A bomb shell of mass `100kg` travelling at `500ms^(-1)` explodes into two fragments of masses in the ratio `2 : 3`. The smaller one flies at an angle of `45^(@)` and the heavier at `60^(@)` with respect to the `X`-axis in the `(x,y)` and `(x,-y)` planes respectively. Calculate their velocities.

Explanation

`m_(1) : m_(2) : : 2 : 3`
`m_(1)=(2)/(5)xx100=40kg` `m_(2)=(3)/(5)xx100=60kg`
Equating the horizontal and vertical components of momentum, we get
`mv=m_(1)v_(1f)costheta_(1)=m_(2)v_(2f)costheta_(2)`
i.e. `100xx500=40v_(1f)cos45^(@)+60v_(2f)cos60^(@)`
`5000=4v_(1f)xx(1)/(sqrt(2))+6v_(2t)xx(1)/(2)` or `2500=sqrt(2)v_(1f)+(3)/(2)v_(2f)`.................`(1)`
Similarly , `0=4v_(1f)sin45^(@)-6v_(2f)sin60^(@)`
i.e.`2v_(1f)xx(1)/(sqrt(2))=3v_(2f)xx(sqrt(3))/(2)` or `sqrt(2)v_(1f)xx(1)/(sqrt(2))=3v_(2f)xx(sqrt(3))/(2)`
or `sqrt(2)v_(1f)=(3sqrt(3))/(2)v_(2t)`..............`(2)`
Using `(2)` in `(1)` we get `sqrt(2)v_(1f)=3(sqrt(2))/(2)xx610`
i.e. `v_(1f)=(3)/(2)sqrt((3)/(2))xx610=sqrt((27)/(8))xx610=610xx1.837`
i.e. `v_(1f)=1120.6ms^(-1)`. The velocities of the two fragments are `610.0ms^(-1)` and `1120.6ms^(-1)`