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A block A of mass m connected with a spring of force constant k is executing SHM. The displacement time equation of the block is x =x_(0) + a sin omega t.An identical block B moving towards negative x-axis with velocity vo collides elastically with block A at time t = 0. Then, |
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Answer» displacement time equation of A after collision will be `x=x_(0) -v_(0) sqrt(m/k) sin omegat` `therefore v=(DX)/(DT) = aomega cos omega g` (for block A) at t=0, `x=x_(0)` and `v= aomega` i.e., block A is at `x =x_(0)` (mean position) and its velocity is `a omega`in positive x-direction. It collides elastically with an identical block B moving towards negative x-direction with velocity `v_(0)`. So, the blocks will INTERCHANGE their velocities i.e., `V_(a) =v_(0)`(in negative x-direction) and `v(B)= aomega`(in positive x-direction) Let A be the new AMPLITUDE of block A, then from conservation of mechanical energy `1/2mv_(A)^(2) =1/2kA^(2)` `therefore A =v_(A) sqrt(m/k) = v_(0) sqrt(m/k)` `therefore` New displacement-time equation for block A will be: `x=x_(0) -v_(0) sqrt(m/k) sin omega t` |
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