An artillery piece of mass \( M_1 \) fires a shell of mass \( M_2 \) horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:
- \( \frac{M_1}{M_1 + M_2} \)
- \( \frac{M_2}{M_1} \)
- \( \frac{M_2}{M_1 + M_2} \)
- \( \frac{M_1}{M_2} \)
The Correct Option is B
Approach Solution - 1
To determine the ratio of the kinetic energy of the artillery to that of the shell, we can use the principles of conservation of momentum and the definition of kinetic energy.
- Conservation of Momentum: \(M_1 \cdot V_1 = M_2 \cdot V_2\) where \( V_1 \) and \( V_2 \) are the velocities of the artillery and shell respectively after firing.
- Kinetic Energy Formula: The kinetic energy of an object is given by: \(\frac{1}{2}mv^2\) where \( m \) is the mass and \( v \) is the velocity of the object.
- From the conservation of momentum, we can express the artillery's velocity in terms of the shell's velocity: \(V_1 = \frac{M_2 \cdot V_2}{M_1}\)
- Kinetic Energy of Artillery: \(K.E._{artillery} = \frac{1}{2}M_1V_1^2 = \frac{1}{2}M_1 \left(\frac{M_2 \cdot V_2}{M_1}\right)^2 = \frac{1}{2} \frac{M_2^2 \cdot V_2^2}{M_1}\)
- Kinetic Energy of Shell: \(K.E._{shell} = \frac{1}{2}M_2V_2^2\)
- Ratio of Kinetic Energies: The ratio of the kinetic energy of the artillery to that of the shell is: \(\text{Ratio} = \frac{\frac{1}{2} \frac{M_2^2 \cdot V_2^2}{M_1}}{\frac{1}{2}M_2 \cdot V_2^2} = \frac{M_2}{M_1}\)
- Conclusion: Thus, the correct answer is \(\frac{M_2}{M_1}\), which is the ratio of the kinetic energy of the artillery to that of the shell.
This solution demonstrates the use of physics concepts such as conservation of momentum and kinetic energy, leading us to the correct answer.
Approach Solution -2
Given that momentum is conserved:
\[ |p_1| = |p_2| \]
Let \( p \) denote the magnitude of the momentum. The kinetic energy (KE) of an object is given by:
\[ KE = \frac{p^2}{2M} \]
Since the momenta of the artillery and the shell are equal, the kinetic energy is inversely proportional to the mass:
\[ KE \propto \frac{1}{M} \]
Thus, the ratio of kinetic energies of the artillery (\( KE_1 \)) and the shell (\( KE_2 \)) is:
\[ \frac{KE_1}{KE_2} = \frac{\frac{p^2}{2M_1}}{\frac{p^2}{2M_2}} = \frac{M_2}{M_1} \]
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