A bomb of mass '5m' at rest explodes into three parts of masses `2m`, `2m` and `m`. After explosion, the equal parts move at right angles with speed v each. Find speed of the third part and total energy released during explosion.

Answer

B

Explanation

Let the two equal parts move along positive x and positive y directions and suppose the velocity of third part is V. From law of conservation of linear momentum,
we have, `p_i=p_f`
`implies 02m(vhati)+2m(vhatj)+mV`
" Solving this equation we have", `V=-2vhati-2vhatj`
`:. "Speed of this particle"
`=|V|or V`
`=sqrt((-2v)^2+(-2v)^2)`
`=2sqrt2v`
Energy released during explosion, `="kinetic energy of all three parts"`
`=1/2(2m)v^2+1/2(2m)v^2+1/2(m)(2sqrt2v)^2`
`=6mv^2`