Explanation Given information: The mass ( m ) of shell is 6 kg . The mass of fragment A is 2 kg . The mass of fragment B is 1 kg . The mass of fragment C is 3 kg . The velocity ( v 0 ) of shell is ( 12 m / s ) i − ( 9 m / s ) j − ( 360 m / s ) k . Calculation: Refer from Figure P14.24 in the textbook. Express the position vector of A as follows: r A = − 1.5 i Express the position vector of B as follows: r B = 4 i + 2j Express the position vector of C as follows: r C = − 3 j Express the position vector of A / D as follows: r A / D = − 1.5 i − 4k r A / D = 1.5 2 + 4 2 = 18.25 = 4.272 Express the position vector of B / D as follows: r B / D = 4 i + 2j − 4 k r B / D = 4 2 + 2 − ( 4 ) 2 = 16 + 4 + 16 = 6 Express the position vector of C / D as follows: r C / D = − 3 j − 4k r C / D = ( − 3 ) 2 − ( 4 ) 2 = 9 + 16 = 5 Find the unit vector along ( λ A ) using the relation as follows: λ A = 1 r A / D ( r A / D ) (1) Substitute 4.272 for r A / D and − 1.5 i − 4 k for r A / D in Equation (1). λ A = 1 4.272 ( − 1.5 i − 4 k ) Find the unit vector along ( λ B ) using the relation as follows: λ B = 1 r B / D ( r B / D ) (2) Substitute 6 for r B / D and 4 i + 2j − 4 k for r B / D in Equation (2). λ B = 1 6 ( 4 i + 2j − 4 k ) Find the unit vector along ( λ C ) using the relation as follows: λ C = 1 r C / D ( r C / D ) (3) Substitute 5 for r C / D and − 3 j − 4 k for r C / D in Equation (3). λ C = 1 5 ( − 3 j − 4 k ) Consider that the change in the elevation due to gravity is neglected. Then, the velocity vectors after the explosion have the unit vector direction

12th Edition

ISBN: 9781264095032

Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 14.1, Problem 14.25P

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Find the speed of each fragment A, B, and C immediately after the explosion.

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Chapter 14.1, Problem 14.24P

Chapter 14.1, Problem 14.26P

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Find the speed of each fragment A , B , and C immediately after the explosion.

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