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Revision
| Question | Answer |
|---|---|
| (a) In the first three seconds Mary’s acceleration is constant. Use the slope of the graph to calculate its size in metres per second squared. | Slope of graph gives acceleration. Hence: a = 30ms-1/3s = 10 ms-2 |
| What distance does Mary fall in the first three seconds? | Distance = area under the graph line = ½ x 3s x 30 ms-1 = 45m |
| (c) As Mary falls she notices that her acceleration decreases until she reaches terminal velocity. At what time has she reached terminal velocity? | Terminal velocity when speed change is zero, this happens at 11s. |
| Explain why you chose this time for when Mary reaches terminal velocity. | Terminal velocity means she will not accelerate to a higher speed. Nor will she decelerate to a lower speed as the air resistance force exactly counterbalances the force of gravity acting downwards on her. |
| Treat moving downwards as moving in a positive direction. On the lines below the graph explain why you have drawn the shapes for each part of the motion, before reaching terminal velocity and after reaching it. | As velocity is represented by slope of d/t graph, there is a curved part before terminal velocity time reached due to acceleration. At terminal velocity, velocity is constant hence acceleration is zero and graph line has constant slope. |
| (f) Using Newton’s Second Law, explain why Mary must have no net force acting on her when she is falling at a constant velocity? | The net force on her is zero Newtons at her constant terminal velocity because she has no net acceleration. Hence by Newton’s second law, if anet = 0, then Fnet = m anet = 0 N. |
| (g) Mary has a mass of 65 kg. Calculate the size of the force of gravity acting on her as she falls. | F = mg = 65kg x 10 N/kg = 650 N |
| (h) On the diagram of Mary sketch vector arrows to represent the forces on her when she is falling at terminal velocity. | two arrows facing the opposite direction,drag and pull |
| (i) Deduce the size of the air resistance force acting on Mary when she falls at terminal velocity. | At terminal velocity net force = 0 hence size of air resistance force = size force of gravity = 650 N |
| (a) An aeroplane takes Mary up to a height of 5000m for her jump. The plane and its contents have a total mass of 750 kg. Use the formula to calculate the aeroplane’s gain in gravitational potential energy in reaching this height. | = 750 kg x 10 N/kg x 5000m = 3.75 x 107 J (37,500,000) |
| (b) What is the minimum work that the engine of the plane has to do to reach this height? | 3.75 x 107 J |
| (c) In reality, the engine has to do more work than this for the plane to reach 5000m. Explain why this is the case. | n order to rise to this height energy is changed in form to Gravitational Potential Energy. |
| (d) If the plane is flying at a speed of 60 ms-1, use the formula Ek = ½mv2 to calculate how much kinetic energy it has. | = ½ x 750 kg x 602 = 1.35 x 106 J |
| (e) If the plane were to have a catastrophic engine failure and ‘fall out of the sky’ while flying at 60 ms-1, ignoring air resistance, how fast would it be going just as it hit the ground? | Deduction that the Ek at ground = Ek at 5000m + Ep lost in fall = 1.35 x 106 J + 3.75 x 107 J = 3.885 x 107 J Followed by calculation of v using rearrangement of Ek = ½mv2 thus: = 322ms-1 |
| (f) On another trip a plane does 2.40 x 107 J of work to reach the jump height. It takes 10 minutes to get to this height. What power must the plane engine have to do this? | P = W/t = 2.4 x 107J (10 x 60s) = 4.00 x 104 W |
| (g) At sea level air pressure has a value of 1.01 x 105. Use the formula P = F/A to obtain units for pressure. | P = F/A means units of pressure = units of Force units of area = N/m2 |
| (h) When Mary is at 5000m air pressure has dropped to 5.00 x 105 Pa. What sized force will a 1 cm x 1 cm area of Mary’s skin feel due to the air? | 1cm2 area = 10-4 m2 area. Hence F = PA = 5.00 x 105 x 10-4 = 50N |
Created by:
lachlanosborne
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