A 72.0-kg shell at rest explodes into two fragments, one with a mass of 2.0-kg and the other with a mass of 5.0-kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?

The given data is listed below as-

• The mass of the shell is, M = 7.0 kg
• The mass of the first fragment is, $m_A=2.0kg$
• The mass of the second fragment is, $m_B=5.0kg$
• Kinetic energy of the heavier fragment gains after the explosion, K = 100 J

The momentum of a particle is the product of the particle’s mass and velocity and is given by-

$\overrightarrow{\mathbf{p}}\mathbf{=}\mathbf{m}\overrightarrow{\mathbf{v}}$

Where m is the mass of the particle and V is the velocity of the particle.

The total momentum of the system is constant since there is no external horizontal force on the system.

${\overrightarrow{\mathrm{p}}}_{\mathrm{total}}=\mathrm{constant}$

The shell was at rest before the explosion

So, ${\overrightarrow{\mathrm{p}}}_{\mathrm{A}}+{\overrightarrow{\mathrm{p}}}_{\mathrm{B}}=0$

${\mathrm{p}}_{\mathrm{A}}=-{\mathrm{p}}_{\mathrm{B}}$………..(1)

Therefore, after the explosion, the lighter fragment moves in +x direction with velocity localid="1665058111598" ${\mathrm{v}}_{\mathrm{A}}$, the heavier fragment moves in -x direction with ${\mathrm{v}}_{\mathrm{B}}$.

Now, ${\overrightarrow{\mathrm{p}}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}$

${\overrightarrow{\mathrm{p}}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}$

Substituting in equation (1)

${\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}$

Now substitute values for $m_A$and $m_B$

For, ${\mathrm{m}}_{\mathrm{A}}=2.0\hspace{0.33em}\mathrm{kg}$ and ${\mathrm{m}}_{\mathrm{B}}=5.0\hspace{0.33em}\mathrm{kg}$

$$\begin{gathered} 2\hspace{0.33em}{\mathrm{v}}_{\mathrm{A}}=5\hspace{0.33em}{\mathrm{v}}_{\mathrm{B}} \\ {\mathrm{v}}_{\mathrm{A}}=2.5\hspace{0.33em}{\mathrm{v}}_{\mathrm{B}} \end{gathered}$$

The kinetic energy is given by

$k=\frac{1}{2}m{v}^2$

For, K = 100 J and $m_B=5.0kg$

$$\begin{gathered} 100=\frac{1}{2}\left(5.0\right){v^2}_B \\ {v}_B=\sqrt{\frac{2\times 100}{5}} \\ {v}_B=6.32m/s \end{gathered}$$

Now, $v_A=2.5v_B$

Therefore,

$$\begin{gathered} v_A=2.5\times 6.32 \\ =15.81m/s \end{gathered}$$

Now, find value of kinetic energy is

$$\begin{gathered} K=\frac{1}{2}\times \left(2.0\right)\times {\left(15.81\right)}^2 \\ =250\mathrm{J} \end{gathered}$$

Thus, the kinetic energy that the fragment gains is 250 J..