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Ch 11: Impulse and Momentum
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 11: Impulse and MomentumProblema 53
Capitolo 11, Problema 53

A 1500 kg weather rocket accelerates upward at 10 m/s². It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

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Step 1: Calculate the velocity of the rocket just before the explosion using the formula for uniformly accelerated motion: \( v = u + at \), where \( u \) is the initial velocity (0 m/s), \( a \) is the acceleration (10 m/s²), and \( t \) is the time (2.0 s).
Step 2: Determine the total momentum of the rocket just before the explosion using the formula \( p = mv \), where \( m \) is the mass of the rocket (1500 kg) and \( v \) is the velocity calculated in Step 1.
Step 3: Apply the principle of conservation of momentum. After the explosion, the total momentum of the system (rocket fragments) must equal the total momentum before the explosion. Let the masses of the fragments be \( m_1 \) (lighter fragment) and \( m_2 \) (heavier fragment), where \( m_2 = 2m_1 \). Write the momentum conservation equation: \( p_{initial} = m_1v_1 + m_2v_2 \).
Step 4: Use the given maximum height of the lighter fragment (530 m) to calculate its velocity just after the explosion. Use the kinematic equation \( v^2 = u^2 + 2as \), where \( v \) is the final velocity (0 m/s at maximum height), \( u \) is the initial velocity, \( a \) is the acceleration due to gravity (-9.8 m/s²), and \( s \) is the displacement (530 m). Solve for \( u \), which represents the velocity of the lighter fragment immediately after the explosion.
Step 5: Substitute the velocity of the lighter fragment (\( v_1 \)) and its mass (\( m_1 \)) into the momentum conservation equation from Step 3 to solve for the velocity of the heavier fragment (\( v_2 \)). Ensure the direction of \( v_2 \) is consistent with the conservation of momentum.

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Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for understanding how the rocket accelerates upward and how the forces change during the explosion. The formula F = ma (force equals mass times acceleration) helps in calculating the forces involved before and after the explosion.
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Intro to Forces & Newton's Second Law

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before an event must equal the total momentum after the event, provided no external forces act on it. In this scenario, the momentum of the rocket just before the explosion must equal the combined momentum of the two fragments immediately after the explosion, allowing us to determine the speed and direction of the heavier fragment.
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Conservation Of Momentum

Kinematics of Projectile Motion

Kinematics involves the study of motion without considering the forces that cause it. In this case, the motion of the lighter fragment after the explosion can be analyzed using kinematic equations to determine its maximum height and the time it took to reach that height. This information is essential for understanding the dynamics of the fragments post-explosion and how they relate to each other.
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Introduction to Projectile Motion
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