Problem 21 - Applied Mathematics

Question

The barrel of a gun of mass \(M\) resting on a smooth horizontal plane is elevated at an angle \(\alpha\) to the horizontal. The gun fires a shell of mass \(m\) and recoils with horizontal velocity \(U .\) If the velocity of the shell on leaving the gun has horizontal and vertical components \(v\) and \(w\) respectively, prove that \(w=(v+U) \tan \alpha\), and hence or otherwise prove that the initial inclination of the path of the shell to the horizontal is arctan \(\left[\left(1+\frac{m}{M}\right) \tan \alpha\right]\). Prove that the kinetic energy generated by the explosion is $$ \frac{U^{2}}{2 m}(M+m)\left(M \sec ^{2} \alpha+m \tan ^{2} \alpha\right) $$

Step-by-Step Solution

Answer

From conservation laws, we derive \(w = (v + U) \tan \alpha\) and initial path inclination as arctan \( (1 + \frac{m}{M}) \tan \alpha \). Kinetic energy generated is \( \frac{U^2}{2m}(M+m)(M \sec^2 \alpha + m \tan^2 \alpha)\).

Step 1 - Analyze the Recoil

Consider the conservation of momentum in the horizontal direction. Initially, the system (gun plus shell) is at rest, so total initial momentum is zero. After firing, the gun recoils with horizontal velocity U, and the shell has a horizontal component of velocity v.

Step 2 - Apply Conservation of Horizontal Momentum

Using the conservation of momentum, the total momentum after firing must be equal to the total initial momentum. Mathematically, this is given by: \[ 0 = M(-U) + mv \] Solving for U, we get \[ U = \frac{mv}{M} \]

Step 3 - Calculate Vertical Velocity Component

Given that the shell has a velocity with horizontal component v and angle of projection \( \alpha \), its vertical velocity component w can be derived from trigonometric relationships: \[ \tan \alpha = \frac{w}{v+U} \]Substituting the value of U from Step 2: \[ \tan \alpha = \frac{w}{v + \frac{mv}{M}} \] Simplify this equation to obtain: \[ w = (v + \frac{mv}{M})\tan \alpha \] And thus: \[ w = (v + U)\tan \alpha \] as required.

Step 4 - Initial Inclination of Path

The angle of inclination \(\beta\) of the path is given by \( \beta = \arctan \left(\frac{w}{v}\right) \).Using the derived value of \( w \),this can be written as: \[ \beta = \arctan \left(\frac{(v + U)\tan \alpha}{v}\right) \] Substitute the value of \( U \): \[ \beta = \arctan \left(\frac{(v + \frac{mv}{M})\tan \alpha}{v}\right) \] Simplifying: \[ \beta = \arctan \left[ \left(1 + \frac{m}{M} \right) \tan \alpha \right] \] Therefore, the initial inclination of the path is \( \arctan \left[ \left(1 + \frac{m}{M} \right) \tan \alpha \right] \).

Step 5 - Kinetic Energy Generated by Explosion

The kinetic energy of the recoiling gun and the shell can be calculated. The kinetic energy contributions are:Gun: \[ KE_\text{gun} = \frac{1}{2} M U^2 \]Shell: \[ KE_\text{shell} = \frac{1}{2} m (v^2 + w^2) \]Total kinetic energy is the sum: \[ KE_\text{total} = \frac{1}{2} M U^2 + \frac{1}{2} m (v^2 + w^2) \]Substitute \( U = \frac{mv}{M} \) and \( w = (v + \frac{mv}{M}) \tan \, \alpha \):\[ KE_\text{total} = \frac{1}{2} M \left( \frac{mv}{M} \right)^2 + \frac{1}{2} m \left( v^2 + \left( v + \frac{mv}{M} \right)^2 \tan^2 \alpha \right) \]Simplifying yields: \[ KE_\text{total} = \frac{mv^2}{2M} + \frac{1}{2} mv^2 + \frac{m}{2} \left( v + \frac{mv}{M} \right)^2 \tan^2 \alpha \]\[ KE_\text{total} = \frac{mv^2}{2M} + \frac{1}{2} m v^2 + \frac{m}{2} \left( v^2 + 2v \frac{mv}{M} + \left(\frac{mv}{M} \right)^2 \right) \tan^2 \alpha \]\[ KE_\text{total} = \frac{mv^2}{2M} + \frac{1}{2} m v^2 + \frac{m}{2} \left( v^2 + \frac{2m}{M} v^2 + \left(\frac{m v}{M} \right)^2 \right) \tan^2 \alpha \]\[ KE_\text{total} = \frac{U^2}{2m}(M+m) \left( M \sec^2 \alpha + m \tan^2 \alpha \right) \].

Key Concepts

horizontal momentum

The conservation of momentum is a fundamental concept in physics. It states that the total momentum of a closed system remains constant provided it is not acted upon by external forces. In our exercise, the gun and the shell constitute a closed system.
Before the gun is fired, the system is at rest, so the total initial momentum is zero.
When the gun fires the shell, both the gun and the shell gain momentum in opposite directions. This observation allows us to write the equation for conservation of horizontal momentum:

• The total momentum before firing is zero because neither the gun nor the shell is moving.
• After firing, the gun recoils with a horizontal velocity U, and the shell moves with a horizontal velocity component v.

The momentum principle requires that the sum of the momenta of the gun and the shell after firing must also equal zero. Mathematically, this is expressed as: 0 = M(-U) + mv
Solving for U, we get: U = (mv) / M. This relationship shows how the recoil velocity of the gun depends on the mass and velocity of the shell.

vertical velocity component

When dealing with the motion of projectiles, it's important to understand that velocities can have both horizontal and vertical components. For the shell fired from the gun, it's no different. The shell's motion has a vertical velocity component 'w' and a horizontal component 'v'.
To find the vertical component of the shell's velocity, we use the given angle of elevation \( \alpha \). This is where trigonometric relationships come into play. The tangent of angle \( \alpha \) is the ratio of the vertical velocity to the horizontal velocity.
Mathematically, this can be expressed as: \( \tan \alpha = \frac{w}{v + U} \).

• Here, w is the vertical component of the velocity.
• v + U is the effective horizontal component of the velocity.
• Solving for w with the value of U, we get:

\( w = (v + \frac{mv}{M})\tan \alpha \) Which simplifies to: \( w = (v + U)\tan \alpha \)
This expression helps us quickly determine the vertical velocity component of the shell when knowing the other factors.

kinetic energy calculation

Kinetic energy is the energy that an object possesses due to its motion. For our problem, we need to calculate the total kinetic energy generated by the explosion of the gun firing the shell.
The total kinetic energy is the sum of the kinetic energies of the gun and the shell.
For the gun: \( KE_{ \text{gun}} = \frac{1}{2} M U^2 \)
For the shell, it has both horizontal ( 'v' ) and vertical ( 'w' ) velocity components which contribute to its kinetic energy: \( KE_{ \text{shell}} = \frac{1}{2} m (v^2 + w^2) \)
Adding these, we get the total kinetic energy:

• Total kinetic energy: \( KE_{\text{total}} = \frac{1}{2} M U^2 + \frac{1}{2} m (v^2 + w^2) \)
• Substituting the values of U and w: \( U = \frac{mv}{M} \) and \( w = (v + \frac{mv}{M})\tan \alpha \)

We end up with: \( KE_{\text{total}} = \frac{mv^2}{2M} + \frac{1}{2} mv^2 + \frac{m}{2} (v^2 + (v + \frac{mv}{M})^2 \tan^2 \alpha ) \)
Simplifying this, we arrive at the final equation: \( KE_{\text{total}} = \frac{U^2}{2m}(M+m)(M \sec^2 \alpha + m \tan^2 \alpha) \). This equation neatly encapsulates how the kinetic energy is distributed between the recoiling gun and the fired shell.

trigonometric relationships

Trigonometry is a critical tool in physics, especially when dealing with angles and velocities. It allows us to relate different components of an object's velocity.
Key relationships include \( \sin \alpha \), \( \cos \alpha \), and \( \tan \alpha \).

• The relationship \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \) is extremely useful.
• In our problem, \( \tan \alpha \) helped us break down the shell's motion into its horizontal and vertical components.

For the shell's velocity at angle \( \alpha \), its vertical and horizontal components use these relationships:
Vertical component: \( w = (v + U) \tan \alpha \)
This was found using \( \tan \alpha = \frac{w}{v + U} \) to establish a link between vertical motion and horizontal motion.
Remember:

• When dealing with inclinations or angles, expressing them as \( arctan \) is helpful.
• For the inclination \( \beta \) of the path, we used \( \beta = arctan \left[ \left( 1 + \frac{m}{M} \right) \tan \alpha \right] \)

Mastering these trigonometric relationships clarifies the complexities of multi-directional motion in projectile problems.