Question 1 (60 marks) Background information Ski-jumping is a popular sport in the Winter Olympics. The ski-jumpers start at the top of a steep slope, called the ‘in-run’, picking up speed as they...

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Question 1 (60 marks) Background information Ski-jumping is a popular sport in the Winter Olympics. The ski-jumpers start at the top of a steep slope, called the ‘in-run’, picking up speed as they move. The slope then has a brief horizontal section with a sudden drop – this is the jump point. The hill then follows a profile similar to that taken by the ski-jumper in the air, who land somewhere on the landing slope section. Finally, the ski-jumper comes to a stop in the out-run section. See the diagram below: Section A – The In-run (20 marks) A ski-jumper starts from rest at the start position, and accelerates down the in-run. We will ignore the effects of friction between the icy slope and the skis, and the effect of air resistance. The total mass of the ski-jumper and their equipment (skis, helmet, etc) is 80kg. The average acceleration for the ski-jumper for the first 3 seconds of the in-run is 4.0 m/s2. 1. What was the kinetic energy of the ski-jumper at the start position? 2. What is the velocity of the ski-jumper after 3.0 seconds? 3. What is the kinetic energy of the ski-jumper after 3.0 seconds? It takes a total of 7.5 seconds to reach the jump point. At the jump point, the ski-jumper has reached a velocity of 27.0 m/s. 4. What is this velocity in units of km/h? 5. What is the average acceleration for the entire in-run, in units of m/s2? 6. What is the kinetic energy of the ski-jumper at the jump point? The Law of Conservation of Energy states that ‘energy cannot be created or destroyed – only converted from one form to another’. 7. Where did the ski-jumper’s kinetic energy come from? 8. Estimate the height difference between the start point and the jump point. 9. If we did not ignore the effects of friction and air resistance, would the start point need to be higher or lower than this estimate, if the ski-jumper was still to have a velocity at the jump point of 27.0 m/s? Section B – The Jump (20 marks) From the moment the ski-jumper leaves the jump point horizontally, at a speed of 27.0 m/s, they are moving freely through the air in what is known as projectile motion. We will continue to ignore the effects of air resistance. 1. Draw a free body diagram showing the forces acting on the ski-jumper. 2. What acceleration does the ski-jumper have during jump phase? 3. Draw a vector diagram showing the horizontal and vertical components of the skijumper’s velocity, 2 seconds after leaving the jump point. The ski-jumper is in the air for 3.2 seconds before landing. The acceleration due to gravity is 9.8 m/s2. 4. What is the horizontal distance between the jump point and the landing point? 5. What is vertical distance between the jump point and the landing point? 6. What is the straight-line distance between the jump point and the landing point? 7. What is the horizontal velocity of the ski-jumper at the instant they land? 8. What is the vertical velocity of the ski-jumper at the instant they land? 9. What is the combined velocity of the ski-jumper at the instant they land? Section C – The Out-Run (20 marks) Assume that upon landing, all of the kinetic energy of the ski-jumper is conserved and the landing does not cause any loss of their total velocity. At the landing point, the slope of the ski hill is 32 degrees to the horizontal, as shown in the diagram. The out-run then slowly reaches its lowest point before becoming an incline. We continue to ignore air resistance and the force of friction between the skis and the icy hill. 1. Draw a free body diagram showing the forces acting ski-jumper just after they have landed. Include values for the forces where you can. 2. What is the value, just after the ski-jumper lands, of the accelerating force which acts parallel to the surface of the ski hill? 3. The lowest part of the out-run is 10m below the landing point. What is the skijumper’s (horizontal) velocity at this point? Immediately after its lowest point, the out-run becomes an inclined surface with a constant slope of 10º. The ski-jumper also begins to apply a drag force against the icy surface with their skis, providing a decelerating force of 600N. 4. By combining this decelerating force with that generated as a result of the incline, calculate the net force on the ski-jumper in the direction parallel to the surface of the ice as they travel up the incline. 5. How long will it take for the skier to come to a complete stop? 6. What will be the vertical difference between the stopping point and the landing point? Question 2 (40 marks) Write a short piece (200-300 words) on one of the following: • How photons of light are created and absorbed by atoms and molecules • The Doppler effect for light and sound waves • The concept of standing waves in musical instruments • Why, to an observer standing on the edge, swimming pools do not look as deep as they really are. Include at least one diagram in your answer.
Answered Same DayJul 14, 2021

Answer To: Question 1 (60 marks) Background information Ski-jumping is a popular sport in the Winter Olympics....

Rajeswari answered on Jul 21 2021
138 Votes
Physics assignment
1. Start position velocity is 0. Hence Kinetic energy = ½ mv2=0
2. Velocity after 3 seconds is v=u+at = 0+3(4) = 12 m/sec
3. KE after 3 seconds = ½ mv2 =
4.
At jump point velocity = 27m/sec = km/hr.
5. Average acceleration = (v-u)/t = (27-0)/7.5 = 3.6 m/sec2
6. At jump point, Kinetic energy = = ½ mv2=80)(272) = 29160 Joules
7. By conservation of energy kinetic energy at jump point = kinetic energy due to motion. While at rest this was potential energy mgh, and while in motion converted to kinetic energy.
8. To find height difference from start point to end point.
By using conservation of energy we have mgh = KE = 29160 at end point
Where h was initial height.
So height from start point to jump point = 29160/mg = 37.194 m.
9. If we did not ignore the effects of friction and air resistance, would the start point need to be higher or lower than this estimate, the ski-jumper could not have a velocity at the jump point of 27.0 m/s because due to friction velocity would be reduced. Friction acts opposite to the direction of motion thus reducing distance travelled and hence velocity.
1.
2. During jump phase, he is not in a position to accelerate so only gravity acts on him in the vertical direction.
3.
Here v is the velocity after 2 seconds.
4. Horizontal distance no gravity plays. So horizontal distance is velocity x time = 3.2 * 27 cos 32 = 73.2713 m
5. For vertical distance we use s = ut + ½ gt^2
Vertical distance = 27sin 32 (3.2) + ½ (9.8) *4=45.7850+19.6=65.385 m
6. Straight line distance = square root of vertical dist^2 + hor dist ^2
=
7. Horizontal velocity at the instant he lands = u+at = 27cos 32 + 0(3.2) = 22.8973 m/sec
8. Vertical velocity at the time of landing = u+at = 27sin32 + 3.22(9.8)
=14.3078+100.352 = 114.66...
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