Problem 108 - University Physics with Modern Physics

Question

A \(20.0-\mathrm{kg}\) projectile is fired at an angle of \(60.0^{\circ}\) above the horizontal with a speed of 80.0 \(\mathrm{m} / \mathrm{s} .\) At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

Step-by-Step Solution

Answer

(a) 1131.2 m; (b) 32000 J released.

Step 1: Break Down the Problem

The problem involves a projectile motion with an explosion at the peak. We need to find the distance the other fragment travels after the explosion and the energy released. The projectile's journey involves calculating the motion before and after the explosion.

Step 2: Determine Initial Velocity Components

Calculate the horizontal and vertical components of the projectile's initial velocity. The initial speed is 80 m/s at an angle of 60°.\[V_{0x} = V_0 \cdot \cos(60^{\circ}) = 80 \cdot 0.5 = 40 \, \mathrm{m/s}\]\[V_{0y} = V_0 \cdot \sin(60^{\circ}) = 80 \cdot \frac{\sqrt{3}}{2} \approx 69.3 \, \mathrm{m/s}\]

Step 3: Time of Flight Pre-Explosion

The vertical motion determines the time to reach the highest point. At the peak, the vertical velocity is zero.Use the formula: \[ t_{peak} = \frac{V_{0y}}{g} \approx \frac{69.3}{9.8} \approx 7.07 \, \mathrm{s} \]

Step 4: Total Horizontal Distance Before Explosion

Compute horizontal distance using horizontal velocity and time to the peak, which is double to account for ascent and descent.\[ x_{horizontal} = V_{0x} \cdot 2 \cdot t_{peak} = 40 \times 2 \times 7.07 \approx 565.6 \, \mathrm{m} \]

Step 5: Analyze the Explosion

After the explosion at the peak, one fragment falls vertically, and the second fragment continues horizontally. Mass halves, so each fragment is 10 kg.

Step 6: Velocity of Fragment 2 (Post-Explosion)

By conservation of momentum in the horizontal direction (since the vertical fragment starts with zero velocity):\[ m_{total} \cdot V_{0x} = m_{frag2} \cdot V_{f2x} \]\[ 20 \cdot 40 = 10 \cdot V_{f2x} \rightarrow V_{f2x} = 80 \, \mathrm{m/s} \]

Step 7: Distance Traveled by Fragment 2

Since fragment 2 continues with 80 m/s horizontally, compute its horizontal travel time from the peak (7.07s to ground):\[ x_{final} = V_{f2x} \cdot t_{descent} \approx 80 \times 7.07 \approx 565.6 \, \mathrm{m} \]Total distance from the firing point:\[ x_{total} = 565.6 + 565.6 = 1131.2 \, \mathrm{m} \]

Step 8: Calculate Energy Released

Calculate the kinetic energy before and after the explosion:**Initial Kinetic Energy**: \[ KE_{initial} = \frac{1}{2} m V^2 = \frac{1}{2} \cdot 20 \cdot 80^2 = 64000 \, \mathrm{J} \]**Final Kinetic Energy**:- Fragment 1 has no velocity.- Fragment 2: \[ KE_{frag2} = \frac{1}{2} \cdot 10 \cdot 80^2 = 32000 \, \mathrm{J} \]Energy released:\[ \Delta E = KE_{initial} - KE_{frag2} = 64000 - 32000 = 32000 \, \mathrm{J} \]

Key Concepts

Kinetic Energy

In physics, kinetic energy refers to the energy possessed by an object due to its motion. It is an essential concept to understand when dealing with projectile motion, especially in explosive scenarios. For any object with mass that is moving, its kinetic energy can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass and \( v \) is the velocity of the object.

In the context of the provided exercise, the initial kinetic energy of the projectile before the explosion is given as 64,000 Joules. This is calculated using the total mass (20 kg) and the velocity (80 m/s) of the projectile just before it reaches the peak of its trajectory.

After the explosion, the kinetic energy situation changes drastically. Only the fragment that continues to move horizontally contributes to the final kinetic energy value, which is significantly reduced to 32,000 Joules due to the halved mass and maintained velocity of 80 m/s. The difference between these two energy states indicates the energy released during the explosion.

Momentum Conservation

Momentum is a vector quantity that represents the product of an object's mass and velocity. It is crucial to understand that in closed systems, momentum is always conserved. This principle of conservation of momentum states that the total momentum before an event must equal the total momentum after the event, provided there is no external force acting on it.

In the projectile motion problem we are examining, the explosion at the peak of the trajectory provides a perfect scenario to apply momentum conservation. Prior to the explosion, the projectile has a horizontal momentum calculated by multiplying its mass (20 kg) and horizontal velocity component (40 m/s). Thus, the total horizontal momentum is 800 kg·m/s.

When the projectile explodes, it splits into two fragments, each of mass 10 kg. One fragment drops vertically with no horizontal velocity, effectively making its horizontal momentum zero. The momentum conservation principle guides the calculation for the velocity of the second fragment that continues in the horizontal direction. Given the initial total momentum, this fragment must travel with a velocity that maintains the 800 kg·m/s momentum, calculated to be 80 m/s horizontally.

Explosion Dynamics

Explosion dynamics in projectile motion involve sudden forceful separation, resulting in changes in motion characteristics and energy distribution. The moments right after the explosion are crucial to understand, especially concerning the movement and energy of resulting fragments.

In the case of the projectile exercise, the explosion occurs at the peak, where the vertical speed is zero. Dynamics involve converting potential energy and the prior kinetic energy into kinetic energy of the fragments. Key factors involved include:

• Fragmentation: The projectile splinters into two pieces of equal mass (10 kg each).
• Trajectory Change: One fragment falls directly downwards, indicating its post-explosion kinetic velocity is non-horizontal.
• Energy Release: Energy before and after the explosion differs, highlighting energy used during the fragmentation.
• Travel Distance: The second fragment maintains kinetic velocity, covering significant distance horizontally.

The essence of explosion dynamics within this problem is encapsulated in managing the balance of energy and motion post-fragmentation, showing both the dramatic nature of explosions and the predictable application of physical laws such as energy conservation and momentum.