Physics and Dream World - Assignment Example

Dream World Assignment If youre traveling in a fast car. What feelings do you experience in each of the following cases. The car a) Drops suddenly: Solution: If the car drops suddenly, my body keeps moving because it has linear momentum. This is related to the Newtons law of momentum: Heavy bodies prefer to keep moving in the same way as its moving right now. b) Rises abruptly after dipping Solution: If the car rises abruptly after dipping, my body keeps moving upward because it has linear momentum. c) Turns sharply Solution: If the car turns sharply, my body keeps moving in the same direction because it has linear momentum. d) Why do you feel theses feeling ( Explain using force diagrams ) Solution: If the car drops suddenly, my body keeps moving because it has linear momentum. This is related to the Newtons law of momentum: Heavy bodies prefer to keep moving in the same way as its moving right now. If the car rises abruptly after dipping, my body keeps moving upward because it has linear momentum. If the car turns sharply, my body keeps moving in the same direction because it has linear momentum. 2. Since the carriage operates solely under gravity and friction is negligible, should every ride be the same?. Why (explain using Ep and Ek analysis)? Solution: Every body moves in accordance to apply a force on it. If gravity force is applied on a carriage there is another force with the same magnitude but negative sign called “normal force”. The sum of both gravity and normal forces are equal to zero. If friction force is negligible thus every ride should be the same at any time. Potential energy over a plane will be constant because is the same height. If there is not friction force, the kinetic energy is constant over the time. THUNDER RIVER RAPID RIDE 3. Suggest why all the waves are not the same height. Solution: The waves are not the same height because there is turbulent water i.e. the river has many overlapping waves in many directions. So there are regions with constructive amplitude and many others with destructive amplitude. The next 3 questions are to be done after the ride 4. (a) determine the angular velocity of the tire Solution: I suggest that every single turn is done every 10 seconds. Thus, the angular velocity is given by: W = 2π / T where T = 10 seconds W = 2π / 10 = 0.6283 rad/sec (b) average speed of the ride (the ride is 75m long) Solution: I suggest the time travel is about 8 seconds, thus the average speed is given by: V = 75 m / ( 8 s ) = 9.375 m/s = 33.75 km/hr 5. Raft A carries a passenger load of 100 Kg and raft B 400 kg. In both rafts the load is spread symmetrically. Will one raft travel significantly faster than the other. Why? Solution: Two rafts with different masses will move at the same acceleration and velocity if same conditions are preserved on both objects. It does not matter how heavy is, they will move at the same rate under the same dynamical conditions. 6. (a) An empty mine carriage of mass 200 kg rolls freely down a 30 smooth slope. Find 1) the acceleration of the carriage down the slope 2) The velocity of the carriage after traveling 3m from rest. Solution: Fx : m g sin 30 = m a Fy : N cos 30 – mg = 0 N = mg / cos 30 m a = m g sin 30 1) The acceleration of the carriage down the slope: a = g sin 30 where g = 9.81 m/s2 a = 9.81 sin(30) a = 4.905 m/s2 2) The velocity of the carriage after traveling 3m from rest: vf2(t) = vi2 + 2 a d Where d = 3 m. vi = 0 m/s g = 9.81 m/s2 a = 9.81 sin(30) m/s2 vf(d = 3m) = ( 0.02 + 2 * 9.81 sin 30 (m/s2) ( 3 m ) ) ½ vf(d = 3m) = 5.4249 m/s (b) The same carriage caring the two students run down the slope but this time experiences a constant resistance of 200 N. Find 1) the net force down the plane 2) the velocity after traveling 3m from rest. Solution: 1) The net force down the plane: The mass of each student is around 65 kg. Fx : m g sin 30 – F = m a Fy : N cos 30 – mg = 0 The net force down the plane Fx is given by: Fx : m g sin(30) - F Where: m = 200 kg + 2 * 65 kg = 330 kg g = 9.81 m/s2 F = 200 N Fx : 330 kg 9.81 m/s2 sin(30) – 200 N Fx : 1418.65 N 2) The velocity after traveling 3m from rest: The net force down plane is Fx = 1418.65 N with a total mass m = 330 kg, thus the acceleration is given by: a = Fx / m a = 1418.65 / 330 = 4.29 m/s2 Thus, the velocity after traveling 3m from rest is given by: vf2(t) = vi2 + 2 a d Where d = 3 m. vi = 0 m/s a = 4.29 m/s2 vf(d = 3m) = ( 0.02 + 2 * 4.29 (m/s2) ( 3 m ) ) ½ vf(d = 3m) = 5.073 m/s ( c) State and explain any differences between the full carriage and the empty carriage. Solution: The acceleration down plane is given by the inclination angle. It does not matter how heavy the object is. However, any force applied on the body will affect its dynamics. Any resistive force will decrease its acceleration down plane. CANNONBALL EXPRESS 7. The track is 3km allowing for station stops, etc. time your journeys and calculate the average speed in km/h. Solution: The average speed is equal to the total distance divide by the total time invert to do it. V_average = 3 km / time 8. The trains maximum speed is 15km/hr. Convert to m/s. Solution: 1 km is equal to 1000 m and 1 hour is equal to 60*60 = 3600 seconds. Thus, 15 km/hr ( 1000 m / 1 km ) ( 1 hours / 3600 seconds ) = 15 * 1000 / 3600 m/s = 4.1666 m/s 9. In what distance will a force of 400 N stop the 30 tone train moving at 4 m/s (use Work done = change in K.E.; W = F * s and K.E. = ½ m v² Solution: In order to solve this problem, we release that Work done = K.E. F * s = ½ m v² Where m = 30 tones = 30 000 kg v = 4 m/s F = 400 N Thus, s = ( ½ m v² ) / F s = (½ 30 000 kg (4 m/s)² ) / (400 N) s = 600 m The train will stop in 600 meters. 10. If the engine is 12 tones and the three carriages weighed 6 tones each. Friction is 15 N per tone and the engine exerts a constant force of 6 000 N. Calculate the acceleration of the train. Solution: F_total: m a = F - Fr Engine constant force, F = 6 000 N Total mass, m = 12 tones + 3 x 6 tones = 30 tones = 30 000 kg Friction force, Fr_m = 15 N per tone Fr = 15 * 30 = 450 N a = ( F – Fr ) / m a = (6000 N - 450 N)/ (30 000 kg) a = 0.185 m/s² VINTAGE FUN 11. The maximum velocity attainable by the MODEL T is 7 km/hr and an acceleration of 0.3 m/s. How far does it travel in (a) 5 s (b) 10 s. Solution: d = ½ a t² + v t a = 0.3 m/s v = 7 km/hr = 1.9444 m/s a) t = 5 s d( t =5 s) = ½ ( 0.3 ) ( 5 )² + ( 1.9444 ) ( 5 ) = 13.472 m b) t = 10 s d( t =10 s) = ½ ( 0.3 ) ( 10 )² + ( 1.9444 ) (10 ) = 34.444 m 12. Traveling at maximum speed, the foot is released from the throtlte. Calculate the retardation which stops the car in 5m. Solution: The final deceleration “a” is given by: vf² = vi² + 2 a d Where vf = 0.0 m/s vi = 7 km/hr = 1.9444 m/s d = 5 m a = (vf2 – vi2) / d a = ( 0.0 - 1.9444) / 5 a = - 0.3888 m/s2 On the other hand time t is given by: a = (df – d0) / t t = (df – d0) / a Where d0 = 0.0 df = 5.0 m a= -0.3888 m/s2 t = ( 0.0 – 5.0)/(-0.3888) t = 12.86 seconds This is retardation time. PADDLE POWER 13. Take the mass of the Captain Stuart and passengers as 35 tones and the cruising speed as 0.8 m/s. Compare the K.E. of the Captain Stuart with that of the shoting gallery pellet of mass 20g traveling at 750 m/s. Solution: The K.E. of Captain Stuart and passengers is given by: Ek = ½ m v² Where m = 35 tones = 35 000 kg v = 0.8 m/s Ek = 0.5 * 35 000 (0.8)2 Ek = 11 200 joules The K.E. of 20g traveling at 750 m/s is given by: Ek = ½ m v² Where m = 20 g = 0.02 kg v = 750 m/s Ek = 0.5 * 0.02 (750)2 Ek = 5 625 joules The kinetic energy of Captain Stuart is around twice the kinetic energy of 20 g traveling at 750 m/s. 14. If is takes 40 m for the Captain Stuart to reach its steady cruising speed from rest, find the net force on the steamer (use Work done = change in K.E.) Solution: We used the fact that: F * s = ½ m v2 F = ( ½ m v) / s Where: m = 35 tones = 35 000 kg v = 0.8 m/s d = 40 m F = (½ * 35 000 * 0.8) / 40 F = 350 N The net force on the steamer. BUMPER TIME 15. Traveling at 3 m/s you steer a BANDITO BUMPER straight into a wall and rebound with a velocity of 1 m/s. If the mass of the car plus you and passengers is 150 kg, calculate the loss of K.E. at impact. Solution: The loss of kinetic energy is given by: K.E. = ½ m (vf2 – vi2) Where: m = 150 kg vf = 1 m/s vi = 3 m/s K.E. = 0.5 * 150 * (1.02 – 32) K.E. = -600 joules. The loss of kinetic energy is 600 joules. 16. Cruising at 3 m/s you hit a BUMBER traveling in the same direction at 2 m/s. However you dont rebound. Both cars continue forward at speed of 2.4 m/s. Calculate the mass of the second car. Solution: Kinetic energy plays roll in the following way: K.E.1 + K.E.2 = K.E.3 Where m1 is the mass of the car 1, m2 is the mass of the second car. v1 and v2 are the velocity of the first and second car respectively. v3 is the final velocity of both cars. ½ m1 v12 + ½ m2 v22 = ½ (m1+m2) v32 m1 v12 + m2 v22 = (m1+m2) v32 m1 v12 + m2 v22 = m1 v32+m2 v32 m1 v12 + m2 (v22 - v32) = m1 v32 m1 (v12 - v32) + m2 (v22 - v32) = 0 m2 = m1 (v12 - v32) / (v32 – v22) The mass of the second car is given by: v3 = 2.4 m/s, v2 = 2 m/s and v1 = 3 m/s. m2 = m1 (32 – 2.42) / (2.42 – 22) m2 = 1.8409 m1 The mass of the second car in terms of the mass of the first car. CYCLONE 17. if Ep = mgh, calculate the gain in Potential Energy of the roller coaster as it is drawn up the initial slop that has a vertical rise = 30 m and mass of each carriage is 500 kg. Solution: There are 6 carriage in the CYCLONE, thus total mass m = 300 kg. Potential energy is given by: Ep = m g h where m = 3000 kg g = 9.81 m/s² h = 30 m Ep = 3000 kg * 9.81 m/s² * 30 m Ep = 882900 kg m²/s² Ep = 882900 joules 18. Find the speed at the lowest point of the slope, just before entering the first loop. (use Gain is Ek = Loss in Ep) Solution: Ek = Ep Ek = ½ m v² Ep = m g h Thus, Ek = ½ m v² = m g h where m is the total mass of the system, i.e. m = 3000 kg. However, we dont need the mass m to get the velocity at the lowest point of the slope. v = ( m g h / ( ½ m ) ) ½ v = ( 2 g h ) ½ g = 9.81 m/s² h = 30 m v = ( 2 * 9.81 m/s² * 30 m )½ v = 24.2610 m/s GIANT DROP 19. Calculate the velocity you are traveling at the bottom of the ride. Solution: The Giant Drop, was officially declared the “tallest, vertical free-fall ride in the world” by the Guinness Book of World Records in their 1999 edition. Standing 119 meters high, The Giant Drop is now a legendary Australian landmark. Kinematic equations that involve free fall and final velocity is given by: vf² = vi² + 2 a d a = 9.81 m/s² d = 119 m vi = 0.0 m/s vf = ( 2 a d )¹/² vf = ( 2 * 9.81 * 119 ) ½ vf = 48.29 m/s TOWER OF TERROR 20. Using the information on the internet (Dreamworld, 2010) calculate: a) The starting height of the track (assuming no friction). Explain your answer. Solution: d = vi * t - ½ * a * t² vi = 161 km/hr = 44.72 m/s Because this the velocity after 6 seconds initial conditions. a = 9.81 m/s² d = (44.72 m/s) t - ½ 9.81 m/s² t² t1 = 0 sec. Because this is the first root of the above equation = 0. t = (44.72 m/s ) / ( ½ * 9.81 m/s² ) t = 9.11 sec. Because this is the second root of the above equation = 0. Maximum distance is at: d = 44.72 m/s– 9.81 m/s² * t = 0 t = ( 44.72 m/s ) / ( 9.81 m/s² ) tcritical = 4.55 sec. d = -9.81 m/s² < 0 (the critical point is maximum) d ( t = 4.55 sec ) = ( 44.72 m/s ) ( 4.55 s ) - ½ 9.81 m/s² (4.55 s)² d ( t = 4.55 sec ) = 101.93 m This is the height of the tower = 101.93 meters b) The speed at the end of the free fall. Solution: Kinematic equations that involve free fall and final velocity is given by: vf² = vi² + 2 a d where vi = 0.0 m/s a = 9.81 m/s² d = 101.93 m vf = ( 2 a d )¹/² vf = ( 2 * 9.81 m/s² * 101.93 m)^1/2 vf = 44.72 m/s This is the final velocity after free fall motion. References: DreamWorld 2010, “Dream World” http://dreamworld.com.au, accessed on 27 October 2010 Stanbrough 2009, “Physics at BHS” http://www.batesville.k12.in.us/physics/index.html accessed on 27 October 2010 Read More