A rubber cord has a cross -sectional area `1 mm^(2)`and total un stretched length`10.0cm`. It is stretched to `12.0 cm` and then released to project a missile of mass 5.0 g. Taking young's modulus Y for rubber as `5.0xx10^(8) N//m^(2)`.Calculate the velocity of projection .

Answer

`20ms^(-1)`

Explanation

Assumption: Total elastic energy of catapult is converted into kinetic energy of missile without any heat loss.
Elastic energy `=1/2xx"Load"xx"Extension"`
Here extension
`/_\L=12.0-10.0=2.0cm=2.0xx10^(-2)m`
From relation
`Y=(FL)/(A/_\L)`
Load, `F=(YA/_\L)/L`
`=(5.0xx10^(8)xx10^(-6)xx2.0xx10^(-2))/(10.0xx10K^(-2))=100N`
if `v` velocity of projection we have
Elastic energy of catapult`=`kinetic energy of missile
i.e. `1/2xx"load"xx"extension"=1/2mv^(2)`
`1/2xx100xx2.0xx10^(-2)=1/2xx(5.0xx10^(-3))v^(2)`
This gives `v^(2)=(100xx10^(-2)xx2)/(5xx10^(-3))=400`
`:. v^(2)=sqrt(4000=20ms^(-1)`