Question

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:

• A. \( \frac{M_1}{M_1 + M_2} \)
• B. \( \frac{M_2}{M_1} \)
• C. \( \frac{M_2}{M_1 + M_2} \)
• D. \( \frac{M_1}{M_2} \)

Answer 1

The Correct Option is B

The ratio of the kinetic energy of the artillery to that of the shell is calculated using the principles of conservation of momentum and the definition of kinetic energy.

1. Conservation of Momentum: \(M_1 \cdot V_1 = M_2 \cdot V_2\) This equation represents the conservation of momentum, where \( M_1 \) and \( M_2 \) are the masses of the artillery and shell, and \( V_1 \) and \( V_2 \) are their respective velocities after firing.
2. Kinetic Energy Formula: The kinetic energy (KE) of an object is defined as: \(\frac{1}{2}mv^2\) where \( m \) is the object's mass and \( v \) is its velocity.
3. From the conservation of momentum equation, the artillery's velocity can be expressed as: \(V_1 = \frac{M_2 \cdot V_2}{M_1}\)
4. Kinetic Energy of Artillery: The kinetic energy of the artillery is: \(K.E._{artillery} = \frac{1}{2}M_1V_1^2 = \frac{1}{2}M_1 \left(\frac{M_2 \cdot V_2}{M_1}\right)^2 = \frac{1}{2} \frac{M_2^2 \cdot V_2^2}{M_1}\)
5. Kinetic Energy of Shell: The kinetic energy of the shell is: \(K.E._{shell} = \frac{1}{2}M_2V_2^2\)
6. Ratio of Kinetic Energies: The ratio of the artillery's kinetic energy to the shell's kinetic energy is computed as: \(\text{Ratio} = \frac{\frac{1}{2} \frac{M_2^2 \cdot V_2^2}{M_1}}{\frac{1}{2}M_2 \cdot V_2^2} = \frac{M_2}{M_1}\)
7. Conclusion: Therefore, the ratio of the kinetic energy of the artillery to that of the shell is \(\frac{M_2}{M_1}\).

This derivation utilizes the physical principles of momentum conservation and kinetic energy definition to arrive at the final ratio.