Question

A body of mass $18\text{kg}$ initially at rest explodes into three fragments with masses in the ratio of $2:1:3$. If smaller fragments fly in mutually perpendicular directions and their velocities are $10\text{m/s}$ and $5\text{m/s}$ respectively, then what is the speed of the heaviest fragment?

• A. $4.71\text{m/s}$
• B. $30\text{m/s}$
• C. $30\sqrt{2}\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is A $4.71\text{m/s}$

The ratio of the fragmented masses is $2:1:3$. Let the common ratio be $x$.
Then, $2x+x+3x=18$
As before explosion, mass of body is $18\text{kg}$,
$6x=18\Rightarrow x=3$

$\therefore$The masses of the three pieces are $6\text{kg},3\text{kg and}9\text{kg}$.

Given :
After explosion, the velocity of the $3\text{kg}$ piece is $10\text{m/s}$ and the velocity of the $6\text{kg}$ piece is $5\text{m/s}$.

Momentum of the $3\text{kg}$ piece is $P_1=3\times 10=30\text{kgm/s}$
Momentum of the $6\text{kg}$ piece is $P_2=6\times 5=30\text{kgm/s}$

Let Momentum of the $9\text{kg}$ piece is $P_3$

As there are no net external force, applying PCLM
$p_i=p_f\Rightarrow 0=p_1+p_2+p_3\Rightarrow p_3=-(p_1+p_2)\Rightarrow |p_3|=|p_1+p_2|$

The magnitude of momentum of the third piece should be equal to the magnitude of resultant of the momentums of the first two pieces.
Hence, $9\times v_3=\sqrt{p_1^2+p_2^2}=\sqrt{{30}^2+{30}^2}=30\sqrt{2}$
$\Rightarrow v_3=\frac{30\sqrt{2}}{9}=4.71\text{m/s}$
Hence the correct option is (a).