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ActivPhysics OnLine Problem 2.4 Rocket Blasts Off Draw free body diagram. Choose upwards : + downwards: -

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Presentation on theme: "ActivPhysics OnLine Problem 2.4 Rocket Blasts Off Draw free body diagram. Choose upwards : + downwards: -"— Presentation transcript:

1 ActivPhysics OnLine Problem 2.4 Rocket Blasts Off Draw free body diagram. Choose upwards : + downwards: -

2 Problem 2.4: Question 1 Rocket Blasts Off Upward thrust on rocket= +5.0 N Gravity force= (m)(g) “ =(0.50kg)(-10.0 m/s 2 )

3 Total force on rocket = 5.0 N – 5.0 N = 0 N

4 Question 2: What is the height of rocket after 8 s if v i = +30 m/s? Total force on rocket =5.0 N–5.0N = 0N s f = s i + v t s f = 0 + (30 m/s) (8 s) = 240 m

5 Question 2: (continued) Describe in words the motion of the rocket. The rocket travels at a constant velocity of +30 m/s for the first 8 s while the engine provides thrust. After 8 s the rocket thrust stops, the rocket slows down but keeps moving upward for 3 more seconds. After 11 s the rocket begins to descend with a negative velocity.

6 Question 3: What is the rocket’s motion until t = 8 s? F th = +20.0 N m = 0.50 kg v i = 0 m/s g = -10 m/s 2 Total force on rocket = 20.0 N -5.0N = 15 N F total = m a = 15.0 N a = F total /m = 15.0 N/ 0.5 kg a = + 30 m/s 2 v = v i + a t v = 0 + (30 m/s 2 ) (8 s) = 240 m/s s f = s i + v i t + ½ (a t 2 ) s 8 = 0 + 0 + ½ (30 m/s 2 ) (8 s) 2 s 8 = 960 m

7 Question 4: What is the rocket’s acceleration after t = 8 s? After rocket engine has finished, Total force on rocket = -5.0N a = -5.0N/0.50 kg = -10 m/s 2 Rocket position at highest point? v = 0 at highest point. v = v i + a t 0 = 240 m/s + (-10 m/s 2 ) t t = 24 s after thrust is finished. s f = s i + v i t + ½ (a t 2 ) s f = 960 m + (240 m/s) (24 s) + ½ (-10 m/s 2 ) (24 s) 2 s f = 3840 m

8 Question 5: Consider the rocket’s motion until t = 8 s. F th = +20.0 Nm = 0.50 kg v i = +30 m/sg = -10 m/s 2 Total force on rocket = 20.0 N -5.0N = 15.0 N a = F total /m = 15.0 N/ 0.5 kg a = 30 m/s 2 v = v i + a t v = 30 m/s + (30 m/s 2 ) (8 s) v = 270 m/s s f = s i + v i t + ½ (a t 2 ) s 8 = 0 + (30 m/s) (8 s) + ½ (30 m/s 2 ) (8 s) 2 s 8 = 1200 m

9 Question 5: Part 2 Find the rocket’s maximum height. a = -10 m/s 2 after thrust is over. At rocket’s maximum height v = 0. v 2 = v i 2 + 2 a (s f - s i ) 0 = (270 m/s) 2 +2 (-10 m/s 2 ) (s f - 1200m) 2 (10 m/s 2 ) (s f - 1200m) = (270 m/s) 2 (s f - 1200m) = (270 m/s) 2 /20 m/s 2 s f = 4845 m


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