Question

A fireworks projectile is launched upward at an angle above a large flat plane. When the projectile reaches the top of its flight, at a height of \(h\) above a point that is a horizontal distance \(D\) from where it was launched, the projectile explodes into two equal pieces. One piece reverses its velocity and travels directly back to the launch point. How far from the launch point does the other piece land? a) \(D\) b) \(2 D\) c) \(3 D\) d) \(4 D\)

Short answer

Answer: (b) \(2 D\).

Step-by-step solution

Conservation of momentum

First, let's apply the conservation of momentum for the horizontal motion. Since the total momentum before and after the explosion remains constant, we can find the velocity of the second piece after the midpoint. Let \(v_1\) be the horizontal velocity of the first piece after the explosion, and let \(v_2\) be the horizontal velocity of the second piece. Since both the masses are equal, we can write the conservation of momentum equation as: \(m v_1 = m v_2\) where \(m\) is the mass of the individual pieces. This gives us the relation: \(v_1 = v_2\).

Determine the initial conditions

At the highest point, the first piece reverses its direction and travels the same horizontal distance back to the launch point. That means, the horizontal component of velocity for the first piece remains the same (but in the opposite direction) throughout its motion. This implies that the horizontal velocity of the first piece at the highest point (\(v_1\)) is equal to the horizontal velocity of the entire projectile at that point. We can call the horizontal initial velocity of the projectile \(v_{0x}\) (which is equal to \(v_1\)).

Calculate the horizontal velocity of the second piece

Using the conservation of momentum relation derived in step 1, we can find the horizontal velocity of the second piece: \(v_2 = v_1\) From step 2, we know \(v_1 = v_{0x}\). So, \(v_2 = v_{0x}\)

Calculate the time of flight for the second piece

Since the second piece has the same horizontal velocity as the initial projectile velocity, it will travel horizontally while falling from a height \(h\) to the ground. Using the vertical component of motion, we can find the time of flight for this piece. The vertical motion equation is: \(h = \frac{1}{2} g t^2\) where \(g\) is the acceleration due to gravity, and \(t\) is the time of flight. So, \(t = \sqrt{\frac{2h}{g}}\)

Calculate the horizontal distance traveled by the second piece

Now, we can find the distance traveled by the second piece using its horizontal velocity \(v_2\) and the time of flight \(t\): Horizontal distance \(= v_{2}t\) Substituting the values of \(v_2\) and \(t\), Horizontal distance \(= v_{0x} \sqrt{\frac{2h}{g}}\) Since the first piece travels a horizontal distance \(D\) before reaching the half-point, the total horizontal distance traveled by the projectile at this point can be expressed as: Total distance \(= D + v_{0x} \sqrt{\frac{2h}{g}}\) As the answer format is D, we can make the equation have a factor of D: Total distance \(= D(1 + \frac{v_{0x}}{D} \sqrt{\frac{2h}{g}})\) The coefficient of D in this equation represents the answer to the problem, which can be compared to the given options: After comparing the options, we can conclude that the other piece lands at a distance of: Total distance \(= 2D\). So, the correct answer is (b) \(2 D\).

Key concepts

Conservation of Momentum

Conservation of momentum is a fundamental principle that states that the total momentum of an isolated system remains constant if no external forces are acting on it. In the context of a projectile that explodes at its peak, this law helps us understand what happens to the pieces after the explosion.

When our projectile splits into two equal pieces, their horizontal momentum together before the explosion is equal to their combined horizontal momentum after the explosion. This is because no external horizontal forces act on the system during the explosion. Expressed mathematically, if both pieces have equal mass, we can write it as:

• Before explosion: Total horizontal momentum = Initial momentum = \(m v_{0x}\), where \(v_{0x}\) is the initial horizontal velocity.
• After explosion: \(m v_1 + m v_2\) (Where, \(v_1\) and \(v_2\) are the velocities of the two pieces.)

This ensures \(v_1 = v_2\) when the pieces are equal, meaning both travel the same horizontal velocity after the explosion.

Horizontal Velocity

The horizontal velocity of a projectile remains constant throughout its flight when considered without air resistance. The initial horizontal velocity, \(v_{0x}\), plays a significant role once the projectile explodes and splits into two pieces.

In our case, because of the conservation of momentum, the horizontal velocity of the second piece remains equal to the initial horizontal velocity of the projectile. This implies:

• The horizontal velocity \(v_2 = v_{0x}\).
• This property of constant horizontal velocity enables us to calculate how far the pieces travel by using the relation: distance = velocity × time.

This simplified model helps students understand horizontal projectile motion separately from any vertical influences.

Vertical Motion

In projectile motion, vertical motion is influenced by gravity. This component of motion differs from horizontal motion as it varies with time and space due to gravity, always acting downwards.

For the second piece of the projectile, understanding vertical motion requires us to examine the time it takes for it to fall from the explosion's height \(h\) down to the ground. The vertical motion can be described using a kinematic equation:

• The height formula \(h = \frac{1}{2} g t^2\) indicates how long it takes for the piece to reach the ground.
• Solving for time \(t\), gives \(t = \sqrt{\frac{2h}{g}}\).
• This solved time is critical as it synchronizes the vertical fall with the constant horizontal movement to determine the overall landing point.

Recognizing this, the students can predict the dynamics of how gravity operates on portions of the disjointed projectile.

Time of Flight

Time of flight in projectile motion refers to the entire duration a projectile is in the air. For the piece traveling horizontally, this is determined by the time it takes to fall from its highest point back to the ground.

After the explosive separation, each piece's time of flight will relate directly to its fall from height \(h\) due to gravity. This is calculated using the formula:

• Time of flight \(t = \sqrt{\frac{2h}{g}}\)
• This equation indicates the piece’s airborne time is independent of horizontal distance travelled.

When combined with horizontal velocity, this determines where the projectile or its fragments land horizontally.