Question

A shell of mass $20kg$at rest explodes into two fragments whose masses are in the ratio $2:3$. The smaller fragment moves with a velocity of $6m{s}^{-1}$. The kinetic energy of the larger fragment is

• A. $96J$
• B. $216J$
• C. $144J$
• D. $360J$

Answer 1

The correct option is A

$96J$

Step 1. Given data

The total mass of the shell is $20kg$(kilograms) and the ratio of two masses is $2:3$.

Therefore, the masses of the two fragments are

$$\begin{gathered} M_1=8kg \\ {M}_2=12kg \end{gathered}$$

Where $M_1$is the mass of the smaller fragment

$M_2$ is the mass of the larger fragment

The velocity of the smaller fragments is $V_1=6m{s}^{-1}$

Step 2. Finding the velocity of the larger fragment

By applying the law of conservation of momentum, we get

$M_1\times V_1=M_2\times V_2$

Where $M_1$is the mass of the smaller fragment

$M_2$ is the mass of the larger fragment

$V_1$ is the velocity of the smaller fragment

$V_2$ is the velocity of the larger fragment

Now, by applying the values in the above formula. we get,

$$\begin{gathered} 8\times 6=12\times V_2 \\ {V}_2=\frac{48}{12} \\ {V}_2=4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.(meterpersecond) \end{gathered}$$

Step 3. The kinetic energy of the larger fragment

We know kinetic energy of larger fragment is given by

$K.E=\frac{1}{2}{M}_2{V_2}^2$

Where, $K.E$is the kinetic energy

$M_2$ is the mass of the larger fragment

$V_2$ is the velocity of the larger fragment

Now, $$\begin{gathered} K.E=\frac{1}{2}\times 12\times {\left(4\right)}^2 \\ =6\times 16 \\ =96J \end{gathered}$$

Hence, the correct option is (A).