Question

A bomb explodes by itself into two fragments of masses $3kg$ and $1kg$. If the total energy of both the fragments is $6.0\times {10}^{4}joule$, then calculate the momentum of smaller fragment (in kg-m/s).

Answer 1

Let $v_1$ and $v_2$ are the velocity of $3kg$ and $1kg$ fragment.

By momentum conservation,

$\Rightarrow 3v_1+1v_2=0$

$\Rightarrow v_1=-\frac{1}{3}{v}_2⟶\left(1\right)$

$\Rightarrow$ Total energy $=\frac{1}{2}{{3v}_1}^2+\frac{1}{2}1\times {v_2}^2=6\times {10}^4$

$=3{\left(\frac{v_2}{3}\right)}^2+{v_2}^2=12\times {10}^4$

$\Rightarrow {{4v}_2}^2=3\times 12\times {10}^4$

$v_2=3\times {10}^{2}m/s$

$\Rightarrow$ Momentum of smaller fragment $=1\times v_2$

$=3\times {10}^{2}kgm/s.$

Hence, the answer is $3\times {10}^{2}kgm/s.$