Question

A 1000.-kg cannon shoots a 30.0-kg shell at an angle of \(25.0^{\circ}\) above the horizontal and a speed of \(500 . \mathrm{m} / \mathrm{s}\) with respect to the ground. What is the recoil velocity of the cannon?

Short answer

A) -13.62 m/s B) 13.62 m/s C) -453.97 m/s D) 453.97 m/s

Step-by-step solution

Identify the initial momentum of the system

Before the cannon is fired, it is stationary along with the shell. Therefore, their initial momentum is zero.

Determine the momentum of the fired shell

To find the momentum of the shell after being fired from the cannon, we can use the equation: $$ \text{momentum} = \text{mass} \times \text{velocity} $$ Since the shell is fired at a \(25.0^{\circ}\) angle above the horizontal, we only need to consider the horizontal component of its velocity. Using trigonometry, we get: $$ v_{\text{horizontal}} = |\vec{v}| \times \cos(\theta) $$ Now, we can plug in the values: $$ v_{\text{horizontal}} = 500 \ \text{m/s} \times \cos(25^\circ) $$ $$ v_{\text{horizontal}} \approx 453.97 \ \text{m/s} $$ Then, calculate the momentum: $$ \text{momentum}_{\text{shell}} = 30.0 \ \text{kg} \times 453.97 \ \text{m/s} $$ $$ \text{momentum}_{\text{shell}} \approx 13619.1 \ \text{kg} \cdot \text{m/s} $$

Apply the conservation of momentum principle

Now that we have the momentum of the fired shell, we can apply the conservation of momentum principle: $$ \text{momentum}_{\text{initial}} = \text{momentum}_{\text{final}} $$ In this case, the initial momentum is zero. Thus: $$ 0 = \text{momentum}_{\text{shell}} + \text{momentum}_{\text{cannon}} $$ Now we will solve for the momentum of the cannon: $$ \text{momentum}_{\text{cannon}} = -\text{momentum}_{\text{shell}} $$ $$ \text{momentum}_{\text{cannon}} \approx -13619.1 \ \text{kg} \cdot \text{m/s} $$

Calculate the recoil velocity of the cannon

Finally, we need to find the recoil velocity of the cannon using its momentum: $$ \text{velocity} = \frac{\text{momentum}}{\text{mass}} $$ Now, plug in the values: $$ v_{\text{recoil}} = \frac{-13619.1 \ \text{kg} \cdot \text{m/s}}{1000 \ \text{kg}} $$ $$ v_{\text{recoil}} \approx -13.62 \ \text{m/s} $$ The recoil velocity of the cannon is approximately -13.62 m/s (the negative sign indicates that the direction of the velocity is opposite to the direction of the fired shell).

Key concepts

Conservation of Momentum

When we think about the recoil velocity of the cannon, we are looking at a clear example of the conservation of momentum in action. The law of conservation of momentum states that in a closed system with no external forces, the total momentum before an event must equal the total momentum after the event.

In the case of our cannon and cannonball, before the cannon is fired, both the cannon and shell have zero momentum because they are not moving. As the cannonball is fired and gains momentum in one direction, the cannon itself must gain an equal amount of momentum in the opposite direction. This explains the recoil or backward movement that we observe in the cannon.

Grasping this concept is crucial because it not only applies to cannons but to any system where objects interact and exert forces on each other, such as cars in a collision or athletes pushing off one another in sports.

Projectile Motion

Projectile motion comes into play when an object is thrown or projected into the air, moving under the influence of gravity while continuing its original motion at the same time. This type of motion is two-dimensional and has both vertical and horizontal components that can be calculated separately.

In our example, the shell is projected at an angle, which means it has an initial horizontal velocity and a vertical velocity. The horizontal velocity affects how far the shell with travel, while the vertical velocity will determine the height of its arc. By understanding projectile motion, students can predict the trajectories of objects and solve for various factors, such as range, maximum height, and time of flight, using principles of kinematics.

Trigonometry in Physics

Trigonometry is a branch of mathematics that's essential for breaking down and understanding forces and movements in physics, especially in two-dimensional motion like projectile motion.

In the case of the cannon firing the shell, we use trigonometry to resolve the shell's velocity into horizontal and vertical components. Knowing the angle of \(25.0^\circ\) and the initial speed, we calculate the horizontal component of the velocity using the cosine function. Such calculations are common in physics problems, allowing us to understand and predict how different forces and velocities impact the movement of objects in different directions.

Momentum

Momentum is a vector quantity that is central to the laws of motion. It is the product of an object's mass and velocity, and it describes the quantity of motion an object has.

In this exercise, when the cannon fires a shell, the momentum of the shell can be calculated using its mass and the horizontal component of its velocity, as derived from trigonometry. The larger an object's momentum, the harder it is to stop. Momentum has both magnitude and direction, which means that the momentum of the cannon after firing is equal in magnitude to the shell's momentum but in the opposite direction. This concept helps explain various phenomena, from vehicle dynamics in engineering to sports where collisions and changes in motion occur.