Solution Midterm 06

# Solution Midterm 06 - Solutions to 105 Midterm Ed Groth...

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Unformatted text preview: Solutions to 105 Midterm Ed Groth, 23-Oct-2006 1. Let F be the lift force on the plane. This is applied at angle θ b from the vertical. The sum of the forces must equal the mass times the acceleration. In the vertical direction we have F cos θ b- mg = 0 or F cos θ b = mg . In the horizontal direction, F sin θ b = mv 2 /r . Divide one by the other to find tan θ b = v 2 /gr or r = v 2 g tan θ b . 2. The thrust is F t =- u ( dM/dt ) If you don’t remember this you should derive the rocket equation. Let v ( t ) be the velocity of the rocket (relative to the ground). Let u be the velocity of the exhaust relative to the rocket. (So u is negative.) Let M ( t ) be the mass of the rocket and let dM ( t ) /dt be the rate of change of the rocket mass (because fuel is being expelled). Consider the interval t-→ t + dt . Conservation of momentum tells us that M ( t ) v ( t ) = ( M + dM )( v + dv ) + (- dM )( v + u + dv ) = Mv + M dv- u dM ....
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Solution Midterm 06 - Solutions to 105 Midterm Ed Groth...

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