Question

A bomb of mass $9$ kg initially at rest explodes into two pieces of mass $3$ kg and $6$ kg. If the kinetic energy of $3$ kg mass is $216$J, then the velocity of $6$kg mass will be?

Answer 1

let the total mass of the bomb be $"4m"$

so $4m=12kg$

$m=\frac{12}{4}$

$m=3kg$

since the ratio of pieces of masses is $1:3$

$m₁=$ mass of smaller piece $=m=3kg$

$m₂=$ mass of larger piece $=3m=3x3=9kg$

$v₁=$ velocity of the smaller piece

$v₂=$ velocity of the larger piece

initially, the bomb was at rest, hence initial total momentum = 0

Using conservation of momentum

Final total momentum = initial total momentum

$m₁v₁+m₂v₂=0$

$mv₁+\left(3m\right)v₂=0$

$v₁+3v₂=0$

$v₁=-3v₂$

Given that : initial kinetic energy $=216$

we know that : kinetic energy $=\left(0.5\right)mv²$ where m = mass , v = speed

so

$\left(0.5\right)m₁v₁²=216$

$\left(3\right)v₁²=432$

$v₁=12\frac{m}{s}$

using eq-1

$v₁=-3v₂$

$12=-3v₂$

$v₂=-4\frac{m}{s}$

The momentum of the larger piece is given as

$P₂=m₂v₂=9(-4)=-36kg\frac{m}{s}$