Question

A 350 -kg cannon, sliding freely on a frictionless horizontal plane at a speed of \(7.5 \mathrm{~m} / \mathrm{s}\), shoots a 15 -kg cannonball at an angle of \(55^{\circ}\) above the horizontal. The velocity of the ball relative to the cannon is such that when the shot occurs, the cannon stops cold. What is the velocity of the ball relative to the cannon?

Short answer

Answer: The velocity of the cannonball relative to the cannon is approximately 286.53 m/s at an angle of 55 degrees above the horizontal.

Step-by-step solution

Understand the given data

We are given the mass of the cannon (350 kg), the mass of the cannonball (15 kg), and the initial velocity of the cannon (\(7.5 \,\text{m/s}\)). The shot occurs at an angle of \(55^{\circ}\) above the horizontal, and the cannon stops after the shot.

Calculate the initial momentum

The initial momentum is completely horizontal, and it is the product of the mass of the cannon and its initial velocity. \(p_{initial} = m_{cannon} \cdot v_{initial} = 350 \,\text{kg} \cdot 7.5\, \text{m/s} = 2625\, \text{kg m/s}\)

Calculate the final momentum vectors

Since the cannon stops after the shot, its final momentum is zero. Now, we have to find the final momentum of the cannonball, with its velocity having components \(v_x\) and \(v_y\). The total momentum of the system must be the same before and after the shot, so we have two equations to solve:$ \begin{cases} m_{ball} \cdot v_x = 2625 \\ m_{ball} \cdot v_y = 0 \end{cases} $

Calculate the final momentum components using the angle

We can use the angle of the shot to relate the components of the final velocity of the cannonball. Using trigonometric functions: \(v_x = v \cdot \cos(55^{\circ})\) , \(v_y = v \cdot \sin(55^{\circ})\)

Solving the momentum equations

From Step 3, we know that \(m_{ball} \cdot v_y = 0\). Since the mass of the cannonball is not zero, it means \(v_y = 0\). Now, we can use the equation \(m_{ball} \cdot v_x = 2625\) to find the horizontal component of the final velocity of the cannonball: \(v_x = \frac{2625}{15} = 175\, \text{m/s}\). Since \(v_y = 0\), we obtain \(v_x = v \cdot \cos(55^{\circ})\). Therefore, \(v = \frac{v_x}{\cos(55^{\circ})} = \frac{175 \,\text{m/s}}{\cos(55^{\circ})} \approx 286.53\, \text{m/s}\)

Conclusion

The velocity of the cannonball relative to the cannon is approximately \(286.53\, \text{m/s}\) at an angle of \(55^{\circ}\) above the horizontal.

Key concepts

Conservation of Momentum

The conservation of momentum is a fundamental principle of physics stating that the total linear momentum of a closed system will remain constant over time, provided no external forces are acting on it. This concept applies to a variety of situations, including collisions and explosions, where different objects in a system exchange momentum but the total system momentum remains unchanged.

In the cannonball momentum problem, we're dealing with a system consisting of a cannon and a cannonball. Initially, the cannon has momentum due to its motion, and the cannonball is stationary within it. When the cannon fires the cannonball, external forces (such as gravity or air resistance) are neglected, assuming a frictionless surface, which effectively makes our system closed. Therefore, the momentum before the shot must equate to the momentum after the shot, satisfying the conservation principle.

The initial momentum is simply the mass of the cannon multiplied by its velocity. After the cannonball is fired, the cannon stops, which means its final momentum is zero. The entire system's momentum is now solely in the motion of the cannonball, which the problem requires us to solve for.

Trigonometric Functions in Physics

Trigonometric functions are indispensable in physics, especially when analyzing phenomena involving angles, such as projectile motion. These functions relate the angles of a triangle to its side lengths and are used to decompose vectors into their orthogonal components – usually horizontal (x-axis) and vertical (y-axis) components.

In our cannonball scenario, the motion of the projectile following its path can be split into two separate motions – one horizontal and one vertical. Using the angle of launch, trigonometric functions help us to find these components. The cosine function gives us the ratio of the adjacent side (horizontal component) to the hypotenuse (resultant velocity), while the sine function gives the ratio of the opposite side (vertical component) to the hypotenuse.

Given the shot's angle and the initial velocity, we can calculate the horizontal and vertical velocities using the cosine and sine of the launch angle, thereby utilizing trigonometric functions to solve for the motion of the projectile.

Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. It is a two-dimensional motion with a vertical acceleration due to gravity and a constant horizontal velocity, assuming no air resistance.

The motion of the cannonball, once fired, is an example of projectile motion. However, in the provided problem, we encounter a somewhat special case where the vertical component of the velocity is zero right after the shot. Normally, in projectile motion, an object's trajectory is parabolic, and its path is determined by both its initial vertical and horizontal velocities, with gravity affecting the vertical motion.

Due to the lack of vertical movement at the moment of firing, the typical projectile motion calculations do not apply directly. Yet, understanding the concept is crucial for more complex problems where both velocity components must be considered and the effects of gravity are included in determining the trajectory of the object.

Momentum Calculation

Momentum calculation is a fundamental task in physics. It involves determining the magnitude and direction of the momentum of an object, which is a product of its mass and velocity. Momentum is a vector quantity, which means it has both magnitude and direction.

In our example, to compute the momentum of the cannon and cannonball, we multiply their respective masses by their velocities. Since the system is conserved and isolated, the horizontal momentum before the shot must equal the horizontal momentum after the shot, which allows us to set up an equation with the unknown velocity of the cannonball. Calculating the final velocity of the cannonball after the cannon stops involves solving this equation and using the angle of the shot to find the velocity components.

The problem simplifies to finding the horizontal component of the ball's velocity since the final vertical component is zero. Dividing the final horizontal momentum by the mass of the cannonball gives us the horizontal velocity, and then, using the angle, the total velocity can be found.