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A block of mass m is placed on a smooth block of mass `M = m` with the help of a spring as shown in the figure. A velocity `v_(0)` is given to upper block when spring is in natural length. Find maximum compression in the spring. A. `sqrt((m)/(2K)) V_(0)`B. `sqrt((m)/(K)) v_(0)`C. `sqrt((2m)/(K)) v_(0)`D. `2 sqrt((m)/(K)) v_(0)` |
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Answer» Correct Answer - A Compression will be maximum when relative velocity between blocks is zero. Means both have Sam velocity apply conservation of momentum `mv_(0) = (M + m) v` `rArr v = (m)/(M + m) v_(0)` Apply conservation of kinetic energy `(1)/(2) mv^(2) + (1)/(2) Mv^(2) + (1)/(2) kx^(2) = (1)/(2) mv_(0)^(2)` `rArr (m^(3)v_(0)^(2))/((M + m)^(2)) + (m^(2)Mv_(0)^(2))/((M + m)^(2)) + kx^(2) = mv_(0)^(2)` `rArr kx^(2) = (m^(3) + m^(2)M)/((M + m)^(2)) v_(0)^(2) = mv_(0)^(2)` `rArr kx^(2) = v_(0)^(2) [(m)/(p) - (m^(2) (M + m))/((M + m)^(2))]` `= v_(0)^(2) [(M_(m) + m^(2) - m^(2))/(M + m)] = (M_(m)v_(0)^(2))/(M + m)` `rArr x = sqrt((Mm)/(k(M + m))) v_(0)` |
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