Answer :
Certainly! Let's go through a detailed, step-by-step solution to find the speeds of the fragments after the explosion.
Step 1: Understand the problem.
- Initial mass of the shell: [tex]\( m \)[/tex].
- Initial velocity of the shell: [tex]\( u \)[/tex].
- Energy generated by the internal explosion: [tex]\( E \)[/tex].
- Masses of the fragments after the explosion: in the ratio [tex]\( m_1 : m_2 \)[/tex]. This means if we denote the masses as [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] respectively, then total mass [tex]\( M = m_1 + m_2 \)[/tex].
Step 2: Conservation of Momentum.
The total momentum before and after the explosion must be conserved. Initially, the momentum of the shell is:
[tex]\[ \text{Initial momentum} = m \cdot u = (m_1 + m_2) \cdot u = (m_1 + m_2)u \][/tex]
Let [tex]\( v_1 \)[/tex] be the velocity of the fragment of mass [tex]\( m_1 \)[/tex] after the explosion and [tex]\( v_2 \)[/tex] be the velocity of the fragment of mass [tex]\( m_2 \)[/tex] after the explosion.
Then the total momentum after the explosion is:
[tex]\[ \text{Final momentum} = m_1 \cdot v_1 + m_2 \cdot v_2 \][/tex]
Since momentum is conserved:
[tex]\[ (m_1 + m_2)u = m_1 v_1 + m_2 v_2 \][/tex]
Step 3: Energy dispersion.
The energy generated by the explosion is [tex]\( E \)[/tex]. This energy is used to change the kinetic energy of the fragments.
Initial kinetic energy of the shell is:
[tex]\[ \frac{1}{2} m u^2 = \frac{1}{2} (m_1 + m_2) u^2 \][/tex]
Final kinetic energy of the fragments is:
[tex]\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
Increase in kinetic energy is equal to the energy released by the explosion:
[tex]\[ \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) - \frac{1}{2} (m_1 + m_2) u^2 = E \][/tex]
Step 4: Solve for the velocities.
From the conservation of momentum:
[tex]\[ (m_1 + m_2)u = m_1 v_1 + m_2 v_2 \][/tex]
From the conservation of energy:
[tex]\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 - \frac{1}{2} (m_1 + m_2)u^2 = E \][/tex]
First, rearrange the momentum equation to isolate one of the velocities (say [tex]\( v_1 \)[/tex]):
[tex]\[ v_1 = \frac{(m_1 + m_2)u - m_2 v_2}{m_1} \][/tex]
Next, substitute [tex]\( v_1 \)[/tex] into the energy equation:
[tex]\[ \frac{1}{2} m_1 \left( \frac{(m_1 + m_2)u - m_2 v_2}{m_1} \right)^2 + \frac{1}{2} m_2 v_2^2 - \frac{1}{2} (m_1 + m_2)u^2 = E \][/tex]
Simplifying this will result in:
[tex]\[ \frac{1}{2} \left( \frac{(m_1 + m_2)u - m_2 v_2}{1} \right)^2 + \frac{1}{2} m_2 v_2^2 - \frac{1}{2} (m_1 + m_2)u^2 = E \][/tex]
Through careful algebraic manipulation and solving, we obtain the velocities:
[tex]\[ v_1 = u + \sqrt{\frac{2 m_2 E}{m_2 (m_1 + m_2)}} = u + \sqrt{\frac{2E}{m_1 + m_2}} \][/tex]
and
[tex]\[ v_2 = u - \sqrt{\frac{2 m_1 E}{m_2 (m_1 + m_2)}} = u - \sqrt{\frac{2E m_1}{m_2 (m_1 + m_2)}} \][/tex]
These equations describe the velocities of the fragments after the internal explosion.
Step 1: Understand the problem.
- Initial mass of the shell: [tex]\( m \)[/tex].
- Initial velocity of the shell: [tex]\( u \)[/tex].
- Energy generated by the internal explosion: [tex]\( E \)[/tex].
- Masses of the fragments after the explosion: in the ratio [tex]\( m_1 : m_2 \)[/tex]. This means if we denote the masses as [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] respectively, then total mass [tex]\( M = m_1 + m_2 \)[/tex].
Step 2: Conservation of Momentum.
The total momentum before and after the explosion must be conserved. Initially, the momentum of the shell is:
[tex]\[ \text{Initial momentum} = m \cdot u = (m_1 + m_2) \cdot u = (m_1 + m_2)u \][/tex]
Let [tex]\( v_1 \)[/tex] be the velocity of the fragment of mass [tex]\( m_1 \)[/tex] after the explosion and [tex]\( v_2 \)[/tex] be the velocity of the fragment of mass [tex]\( m_2 \)[/tex] after the explosion.
Then the total momentum after the explosion is:
[tex]\[ \text{Final momentum} = m_1 \cdot v_1 + m_2 \cdot v_2 \][/tex]
Since momentum is conserved:
[tex]\[ (m_1 + m_2)u = m_1 v_1 + m_2 v_2 \][/tex]
Step 3: Energy dispersion.
The energy generated by the explosion is [tex]\( E \)[/tex]. This energy is used to change the kinetic energy of the fragments.
Initial kinetic energy of the shell is:
[tex]\[ \frac{1}{2} m u^2 = \frac{1}{2} (m_1 + m_2) u^2 \][/tex]
Final kinetic energy of the fragments is:
[tex]\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
Increase in kinetic energy is equal to the energy released by the explosion:
[tex]\[ \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) - \frac{1}{2} (m_1 + m_2) u^2 = E \][/tex]
Step 4: Solve for the velocities.
From the conservation of momentum:
[tex]\[ (m_1 + m_2)u = m_1 v_1 + m_2 v_2 \][/tex]
From the conservation of energy:
[tex]\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 - \frac{1}{2} (m_1 + m_2)u^2 = E \][/tex]
First, rearrange the momentum equation to isolate one of the velocities (say [tex]\( v_1 \)[/tex]):
[tex]\[ v_1 = \frac{(m_1 + m_2)u - m_2 v_2}{m_1} \][/tex]
Next, substitute [tex]\( v_1 \)[/tex] into the energy equation:
[tex]\[ \frac{1}{2} m_1 \left( \frac{(m_1 + m_2)u - m_2 v_2}{m_1} \right)^2 + \frac{1}{2} m_2 v_2^2 - \frac{1}{2} (m_1 + m_2)u^2 = E \][/tex]
Simplifying this will result in:
[tex]\[ \frac{1}{2} \left( \frac{(m_1 + m_2)u - m_2 v_2}{1} \right)^2 + \frac{1}{2} m_2 v_2^2 - \frac{1}{2} (m_1 + m_2)u^2 = E \][/tex]
Through careful algebraic manipulation and solving, we obtain the velocities:
[tex]\[ v_1 = u + \sqrt{\frac{2 m_2 E}{m_2 (m_1 + m_2)}} = u + \sqrt{\frac{2E}{m_1 + m_2}} \][/tex]
and
[tex]\[ v_2 = u - \sqrt{\frac{2 m_1 E}{m_2 (m_1 + m_2)}} = u - \sqrt{\frac{2E m_1}{m_2 (m_1 + m_2)}} \][/tex]
These equations describe the velocities of the fragments after the internal explosion.
