2)` But `x_(1) + x_(2) = x_(0) and m_(1) = m_(2)x_(2)` or `3 x_(1) = 6 x_(2)` or `x_(1) + 2 x_(2)` ` x_(1) + x_(2) = x_(0) or x_(1) + (x_(1))/(2) = x_(0)` or `(3)/(2) x_(1) = x_(0) implies x_(1) = (2)/(3) x_(0) = (2)/(3) xx 6 = 4 cm` c. Appling conservation principal of momentum `3 xx v_(1) - 6 v_(2) = 0` `3 v_(1)=- 6 v_(2)` `v_(1) = 2 v_(2)` Appling mechaincal energy conservation , we get `(1)/(2) xx 3 xx v_(1)^(2) + (1)/(2) xx 6 xx v_(2)^(2) = (1)/(2) kx_(0)^(2)` or `(3)/(2) v_(1)^(2) + 3v_(2)^(2) = (1)/(2) xx 800 xx (0.06)^(2)` or `(3)/(2) (2 v_(2))^(2) + 3v_(2)^(2) = 400 xx 36 xx 10^(-4)` or `v_(2) = (1.2)/(3) = 0.4 m//s` `:. P_(2 max) = 6 xx 0.4 = 2.4 kg m//s` |
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