Two identical steel cubes (masses 50 g , side 1 cm ) collide head -on face to face with a space of 10 cm/s each . Find the maximum compression of each .Young's modulus for steel= Y = 2xx10^(11) N//m^(2)

Two identical steel cubes (masses 50 g , side 1 cm ) collide head -on

1.	Two identical steel cubes (masses 50 g , side 1 cm ) collide head -on face to face with a space of 10 cm/s each . Find the maximum compression of each .Young's modulus for steel= Y = 2xx10^(11) N//m^(2)
Answer» Solution :Let m = 50 g = `50 xx 10^(-3)` kg L = 1 CM = 0.01 m v =10 cm/s = 0.1 m/s `Y = 2xx10^(11) N//m^(2)` Here, KE will be converted to PE `F = (YADeltaL)/l ` (Hooke.s LAW ) ` :. "Also " , F = DeltaL ` (K = spring constant) ` :. k = Y A/L = YL "" [ :. A =L^(2)] ` Intial `KE = 2XX 1/2 mv^(2) = 5xx10^(-4) J ` . Final `PE = s xx 1/2 k(DeltaL)^(2) = k(DeltaL)^(2)` ` :. k(DeltaL)^(2) = 5xx10^(-4) = sqrt((5xx10^(-4))/(2xx10^(11)xx0.1))` `= 1.58 xx10^(-7) m `