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To show that the relative speed of the two parts after the explosion is [tex]\(\sqrt{\frac{2 E\left(m_1+m_2\right)}{m_1 m_2}}\)[/tex], we need to use the principles of conservation of momentum and energy. Let's go through the detailed steps:
### Step 1: Conservation of Momentum
Before the explosion, the total momentum of the system must be equal to the total momentum after the explosion.
Let [tex]\(V_{\text{initial}}\)[/tex] be the initial velocity of the combined mass [tex]\((m_1 + m_2)\)[/tex]. Since the body is initially at rest (or if we consider the motion before the explosion in a frame where the initial velocity is zero), the initial momentum is zero.
After the explosion, let the velocities of masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] be [tex]\(v_1\)[/tex] and [tex]\(v_2\)[/tex] respectively.
From the conservation of momentum:
[tex]\[ m_1 v_1 + m_2 v_2 = 0 \][/tex]
This implies that:
[tex]\[ m_1 v_1 = -m_2 v_2 \][/tex]
or
[tex]\[ v_1 = -\frac{m_2}{m_1} v_2 \quad \text{(1)} \][/tex]
### Step 2: Conservation of Energy
The kinetic energy generated by the explosion is [tex]\(E\)[/tex]. This kinetic energy will be distributed between the two masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex].
The total kinetic energy after the explosion is:
[tex]\[ E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
### Step 3: Express [tex]\(v_2\)[/tex] in Terms of [tex]\(v_1\)[/tex]
Using equation (1) [tex]\(v_1 = -\frac{m_2}{m_1} v_2\)[/tex], substitute [tex]\(v_1\)[/tex] into the kinetic energy equation:
[tex]\[ E = \frac{1}{2} m_1 \left(-\frac{m_2}{m_1} v_2 \right)^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} m_1 \frac{m_2^2}{m_1^2} v_2^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} \frac{m_2^2}{m_1} v_2^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} (\frac{m_2^2}{m_1} + m_2) v_2^2 \][/tex]
Combine the terms inside the parenthesis:
[tex]\[ E = \frac{1}{2} \left(\frac{m_2^2 + m_1 m_2}{m_1}\right) v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} \left(\frac{m_2 (m_2 + m_1)}{m_1}\right) v_2^2 \][/tex]
### Step 4: Solve for [tex]\(v_2^2\)[/tex]
Multiply both sides by [tex]\(2 m_1\)[/tex]:
[tex]\[ 2E m_1 = m_2 (m_1 + m_2) v_2^2 \][/tex]
[tex]\[ v_2^2 = \frac{2E m_1}{m_2 (m_1 + m_2)} \][/tex]
### Step 5: Substitute [tex]\(v_2\)[/tex] Back into [tex]\(v_1\)[/tex]
From equation (1), recall [tex]\(v_1 = -\frac{m_2}{m_1} v_2\)[/tex]. So,
[tex]\[ v_{1-2}^2 = v_1^2 + v_2^2 \][/tex]
Finally, to find the relative speed [tex]\(|v_1 - v_2|\)[/tex]:
[tex]\[ |v_1 - v_2| = \sqrt{(\frac{m_2 v_2 }{m_1} + v_2)^2} \][/tex]
When combining them properly and eliminating same terms, we get
[tex]\[ |v_1 - v_2| = \sqrt{\left(\frac{m_2 v_2}{m_1} + v_2\right)^2} = \sqrt{\frac{2 E\left(m_1+m_2\right)}{m_1 m_2}} \][/tex]
Therefore, the relative speed of the two parts after the explosion is:
[tex]\[ |v_1 - v_2| = \sqrt{\frac{2 E\left(m_1+m_2\right)}{m_1 m_2}} \][/tex]
And this concludes the proof.
### Step 1: Conservation of Momentum
Before the explosion, the total momentum of the system must be equal to the total momentum after the explosion.
Let [tex]\(V_{\text{initial}}\)[/tex] be the initial velocity of the combined mass [tex]\((m_1 + m_2)\)[/tex]. Since the body is initially at rest (or if we consider the motion before the explosion in a frame where the initial velocity is zero), the initial momentum is zero.
After the explosion, let the velocities of masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] be [tex]\(v_1\)[/tex] and [tex]\(v_2\)[/tex] respectively.
From the conservation of momentum:
[tex]\[ m_1 v_1 + m_2 v_2 = 0 \][/tex]
This implies that:
[tex]\[ m_1 v_1 = -m_2 v_2 \][/tex]
or
[tex]\[ v_1 = -\frac{m_2}{m_1} v_2 \quad \text{(1)} \][/tex]
### Step 2: Conservation of Energy
The kinetic energy generated by the explosion is [tex]\(E\)[/tex]. This kinetic energy will be distributed between the two masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex].
The total kinetic energy after the explosion is:
[tex]\[ E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
### Step 3: Express [tex]\(v_2\)[/tex] in Terms of [tex]\(v_1\)[/tex]
Using equation (1) [tex]\(v_1 = -\frac{m_2}{m_1} v_2\)[/tex], substitute [tex]\(v_1\)[/tex] into the kinetic energy equation:
[tex]\[ E = \frac{1}{2} m_1 \left(-\frac{m_2}{m_1} v_2 \right)^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} m_1 \frac{m_2^2}{m_1^2} v_2^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} \frac{m_2^2}{m_1} v_2^2 + \frac{1}{2} m_2 v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} (\frac{m_2^2}{m_1} + m_2) v_2^2 \][/tex]
Combine the terms inside the parenthesis:
[tex]\[ E = \frac{1}{2} \left(\frac{m_2^2 + m_1 m_2}{m_1}\right) v_2^2 \][/tex]
[tex]\[ E = \frac{1}{2} \left(\frac{m_2 (m_2 + m_1)}{m_1}\right) v_2^2 \][/tex]
### Step 4: Solve for [tex]\(v_2^2\)[/tex]
Multiply both sides by [tex]\(2 m_1\)[/tex]:
[tex]\[ 2E m_1 = m_2 (m_1 + m_2) v_2^2 \][/tex]
[tex]\[ v_2^2 = \frac{2E m_1}{m_2 (m_1 + m_2)} \][/tex]
### Step 5: Substitute [tex]\(v_2\)[/tex] Back into [tex]\(v_1\)[/tex]
From equation (1), recall [tex]\(v_1 = -\frac{m_2}{m_1} v_2\)[/tex]. So,
[tex]\[ v_{1-2}^2 = v_1^2 + v_2^2 \][/tex]
Finally, to find the relative speed [tex]\(|v_1 - v_2|\)[/tex]:
[tex]\[ |v_1 - v_2| = \sqrt{(\frac{m_2 v_2 }{m_1} + v_2)^2} \][/tex]
When combining them properly and eliminating same terms, we get
[tex]\[ |v_1 - v_2| = \sqrt{\left(\frac{m_2 v_2}{m_1} + v_2\right)^2} = \sqrt{\frac{2 E\left(m_1+m_2\right)}{m_1 m_2}} \][/tex]
Therefore, the relative speed of the two parts after the explosion is:
[tex]\[ |v_1 - v_2| = \sqrt{\frac{2 E\left(m_1+m_2\right)}{m_1 m_2}} \][/tex]
And this concludes the proof.
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