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Two identical steel cubes (masses 50g, s...

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

Text Solution

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`m=50g=50xx10^(-3)kg`
Side=L=1cm=0.01m
Young's modulus =Y=`2xx10^(11)N//m^(2)`
Maximum compression `DeltaL=?`
In this case all KE will be converted to PE
By Hooke's Law, `(F)/(A)=Y=(DeltaL)/(L)`
Where, A is the surface area and L is length of the side of the cube. If k is spring of compression constant, then
Force F=`k DeltaL`
`k=Y(A)/(l)=YL`
`Initial KE=2xx(1)/(2)mv^(2)=5xx10^(-4)J`
Final PE=`2xx(1)/(2) k(DeltaL)^(2)`
`DeltaL=sqrt((KE)/(k))=sqrt((KE)/(YL))=sqrt((5 xx10^(-4))/(2xx10^(11)xx0.1))=1.58xx10^(7)m (PE=KE)`
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