Question

A firecracker is tossed straight up into the air. It explodes into three pieces of equal mass just as it reaches the highest point. Two pieces move off at \(120 \mathrm{m} / \mathrm{s}\) at right angles to each other. How fast is the third piece moving?

Short answer

The speed of the third piece is approximately \(169.7 \, \mathrm{m/s}\).

Step-by-step solution

Understand the conservation of momentum

At the highest point, the vertical component of the velocity of the firecracker is zero. Since the firecracker explodes at the peak, the system's center of mass must be conserved. Therefore, the total momentum before and after the explosion must be zero because the initial momentum is zero.

Set up the momentum equation

Let the mass of each piece be \(m\). Consider a coordinate system where two pieces have velocities \(v_1 = 120\, \mathrm{m/s}\) along the x-axis and \(v_2 = 120\, \mathrm{m/s}\) along the y-axis, respectively. Let's denote the velocity of the third piece as \(v_3\) at an angle \(\theta\). According to the conservation of momentum, the vector sum of the momenta of all three pieces must be zero.

Write the momentum equations

Since momentum in each direction must be zero, we can set up our equations:\[m \cdot v_1 + m \cdot v_{3x} = 0\quad\text{(x-direction)}\]\[m \cdot v_2 + m \cdot v_{3y} = 0\quad\text{(y-direction)}\]Given \(v_1 = 120\, \mathrm{m/s}\) and \(v_2 = 120\, \mathrm{m/s}\),\[m \cdot 120 + m \cdot v_{3x} = 0\]\[m \cdot 120 + m \cdot v_{3y} = 0\]

Solve for the third piece's velocity components

Solving the x-equation:\[v_{3x} = -120 \, \mathrm{m/s}\]Solving the y-equation:\[v_{3y} = -120 \, \mathrm{m/s}\]

Calculate the magnitude of the third piece's velocity

The magnitude of the velocity is found using Pythagorean theorem:\[v_3 = \sqrt{v_{3x}^2 + v_{3y}^2}\]Substitute in the values:\[v_3 = \sqrt{(-120)^2 + (-120)^2} = \sqrt{14400 + 14400} = \sqrt{28800} = 120 \sqrt{2} \, \mathrm{m/s}\]

Key concepts

Explosion Physics

When a firecracker explodes at its peak height, it's a classic example of explosion physics. Here, the firecracker starts with zero velocity since it was tossed straight up, and the explosion occurs at the highest point where its vertical velocity is zero. This sets the stage for understanding momentum conservation. In explosion physics, the explosion of the firecracker can be likened to a small-scale version of energy release, similar to much larger explosions like those in stars or fireworks. The aim is to understand how the firecracker breaks into three pieces and their subsequent motion.
While the explosion imparts energy, it's essential to realize that momentum is conserved. This means that the system's total momentum vector remains zero because it starts at zero. Each piece of the firecracker gains a new velocity, keeping the vector sum of their momenta zero. Thus, even though kinetic energy is gained by the pieces, momentum conservation is the guiding principle in determining motion post-explosion.

Momentum Equation

The principle of the conservation of momentum is crucial in solving problems involving explosions. Momentum is defined as the product of mass and velocity (\( p = m \cdot v \)). In this scenario, the firecracker breaks into three equal pieces, each with a distinct velocity.
The system's momentum pre-explosion is zero, which means post-explosion, the total must also sum to zero. This requirement gives rise to our momentum equations. Assume that two pieces move perpendicular to one another with known velocities, say along the x and y axes. Thus, the momentum along each axis can be challenged to equal the third piece’s momentum. Through the two main axes, we write momentum balance equations:

• For the x-direction: \[ m \cdot v_1 + m \cdot v_{3x} = 0 \]
• For the y-direction: \[ m \cdot v_2 + m \cdot v_{3y} = 0 \]

Importantly, these equations help determine the velocity components of the third piece. Uniformly solving these shows that each component of this piece's velocity is opposite to that of the already known pieces due to the equal and opposite nature of the derived equations.

Pythagorean Theorem

Finding the magnitude of the third firecracker piece's velocity involves applying the Pythagorean theorem. This theorem is rooted in geometry and is commonly used in physics to determine a resultant vector when other vectors are perpendicular. In our context, the theorem is expressed as \( c = \sqrt{a^2 + b^2} \), where \( c \) is the hypotenuse or resultant vector, and \( a \) and \( b \) are components of this vector.
For this exercise, we've found the velocity components of the third piece after solving the momentum equations: \( v_{3x} = -120 \, \mathrm{m/s} \) and \( v_{3y} = -120 \, \mathrm{m/s} \).
Applying the Pythagorean theorem, we calculate:
\[ v_3 = \sqrt{(-120)^2 + (-120)^2} = \sqrt{14400 + 14400} = \sqrt{28800} = 120 \sqrt{2} \; \mathrm{m/s} \]
Thus, the Pythagorean theorem allows us to determine the vector magnitude or the overall speed of the third piece, effectively combining its two perpendicular velocity components into one unified velocity.