sol8_1031 - HW Solutions 8 8.01 MIT Prof Kowalski Momentum...

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HW Solutions # 8 - 8.01 MIT - Prof. Kowalski Momentum and Collisions. 1 ) 8.55 Please review section 8.6 rocket Propulsion in the book. a ) The average thrust is impulse devided by time: F ave = J t (1) So the ration of the average thrust to maximum thrust is: F ave F max = J t F max = 10 13 . 3 × 1 . 7 = 0 . 442 (2) b ) Using the average force in equation (8.38): v ex = F t m = J m = 10 0 . 0125 = 800 m/s (3) c ) Using the result of part b in equation (8.40) - the sole equation of ”rocket science”: v v 0 = v ex ln( m 0 m ) (4) With m = m 0 m and v 0 = 0 we have: v = v ex ln( m 0 m 0 m ) = 800 ln( 0 . 0258 0 . 258 0 . 0125 ) = 530 m/s (5) 1
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2 ) 8.73 Please refer to figure 8.41 p.322. Using energy method including work: K L 1 + U L 1 + W other = E L 1 + W other = E L 2 = K L 2 + U L 2 (6) 1 2 mv 2 L 1 + mgy L 1 + W other = 1 2 mv 2 L 2 + mgy L 2 (7) I will measure the gravitation potential energy with respect to the horizontal line passing the bottom of the bowl.No non-conservative force is present so W other = 0. v L 1 = 0 y L 1 = R y L 2 = 0 (8) 0 + mgR = 1 2 mv 2 L 2 + 0 v L 2 = 2 gR (9) At the bottom, due to momentum conservation law total momentum before and after sticking together is the same: −→ P = −→ P (10) Where I used ” ” to denote the momentum just after the colli- sion. The momentum at the bottom of the bowl is horizontal so we need only the component of the above vector equation in horizontal direction: So we have mv L 2 + 0 = ( m + m ) v = 2 mv v = v L 2 2 (11) v is the velocity of the total mass 2m of the two boxes. Use again
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