91Ó°ÊÓ

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-

$\stackrel{\mathbf{→}}{\mathbf{p}}\mathbf{=}\mathbf{m}\stackrel{\mathbf{→}}{\mathbf{v}}$

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.

${\stackrel{→}{\mathrm{p}}}_{\mathrm{total}}=\mathrm{constant}$

The shell was at rest before the explosion

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

${\mathrm{p}}_{\mathrm{A}}=-{\mathrm{p}}_{\mathrm{B}}$…â¶Ä¦â¶Ä¦..(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, ${\stackrel{→}{\mathrm{p}}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}$

${\stackrel{→}{\mathrm{p}}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}$

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, ${\mathrm{m}}_{\mathrm{A}}=2.0 \mathrm{kg}$ and ${\mathrm{m}}_{\mathrm{B}}=5.0 \mathrm{kg}$

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

The kinetic energy is given by

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

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

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

Now, $v_A=2.5v_B$

Therefore,

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

Now, find value of kinetic energy is

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

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