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Ch. 09 - Linear Momentum
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 09 - Linear MomentumProblem 17a
Chapter 9, Problem 17a

A 195-kg projectile, fired with a speed of 116 m/s at a 60.0° angle, breaks into three pieces of equal mass at the highest point of its arc (where its velocity is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally. Determine the velocity of the third fragment immediately after the explosion.

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Step 1: Analyze the problem and identify the key principles. The projectile breaks into three equal fragments at the highest point of its arc. At this point, the velocity of the projectile is horizontal, and the total momentum of the system is conserved in both the horizontal and vertical directions. Additionally, the energy released in the explosion can be determined by comparing the kinetic energy before and after the explosion.
Step 2: Calculate the horizontal velocity of the projectile at the highest point. The horizontal velocity remains constant throughout the motion and is given by the horizontal component of the initial velocity: \( v_x = v_0 \cos(\theta) \), where \( v_0 = 116 \; \text{m/s} \) and \( \theta = 60.0^\circ \).
Step 3: Apply the principle of conservation of momentum in the horizontal direction. Before the explosion, the total horizontal momentum is \( p_x = m v_x \), where \( m = 195 \; \text{kg} \). After the explosion, the total horizontal momentum is the sum of the horizontal momenta of the three fragments. Two fragments have known velocities, and the velocity of the third fragment can be determined by solving for the unknown horizontal momentum contribution.
Step 4: Apply the principle of conservation of momentum in the vertical direction. Before the explosion, the vertical momentum is zero because the velocity is purely horizontal. After the explosion, the vertical momentum of the system must also sum to zero. Use this condition to solve for the vertical velocity component of the third fragment, given that one fragment moves vertically downward with a known speed.
Step 5: Calculate the energy released in the explosion. The total kinetic energy before the explosion is \( KE_{\text{before}} = \frac{1}{2} m v_x^2 \). After the explosion, calculate the total kinetic energy of the three fragments using \( KE_{\text{after}} = \sum \frac{1}{2} m_i v_i^2 \), where \( m_i \) and \( v_i \) are the masses and speeds of the fragments. The energy released is the difference: \( \Delta E = KE_{\text{after}} - KE_{\text{before}} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this scenario, when the projectile breaks into three pieces, the momentum before the explosion must equal the total momentum of the fragments immediately after the explosion. This concept is crucial for determining the velocity of the third fragment.
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Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v², where m is mass and v is velocity. In this problem, understanding how kinetic energy is distributed among the fragments before and after the explosion is essential for calculating the energy released during the explosion.
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Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to gravitational forces. The projectile's trajectory can be analyzed by separating its horizontal and vertical components. In this case, recognizing that the projectile reaches its highest point with a horizontal velocity allows for the application of momentum conservation principles to find the velocities of the fragments.
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