A bomb explodes and splits up into three fragments.
Two fragments, each of mass 200g, move away from each other making an angle `120^@` , at a speed of `100 m*s^(-1)`. Find the direction and velocity of the third fragment whose mass is 500 g. Also find out the energy released in explosion.

Explanation

Fig.1.45 shows the velocity of fragment A and B along OA and OB .A and B have equal mass and speed. From the law of conservation of linear momentum, the third piece must move along OD, in the direction opposite to the resultant of OA and OB.

If the velocity of the third piece is v, then taking the component along the line CD in CGS system.
`500v =200xx10^4 cos 60^@ +200xx10^4 cos 60^@`
`or, v=(200xx10^4)/(500)=4xx10^3 cm *s^(-1) =40 m*s^(-1)`.
Hence, the velocity of the third fragment is `40 m*s^(-1)`. it moves so as to make an angle `120^@` with each of OA and OB. The energy released due to explosion, is the kinetic energy the three fragment.
`therefore` Energy released
`=1/2 xx200xx(10^4)^2+1/2 xx200xx(10^4)^2+1/2xx500xx(4xx10^3)^2=2400xx10^7 ergs=2400 J`.