Question

A shell is fired from a gun with a muzzle velocity of \(466 \mathrm{~m} / \mathrm{s}\), at an angle of \(57.4^{\circ}\) with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming level terrain?

Short answer

The other fragment of the shell lands at a distance equal to twice the horizontal distance before the explosion. To find the specific value, finish the calculations as explained in the steps.

Step-by-step solution

Determine the vertical and horizontal components of the initial velocity

We can resolve the initial velocity of 466 m/s into two components: horizontal and vertical. The horizontal component (v_{x0}) can be found using the equation \(v_{x0} = v_0 \cdot cos(\Theta)\), and the vertical component (v_{y0}) can be found using the equation \(v_{y0} = v_0 \cdot sin(\Theta)\) where \(\Theta\) is the angle of projection, 57.4 degrees.

Calculate the time to reach the greatest height

At the highest point of the trajectory, the vertical speed is zero. Using the kinematic equation, \(v_{y} = v_{y0} - g \cdot t\), where g is the acceleration due to gravity (9.81 m/s²), time (t) to reach the highest point can be calculated as \(t = v_{y0} / g\).

Calculate the horizontal distance before the explosion

The horizontal distance (d) before the explosion can be calculated using the simple formula \(d = v_{x0} \cdot t\), as in the x-direction there is no acceleration.

Calculate the horizontal distance after the explosion

After the explosion, due to conservation of momentum, the second fragment continues with the same horizontal speed. Therefore, we calculate the total time it takes for the fragment to hit the ground, and then calculate the total horizontal distance it covers using the formula \(d = v_{x0} \cdot T\), where T is the total time (time to go up and come down).

Key concepts

Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. In projectile motion, kinematics helps us understand how an object moves through space under the influence of gravity. An excellent example is a shell fired from a gun.

In projectile motion, the object's path is a parabola. The initial velocity of the shell is crucial, and it can be broken down into horizontal and vertical components. This decomposition helps predict the shell's behavior throughout its trajectory.

When analyzing the motion's vertical aspect, gravity is the only acting force, causing the shell to accelerate downwards. The horizontal component is unaffected by gravity, allowing us to simplify calculations. We use kinematic equations to solve problems of time, velocity, and distance both vertically and horizontally.

Conservation of Momentum

The conservation of momentum is a fundamental principle in physics that applies even during explosions. Momentum, a product of mass and velocity, remains constant if no external forces act on the system. This principle becomes essential when considering the behavior of a projectile that explodes mid-air.

In the scenario where a shell explodes at the top of its trajectory, momentum conservation allows us to predict the motion of resulting fragments. Before the explosion, only the projectile's horizontal momentum needs consideration, as the vertical momentum is zero at the highest point.

When the shell explodes into two equal fragments, one fragment falls vertically with zero speed, while the other continues with the original horizontal velocity. The unchanged horizontal speed is crucial for determining how far the fragment lands from the gun. This is possible because of the conservation of horizontal momentum, as it allows us to assume no changes before or immediately after the explosion.

Horizontal and Vertical Components

Breaking down a projectile’s motion into horizontal and vertical components makes it simpler to understand and solve problems related to projectile motion. These components allow for separate analysis of motion influenced by different forces.

The horizontal component of motion refers to the constant velocity experienced by the projectile, as no horizontal forces (ignoring air resistance) act upon it. To find the initial horizontal velocity (\(v_{x0}\)), we use the equation \(v_{x0} = v_0 \cdot \cos(\Theta)\). This component can be used to calculate horizontal distances the projectile will travel under unchanged conditions.

The vertical component, determined by gravity, affects how high and how long the projectile moves up and down. The initial vertical velocity \(v_{y0}\) is calculated with \(v_{y0} = v_0 \cdot \sin(\Theta)\). While analyzing vertical movements, it's important to note the time to reach the highest point, where vertical velocity becomes zero before descending back downward.

By looking at both components individually, predicting the full trajectory of a projectile becomes much more manageable.