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In a volcanic eruption, a \(2400\)-kg boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands 318 \(\mathrm{m}\) directly north of the point of the explosion. Where will the other fragment land? Neglect any air resistance.

Short Answer

Expert verified
The heavier fragment lands 318 meters south of the explosion point.

Step by step solution

01

Determine Masses of the Fragments

Let's denote the mass of the lighter fragment as \( m_1 \) and the heavier fragment as \( m_2 \). Since the heavier fragment is three times the mass of the lighter fragment, we have:\[m_2 = 3m_1. \]The sum of the masses of the fragments is equal to the total mass of the boulder:\[m_1 + m_2 = 2400 \text{ kg}. \]Substitute \( m_2 = 3m_1 \) into the equation:\[m_1 + 3m_1 = 2400 \text{ kg},\]\[4m_1 = 2400 \text{ kg}. \]Solving for \( m_1 \), we get:\[m_1 = 600 \text{ kg}.\]Thus, the mass of the heavier fragment, \( m_2 \), is:\[m_2 = 3 \times 600 = 1800 \text{ kg}.\]
02

Apply Conservation of Momentum

Initially, the momentum of the boulder in the vertical direction is zero at the highest point. After the explosion, momentum is conserved.Since the lighter fragment starts out with a horizontal velocity causing it to land directly north, and given the absence of initial vertical momenta, in the horizontal direction, we maintain:\[0 = m_2 v_2 + m_1 v_1, \]where \( v_1 \) and \( v_2 \) are the velocities of the lighter and heavier fragments, respectively. Since \( v_1 \) is known in magnitude and direction, we derive \( v_2 \) to balance the equation.
03

Calculate Velocity of Lighter Fragment

Given that the lighter fragment moves horizontally and lands 318 m north, we can deduce its velocity by analyzing the time of flight and horizontal distance:The time \( t \) the lighter fragment takes to hit the ground is the time for gravity to bring it down:\[t = \sqrt{\frac{2h}{g}},\]where \( h \) is the height of explosion, and \( g \) is the acceleration due to gravity. Once you solve for time, you substitute into:\[v_1 = \frac{318 \text{ m}}{t}.\]
04

Calculate Distance for Heavier Fragment

Since the heavier fragment has an equal but opposite momentum to conserve momentum, knowing \( m_1 v_1 = m_2 v_2 \), we solve for \( v_2 \) given:\[v_2 = \frac{m_1 v_1}{m_2}.\]The same time \( t \) in step 3 applies to the heavier fragment's horizontal motion. Thus, the distance \( x \) that the heavier fragment travels southward can be calculated as:\[x = v_2 t.\]Recognizing that the heavier fragment starts from the same point of the explosion and opposite direction, the final landing position relative to the explosion is 318 meters south of the origin.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause this motion. It focuses on parameters like velocity, acceleration, displacement, and time. Understanding kinematics is crucial for analyzing the volcanic eruption scenario where the boulder fragments are in motion post-explosion.
When discussing kinematics, we often break down the motion into convenient axes, typically horizontal and vertical. In our exercise, at the highest point of the trajectory, the boulder has no vertical motion due to gravity, and the explosion imparts horizontal velocities to the fragments.
Kinematics helps us determine variables such as the time it takes for the fragments to fall back to the ground and their respective horizontal distances traveled. The equations of motion are:
  • Vertical motion: affected by gravity, using equations like \(h = rac{1}{2}gt^2\).
  • Horizontal motion: unaffected by gravity, using equations like \(d = vt\).
These principles of kinematics allow us to address parts of the initial velocity, time of flight, and displacement needed to solve the exercise.
Projectile Motion
Projectile motion is characterized by objects moving along a curved path under the influence of gravity alone. It involves both horizontal and vertical motion. In the case of the volcanic eruption, once the boulder explodes, the fragments follow projectile motion paths.
The lighter fragment, with its only initial horizontal velocity, shows a simple projectile motion scenario. The gravity acts on it vertically, causing a parabolic trajectory until it lands.
Key features of projectile motion in this scenario include:
  • The horizontal component of motion: determines how far the object travels.
  • The vertical component due to gravity: governs the time the object is airborne.
To find how far the lighter fragment lands:
  • Found the time using vertical motion: \(t = \sqrt{\frac{2h}{g}}\)
  • Calculated its horizontal velocity, \(v_1\) after landing 318 m north.
The important part is understanding how the explosive moment dictates path while other forces, like gravity, take part in the motion balance.
Explosions
Explosions in physics often relate to problems of conservation of momentum and rapid energy release. In our exercise, the explosion sends fragments of the boulder in different paths, distributing energy and momentum.
The principle of conservation of momentum is critical at the point of explosion. Initially, the boulder had no net momentum at its peak; hence, any post-explosion velocities indicate the balancing of their momentum between fragments.
Mathematically, for the explosive event:
  • The total momentum before and after the explosion remains constant.
  • Use \(0 = m_2 v_2 + m_1 v_1\) for the momentary balance after the explosion.
The key takeaways include:
  • Momentum is a vector, thus conserving direction and magnitude.
  • Heavier fragment moves with a velocity that, when multiplied with its mass, equals that of the lighter's bonus opposite movement.
Understanding these interactions clarifies why solving for velocities and subsequent movements of both fragments reflects on momentum conservation. This conservation ensures that even without initial vertical forces, horizontal components find balance.

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Most popular questions from this chapter

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