A spaceship while flyting inspace, splits up into three equal parts, due to an explosion. One fragment keeps moving in the same direction, the other two fly of at 60^@ to the original direction , on either side. If the energy released due to the explosion is twice the kinetic energy of the spaceship, find the kinetic energy of each of the fragment.

A spaceship while flyting inspace, splits up into three equal parts, d

1.	A spaceship while flyting inspace, splits up into three equal parts, due to an explosion. One fragment keeps moving in the same direction, the other two fly of at 60^@ to the original direction , on either side. If the energy released due to the explosion is twice the kinetic energy of the spaceship, find the kinetic energy of each of the fragment.
Answer» Solution :Letthe mass of thespaceship be 3m and initial speed be U. Hence,mass of eachfragment =m . LET their VELOCITY be `v_1,v_2 and v_3` after the explosion [Fig.1.46]. Applying the CONSERVATION of momentum law along the direction perpendicular to the original direction of motion, `0= mv_2 sin 60^@=mv_3 sin 60^@` `or, mv_2sin 60^@ =mv_3 sin 60^@` `or, v_2=v_3=v` (say) Applying the law of conservation of linear momentum along the original direction of motion , `3mu=mv_1+2mv cos 60^@` `or, 3u=v_1+v` KINETIC energy of the spaceship before explosion `E=1/2xx3mu^2` Energy released during the explosion, `E_r=1/2 m(v_1^2+2v^2)-1/23mu^2` According to the problem, `E_r=2E or, 1/2m(v_1^2+2v_^2)-1/23mu^2=2xx1/23mu^2` `or, v_1^2+2v^2-3u^2=6u^2` ` or, v_1^2+2v^2=9u^2`......(2) Solving (1) and (2) ,`v_1=u and v=2u` Hence, kinetic energy of the first fragment after explosion `=1/2 mv_1^2=1/2mu_2=1/22/3E=1/3E` Kinetic energy of each of the other two fragement `=1/2 mv^2=1/2 m(2u)^2=2mu^2=22/3E=4/3E`