Explanation Given data: Mass of the weather rocket ( m ) is 1500 kg. The rocket accelerates ( a y ) upward at 10.0 m s 2 . The rocket explodes 2.00 s after liftoff and breaks into two fragments. Mass of the heavier fragment is twice the lighter fragment. Therefore, consider the mass of the lighter fragment ( m L ) is 500 kg, and the heavier fragment ( m H ) is 1000 kg. The height reached by the lighter fragment y 3 is 530 m. Formula used: Refer to the (2.11) in the textbook, and modify the expression of kinematic equation for final velocity of a particle in vertical direction as follows. ( v y ) f = ( v y ) i + a y Δ t (1) Here, ( v y ) i is the initial velocity of the particle, a y is the acceleration of the particle, and Δ t is the duration of the acceleration period. Refer to the (2.12) in the textbook, and modify the expression of kinematic equation for final position of a particle in vertical direction as follows. y f = y i + ( v y ) i Δ t + 1 2 a y ( Δ t ) 2 (2) Here, y i is the initial position of the particle. Refer to the (2.13) in the textbook, and modify the expression of kinematic equation as follows. ( v y ) f 2 = ( v y ) i 2 + 2 a y Δ y (3) Here, Δ y is the distance between the particle’s initial and final position. From the given data, the rocket breaks into two fragments. As the two fragments stick together before the explosion, the collision is considered as inelastic collision. Write the expression for momentum of conservation for the given system as follows. m L ( v L ) f + m H ( v H ) f = ( m L + m H ) ( v ) i (4) Here, m L is the mass of the lighter fragment, m H is the mass of the heavier fragment, ( v L ) f is the final velocity of the lighter fragment, ( v H ) f is the final velocity of the heavier fragment, and ( v ) i is the initial velocity of the rocket (two fragments stick together before the collision). Explanation: From the given data, draw the constructive diagram for the given system as shown in Figure 1. In Figure 1, ( v y ) 0 is the velocity of the rocket at initial position, which is considered as 0 m s , y 0 is the height at which the rocket stays at initial position, t 0 is the instant at initial position of the rocket, a y is the acceleration of the rocket before it breaks, ( v y ) 1 is the velocity of the rocket at the instant of t 1 (time at the rocket breaks into two fragments), y 1 is the height at which the rocket breaks into two fragments, t 1 is the instant at which the rocket breaks into two fragments, y 2 is the height at which the lighter fragment at its initial position, t 2 is the instant at which the lighter fragment at its initial position, ( v L y ) 2 is the velocity of the lighter fragment just after the explosion, y 3 is the height at which the lighter fragment at its final position, t 3 is the instant at which the lighter fragment at its final position, ( v H y ) 2 is the velocity of the heavier fragment just after the explosion, which is to be determined, and ( v L y ) 2 is the velocity of the lighter fragment when it reaches to its final position, which is considered as 0 m s . Calculation of velocity of the rocket at the instant of t 1 [ ( v y ) 1 ] : Modify the expression in equation (1) for velocity of the rocket at the instant of t 1 as follows. ( v y ) 1 = ( v y ) 0 + a y Δ t Substitute 0 m s for ( v y ) 0 , 10.0 m s 2 for a y , and 2.00 s for Δ t , ( v y ) 1 = 0 m s + ( 10.0 m s 2 ) ( 2.00 s ) = 20 m s Calculation of height at which the rocket breaks into two fragments ( y 1 ) : Modify the expression in equation (2) to find the height at which the rocket breaks into two fragments as follows. y 1 = y 0 + ( v y ) 0 Δ t + 1 2 a y ( Δ t ) 2 Substitute 0 m for y 0 , 0 m s for ( v y ) 0 , 10

4th Edition

ISBN: 8220106755235

Author: Field

Publisher: PEARSON

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Question

Chapter 9, Problem 67GP

To determine

To find: The speed and direction of the heavier fragment just after the explosion.

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Chapter 9, Problem 66GP

Chapter 9, Problem 69GP

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