InterviewSolution
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InterviewSolution
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A spring is held correspessed so that its stored energy is `2.4 J`. Its ends are in constant with masses `1 g and 48 g` placed on a frictionless table. When the spring in released, the heaveier mass will acquire a speed of:A. `(2.4)/(49)` ms^(-1)`B. `(2.4 xx 48)/(49) ms^(-1)`C. `mu mgh//sin theta`D. `(10^(6))/(7)`c ms^(-1)` |
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Answer» Correct Answer - C `(1)/(2)m_(1)v_(1)^(2) + (1)/(2) m_(2) v_(2)^(2) = 2.4` or `m_(1)v_(1)^(2) + m_(2) v_(2)^(2) = 4.8`…(i) Now `m_(1)v_(1)= m_(2)v_(2)` or `v_(1) = 48v_(2)` Using (i) `(1)/(1000) (48v_(2))^(2) + (48)/(1000) v_(2)^(2) = 4.8` or `v_(2) = (10)/(7) m//s = (10^(2))/(7) cm//sec`. |
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