Question

Some kids are playing a dangerous game with fireworks. They strap several firecrackers to a toy rocket and launch it into the air at an angle of \(60^{\circ}\) with respect to the ground. At the top of its trajectory, the contraption explodes, and the rocket breaks into two equal pieces. One of the pieces has half the speed that the rocket had before it exploded and travels straight upward with respect to the ground. Determine the speed and direction of the second piece.

Short answer

To find the speed and direction of the second piece, use the following formulas: Speed: \(v_2 = \sqrt{(2v_{0x})^2 + (\frac{3}{2}v_{0y})^2}\) Direction: \(\phi = \arctan(\frac{3 \sin60^{\circ}}{4 \cos 60^{\circ}})\) Please note that the final answers depend on the initial speed \(v_0\) which needs to be given or determined through other information.

Step-by-step solution

Identify the givens and unknowns

We are given: 1. The launch angle, \(\theta = 60^{\circ}\) 2. The condition of one piece moving straight upward. We need to find the speed and direction of the second piece.

Calculate the initial velocity components

Let \(v_{0x}\) be the initial horizontal velocity and \(v_{0y}\) be the initial vertical velocity. We can find these values using trigonometry. Since the velocity vector makes a \(60^{\circ}\) angle with the horizontal ground, we have: \(v_{0x} = v_{0} \cos 60^{\circ}\) \(v_{0y} = v_{0} \sin 60^{\circ}\)

Apply conservation of momentum

When the rocket explodes, the law of conservation of momentum states that: Total momentum before explosion = Total momentum after the explosion \(\vec{p}_{before} = \vec{p}_{after}\) \(m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2} = (m_{1} + m_{2})\vec{v}_{0}\) Since both pieces have equal mass, let \(m_{1} = m_{2} = \frac{1}{2}m\), where \(m\) is the mass of the entire rocket. Plugging this in, we get: \(\frac{1}{2}m \vec{v}_{1} + \frac{1}{2}m \vec{v}_{2} = m \vec{v}_{0}\) We can divide both sides by \(\frac{1}{2}m\), to simplify the equation: \(\vec{v}_{1} + \vec{v}_{2} = 2 \vec{v}_{0}\)

Separate the vector components

Now, let's separate the components of the vectors: For X components: \(v_{1x} + v_{2x} = 2v_{0x}\) For Y components: \(v_{1y} + v_{2y} = 2v_{0y}\)

Use the given condition

Since one piece goes straight upward, its horizontal component is zero, i.e., \(v_{1x} = 0\). Its vertical component has half the speed of the rocket before the explosion, so \(v_{1y} = \frac{1}{2} v_{0y}\).

Calculate the second piece's velocity components

Using these conditions in the momentum conservation equations, we get: \(v_{2x} = 2v_{0x}\) \(v_{2y} = 2v_{0y} - v_{1y} = 2v_{0y} - \frac{1}{2} v_{0y} = \frac{3}{2}v_{0y}\)

Calculate the speed and direction of the second piece

The speed \(v_2\) of the second piece can be found using the Pythagorean theorem: \(v_{2} = \sqrt{v_{2x}^2 + v_{2y}^2} = \sqrt{(2v_{0x})^2 + (\frac{3}{2}v_{0y})^2}\) The direction \(\phi\) can be found using the arctangent function: \(\phi = \arctan(\frac{v_{2y}}{v_{2x}}) = \arctan(\frac{\frac{3}{2}v_{0y}}{2v_{0x}}) = \arctan(\frac{3 \sin60^{\circ}}{4 \cos 60^{\circ}})\) Therefore, the speed and direction of the second piece are: \(v_2 = \sqrt{(2v_{0x})^2 + (\frac{3}{2}v_{0y})^2}\) \(\phi = \arctan(\frac{3 \sin60^{\circ}}{4 \cos 60^{\circ}})\) Please note that the final answers depend on the initial speed \(v_0\) which needs to be given or determined through other information.

Key concepts

Conservation of Momentum in Explosions

When we witness fireworks illuminating the sky, fundamental physics principles are at play. Among these is the conservation of momentum, particularly noticeable during the explosion of a firework or, in our textbook problem, a rocket splitting into parts mid-air. According to the law of conservation of momentum, the total momentum of a closed system is constant if no external forces are acting upon it.

In our problem, the kids' rocket before the explosion has a certain momentum, which is equal to its mass multiplied by its velocity. When it explodes at the peak of its trajectory and breaks into two pieces, the sum of the momenta of both pieces must equal the momentum of the rocket just before the explosion. Even though the rocket's internal structure has changed due to the explosion, the system's total momentum remains unchanged. This principle allows us to solve for the unknown velocities of the rocket pieces post-explosion.

For students learning about momentum conservation, it's essential to remember that this only holds true in the absence of external forces – or when such forces balance out, making them negligible for the system in question.

Projectile Motion

The path taken by an object thrown into space is what we call projectile motion. This type of motion has two components: horizontal and vertical. What's intriguing about projectile motion, as illustrated by the rocket in our scenario, is that these two components are independent of each other.

While gravity affects the vertical component, causing the rocket to slow down, reach an apex, and then accelerate downward, the horizontal component remains constant if we ignore air resistance. The rocket's angle of launch, in this case, 60 degrees, significantly influences its trajectory, determining how far and how high it will go.

The initial angles given by trigonometric functions enable us to decompose the rocket's speed into horizontal and vertical components, a crucial step toward understanding the subsequent motion of the two pieces after the explosion. When dealing with projectile motion problems, always keep in mind the independence of the horizontal and vertical motions and how gravity solely impacts the vertical component.

Vector Components

Vectors play a pivotal role in physics; they are quantities that have both magnitude and direction. Our textbook problem requires decomposing the initial velocity of the rocket into horizontal and vertical components. This is done using trigonometry – a core aspect of vector analysis.

By utilizing the cosine function for the horizontal component and the sine function for the vertical, students can calculate the respective speeds. Why is this important? Because after the explosion, the rocket's pieces will have their own velocities that are also vectors, which can be analyzed component-wise. This strategy gives us the power to dissect complex movements into more manageable calculations.

Understanding vector components is fundamental, not just in projectile motion but in a vast array of physics problems where directional quantities come into play, and it is crucial to visualize and calculate them accurately.

Trigonometry in Physics

The field of trigonometry is a branch of mathematics that is deeply intertwined with physics. It aids in describing relationships involving angles and lengths in right-angled triangles, which are prevalent in physics problems like projectile motion. In the case of our exploding rocket, trigonometry was employed to find the initial horizontal and vertical velocity components of the rocket using the given launch angle.

Moreover, trigonometric functions are utilised again to calculate the direction of the second piece post-explosion. The function arctangent, or 'arctan', allows us to find the angle at which the piece moves relative to the horizontal. Mastery of trigonometric principles is thus indispensable when solving problems that involve angles, such as those found in vector decomposition, wave analysis, and rotational dynamics. The ability to fluidly apply trigonometry principles greatly enhances a student's problem-solving toolkit in physics.