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(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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