A bomb at rest explodes into two fragments of masses `3.0kg` and `1.0kg`. The total kinetic energy of the fragments is `6xx10^(4)J`. Calculate (i) , kinetic energy of the bigger fragment (ii) momentum of the smaller fragment.

Explanation

If `upsilon_(1), upsilon_(2)` are velocities of fragments of masses `m_(1)=3kg` and `m_(2)=1kg`, then according to the principle of conservation of linear momentum,
`m_(1)upsilon_(1)+m_(2)upsilon_(2)`
`3upsilon_(2)+1upsilon_(2)=0` or ` (upsilon_(1))/(upsilon_(2))=-(1)/(3)`
`(E_(1))/(E_(2))=((1)/(2)m_(1)upsilon_(1)^(2))/((1)/(2)m_(2)upsilon_(2)^(2))=(3)/(1)(-(1)/(3))^(2)=(1)/(3)`
As total `K.E.=E=E_(1)+E_(2)=6xx10^(4)J`
`:. E_(1)=(6xx10^(4)xx1)/((1+3))=1.5xx10^(4)J`
`E_(2)=6xx10^(4)-1.5xx10^(4)=4.5xx10^(4)J`
Linear momentum of smaller fragment
`P_(2)=sqrt(2mE_(2))=sqrt(2xx1xx4.5xx10^(4))`
`3xx10^(2)kg ms^(-1)`
As the direction is opposite
`:. P_(2)=-300kg ms^(-1)`