Explanation Given information: The velocity ( v 0 ) of space vehicle is ( 1500 ft / s ) k . Calculation: Write the position vectors as follows: r A = 250 i + 250j + 2250k r B = 600 i + 1300j + 3200k r C = − 475 i − 950 j + 1900k There is no external force linear momentum is conserved. ( m A + m B + m C ) v 0 = m A v A + m B v B + m C v C v A = m A + m B + m c m A v 0 − m B m A v B − m C m A v C (1) Substitute 150 lb for m A , 300 lb for m B , and 450 lb for m C in Equation (1). v A = 150 + 300 + 450 150 v 0 − 300 150 v B − 450 150 v C = 6 v 0 − 2 v B − 3 v C (2) Substitute ( 1500 ft / s ) k for ( v 0 ) , ( 500 ft / s ) i + ( 1100 ft / s ) j + ( 2100 ft / s ) k for v B , and [ − 400 i + ( v C ) y j + ( v C ) z k ] in Equation (2). v A = 6 ( ( 1500 ) k ) − 2 ( ( 500 ) i + ( 1100 ) j + ( 2100 ) k ) − 3 ( [ − 400 i + ( v C ) y j + ( v C ) z k ] ) = − 3 ( v C ) y j − 3 ( v C ) z k + 200i − 2200j + 4800k The velocity of ( v A ) x is 200 . The velocity of ( v C ) y is − 3 ( v C ) y − 2200 . The velocity of ( v A ) z is − 3 ( v C ) z + 4800 . Express the conservation of angular momentum about O . ( H o ) 2 = ( H o ) 1 As the vehicle passes through the origin ( H o ) 1 = 0 ( H o ) 2 = r A × ( m A v A ) + r B × ( m B v B ) + r C × ( m C v C ) = 0 Divided by m A above Equation: r A × v A + m B m A r B × v B + m C m C r C × v C = r A × v A + 2 r B × v B + 3 r C × v C (3) Substitute 6 v 0 − 2 v B − 3 v C for v A in Equation (3). r A × v A + m B m A r B × v B + m C m C r C × v C = r A × ( 6 v 0 − 2 v B − 3 v C ) + 2 r B × v B + 3 r C × v C = r A × ( 6 v 0 − 2 v B − 3 v C ) + 2 r B × v B + 3 r C × v C = 3 ( r C − r A ) × v C + 6 r A × v 0 + 2 ( r B − r A ) × v B (4) Substitute 250 i+250j+2250k for r A , 600 i+1300j+3200k for r B , and − 475 i − 950 j+1900k for r C in Equation (4)

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Author: BEER

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Chapter 14.2, Problem 14.46P

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Find the velocity of space vehicle’s part A .

Solution Summary: The author calculates the velocity of space vehicle's part A. The velocity is (v_0).

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