1.

In the above question the energy before and after the collision will be :

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Solution :Here energy before the collision will be :
`E=(1)/(2)Iomega^(2)+(1)/(2)MV^(2)`
`=(1)/(2)xx(1)/(2)xx2xx0.04xx9+(1)/(2)xx0.5xx25`
`=0.18+6.25=6.43J`
Also energy after collision
`E_(C)=(1)/(2)xx((M)/(2)+m)R^(2)omega^(2)`
`=(1)/(2)[(2)/(2)+0.5]xx0.04xx(10.3)^(2)`
`=(1)/(2)xx1.5xx0.04xx106.09`
= 3.18 J


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