Unit 3 Motion in 1 & 2 Dimensions Revision Questions

Unit 3 Motion in 1 & 2 Dimensions Revision Questions
VCE Physics Unit 3 Motion in 1 & 2 Dimensions Revision Questions

Motion - Revision Questions Question type:
Equations of Motion In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration. Q: Calculate the acceleration of the car for the first 400 m. A: Firstly, list information: u = 0 v = ? a = ? s = 400 m t = 19 s Choose the appropriate equation: s = ut + ½at2 400 = 0 + ½a (19)2 a = 2.22 ms-2

Sketch Graphs In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration. The graphs A to F below should be used to answer the questions below. The horizontal axis represents time and the vertical axis could be velocity or distance. A B C D E F Q: Which of the graphs, A to F, represents the velocity time graph for the entire journey ? A: Graph B Q: Which of the graphs, A to F, best represents the distance time graph of the car for the entire journey ? A: Graph E

Average Speed In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration. Q: Calculate the average speed for the entire journey, covering both the accelerating and braking sections. Need to know u, the initial speed for the braking section which equals the final speed for the accelerating section. A: Average Speed = Total Distance Total Time For the accelerated part of journey: Distance = 400 m Time = 19 sec For the braking part of journey: Distance = needs to be calculated Time = 5.1 sec For accelerating section: u = 0 v = ? a = 2.2 ms-2 s = 400 m t = 19 s v = u + at = 0 + (2.2)(19) = 41.8 ms-1 Now can calc s Need to calc acc v = u + at 0 = a(5.1) a = ms-2 s = ut + ½at2 = (41.8)(5.1) + ½(-8.2)(5.1)2 = = m To get braking distance, use Eqns of Motion u = ? v = 0 a = ? s = ? t = 5.1s Braking list becomes u = 41.8 ms-1 v = 0 a = ? s = ? t = 5.1s List does not contain enough info the calculate s Still not enough info Total Distance = = m Total Time = = 24.1 s So, Average Speed = 506.6/24.1 = 21 ms-1

Momentum/Impulse In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the driver’s head to come to a complete stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag. Q: Calculate the magnitude of the average contact force that the air bag exerts on the driver’s head during this collision. A: Impulse = Change in Momentum FΔt = Δ(mv) So F = Δ(mv)/Δt = (7.0)(8.0)/(1.6 x 10-1) = 350 N

Air Bags/Crumple Zones In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the driver’s head to come to a complete stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag. Q: Explain why the driver is less likely to suffer a head injury in a collision with the air bag than if his head collided with the car dashboard, or other hard surface. A: The change in momentum suffered by the driver’s head is a FIXED quantity no matter how his head is brought to rest. Therefore the product of F and t (ie Impulse) is also a fixed quantity. However the individual values of F and t may be varied as long as their product always remains the same. The air bag increases the time over which the collision occurs, therefore reducing the size of the force the head must absorb so reducing the risk of injury. The air bag also spreads the force over a larger area, reducing injury risk. Without the air bag the driver’s head may hit a hard surface decreasing the time to stop his head and necessarily increasing the force experienced and thus the likelihood of injury. In addition the force will be applied over a much smaller area increasing the likelihood of severe injury

Momentum A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one. X Y 6.0 ms-1 10 tonnes 5 tonnes stationary Before Collision v ms-1 After Collision Q: Calculate the final speed of the joined railway trucks after collision. A: In ALL collisions Momentum is conserved. So Mom before collision = Mom after collision Mom before = (10 x 103)(6.0) + (5 x 103)(0) Mom after = (15x 103)(v) So v = 60,000/15,000 = 4.0 ms-1

Impulse A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one at a speed of 4.0 ms-1. X Y 6.0 ms-1 10 tonnes 5 tonnes stationary Before Collision 4.0 ms-1 After Collision Q: Calculate the magnitude of the total impulse that truck Y exerts on truck X A: Impulse = Change in Momentum Truck X’s change in momentum = Final momentum – Initial Momentum = (10 x 103)(4.0) – (10 x 103)(6.0) = -2.0 x 104 kgms-1 The mechanism for this change in momentum is the impulse supplied by Truck Y So, I = 2.0 x 104 Ns

Inelastic Collisions A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one at 4.0 ms-1. X Y 6.0 ms-1 10 tonnes 5 tonnes stationary Before Collision 4.0 ms-1 After Collision Q: Explain why this collision is an example of an inelastic collision. Calculate specific numerical values to justify your answer. A: In Inelastic collisions Momentum is conserved BUT Kinetic Energy is not. For this collision pBEFORE = (10 x 103)(6.0) + (5 x 103)(0.0) = 6 x 104 kgms-1 pAFTER = (15 x 103)(4.0) = 6 x 104 kgms-1 KEBEFORE = ½mv2 = ½(10 x 103)(6.0)2 = 1.8 x 105 J KEAFTER = ½mv2 = ½(15 x 103)(4.0)2 = 1.2 x 105 J Thus Mom IS conserved but KE is NOT conserved, therefore this is an inelastic collision

A cyclist is towing a small trailer along a level bike track (Figure 1). The cyclist and bike have a mass of 90 kg, and the trailer has a mass of 40 kg. There are opposing constant forces of 190 N on the rider and bike, and 70 N on the trailer. These opposing forces do not depend on the speed of the bike. The bike and trailer are initially travelling at a constant speed of 6.0 m s-1. Question 1 What driving force is being exerted on the road by the rear tyre of the bicycle?

Newton’s Laws A seaplane of mass 2200 kg takes off from a smooth lake as shown. It starts from rest, and is driven by a CONSTANT force generated by the propeller. After travelling a distance of 500 m, the seaplane is travelling at a constant speed, and then it lifts off after travelling a further 100 m. The total force opposing the motion of the seaplane is not constant. The graph shows the TOTAL FORCE OPPOSING THE MOTION of the seaplane as a function of the distance travelled. Q: What is the magnitude of the net force acting on the seaplane after it has travelled a distance of 500 m from the start ? A: At d = 500 m the plane is travelling at CONSTANT VELOCITY, So ΣF = 0

Newton’s 2nd Law Q: What is the magnitude of the seaplane’s acceleration at the 200 m mark ? A: At d = 500 m the seaplane is subject to 0 net force (see previous question). Thus, Driving Force = Opposing Force = 10,000 N (read from graph). At d = 200 m the total opposing force = 2000 N (read from graph) So ΣF = 10, = 8000 N Now, we know that ΣF = ma So, a = ΣF/m = 8000/2200 = 3.64 ms-2

Work Q: Estimate the work done by the seaplane against the opposing forces in travelling for a distance of 500 m. A: Work = Force x Distance = Area under F vs d graph. Area needs to be calculated by “counting squares”. Each square has area = 2000 x 100 = 2 x 105 J Total number of squares (up to d = 500 m) = 9 whole squares (x) + 6 part squares (p) = 12 whole squares. So Work done = (12) x (2 x 105) = 2.4 x 106 J p x x x

Vectors Adam is testing a trampoline. The diagrams show Adam at successive stages of his downward motion. Figure C shows Adam at a time when he is travelling DOWNWARDS and SLOWING DOWN. A: Acc is UPWARD. In order to meet the requirements set - travelling downward BUT slowing down, he must be decelerating ie. Accelerating in a direction opposite to his velocity. Thus acc is upward. Q: What is the direction of Adam’s acceleration at the time shown in Figure C ? Explain your answer.

Force Diagrams Q: On Figure C draw arrows that show the TWO INDIVIDUAL forces acting on Adam at this instant. LABEL EACH ARROW with the name of the force and indicate the RELATIVE MAGNITUDES of the forces by the LENGTHS of the arrows you draw. A: The two forces are: (a) Weight Force (W) acting through Adam’s centre of mass (b) Normal Reaction Force (N) acting from the point of contact b/w Adam’s feet and the air bag. N.B. N >W to meet requirements that acc be upwards at this time NOTE: The two forces are drawn offset to show the point of origin of each, they should be coplaner ie., sit one on top of the other. N W

Projectile Motion A car takes off from a ramp and the path of its centre of mass through the air is shown below. First model the motion of the car assuming that air resistance is small enough to neglect. Q: Which of the directions (A - H), best shows the VELOCITY of the car at X ? A: Direction C. Q: Which of the directions (A - H), best shows the VELOCITY of the car at Y ? A: Direction D A B C D E F G H X Y Direction of Motion

Projectile Motion Q: Which of the directions (A - H), best shows the ACCELERATION of the car at X ? A: Direction E. Q: Which of the directions (A - H), best shows the ACCELERATION of the car at Y ? Now suppose that AIR RESISTANCE CANNOT BE NEGLECTED. A: Direction F. A B C D E F G H X Y Direction of Motion

Newton’s 3rd Law The figure shows a cyclist with the bicycle wheels in CONTACT with the road surface. The cyclist is about to start accelerating forward. a FTR FRT Q: Explain, with the aid of a clear force diagram, how the rotation of the wheels result in the cyclist accelerating forwards. A: The wheels rotate in the direction shown. The force labelled FTR is the force the tyre exerts on the road. This force is directed in the opposite direction to the acceleration and thus cannot be the force producing that acceleration. The force labelled FRT is the Newton 3 reaction force arising from the action of FTR. It is this force (directed in the same direction as the acceleration) that actually produces the acceleration of the bike and rider.

Centripetal Force Mark Webber and his Formula 1 racing car are taking a corner at the Australian Grand Prix. A camera views the racing car head on at point X on the bend where it is travelling at constant speed. At this point the radius of curvature is 36.0 m. The total mass of the car and driver is 800 kg. Q: On the diagram showing the camera’s view of the racing car, draw an arrow to represent the direction of the NET force acting on the racing car at this instant. FC The magnitude of the horizontal force on the car is 6400 N. Q: Calculate the speed of the car. A: FC = mv2/R So v =√FCR/m = √(6400)(36.0)/(800) = 17 ms-1

Centripetal Force Q: Referring to the racing car from the previous slide, explain: (a) Why the car needs a horizontal force to turn the corner. (b) Where this force comes from. A: (a) Newton 1 states all objects will continue in a straight line unless acted upon by a net force. In order for the car to travel in a circular path a constant force directed at right angles to the direction of motion (toward the centre of the circle) must exist. (b) The horizontal force arises from the frictional force exerted BY THE ROAD ON THE TYRES and is directed toward the centre of the circle the car is travelling in.

Projectile Motion A bushwalker is stranded while walking. Search and rescue officers drop an emergency package from a helicopter to the bushwalker. They release the package when the helicopter is a height (h) above the ground, and directly above the bushwalker. The helicopter is moving with a velocity of 10 ms–1 at an angle of 30° to the horizontal, as shown in Figure 1. The package lands on the ground 3.0 s after its release. Ignore air resistance in your calculations. Figure 1 Q: What is the value of h in Figure 1? A: Can approach in a number of ways. One way is: Find how long it takes to rise and fall back to same vertical height (h) Then subtract from total time to give a time for vertical fall under gravity with an initial velocity of 10 ms-1 directed at 30o below the horizontal

Projectile Motion Figure 1 1. Time to get back to h Since this part of the journey is symmetrical time to reach max height = ½ time for this part of journey. h 2. Value of h again analyse vertical motion Analyse vertical motion +ve u = 10 Sin 30o v = ? a = 10 ms-2 s = h t = 2.0 s s = ut + ½ at2 h = (5)(2) + ½ (10)(4) h = 30 m u = - 10 Sin 30o ms-1 v = 0 a = 10 ms-2 s = ? t = ? +ve v = u + at 0 = -10 Sin t t = 5/10 = 0.5 sec. So, total time to get back to h = 1.0 s

Projectile Motion Figure 1 Q: Assuming that the helicopter continues to fly with its initial velocity, where is it when the package lands? Which one of the statements below is most correct? A. It is directly above the package. B. It is directly above a point that is 15 m beyond the package. C. It is directly above a point that is 26 m beyond the package. D. It is directly above a point that is 30 m from the bushwalker. Both the package and the helicopter have the same (constant) horizontal velocity, so the package will always be directly below the helicopter. So, alternative A is correct

Sketch Graphs Q: Which of the graphs below best represents the speed of the package as a function of time? The key to the question is SPEED, this is the sum of both vertical and horizontal. Speed never falls to zero (there is always a horizontal component), so C and D are out A shows the object’s acceleration falling as it nears the ground – not so ! So B must be the correct answer

Momentum A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a truck that is stationary at a set of traffic lights. After the collision they are stuck together and move off with a speed of 2.0 ms–1 Q: How much momentum did the car transfer to the truck? A: Mom is ALWAYS conserved. Mom of car before = (1000)(5) = 5000 kgms-1 Mom of car after = (1000)(2) = 2000 kgms-1 Since mom is conserved Mom loss by car = Mom gain by truck So, Mom transferred to truck = 3000 kgms-1 Q: What is the mass of the truck? A: p = mv so, m = p/v = 3000/2 = 1500 kg

Impulse A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a truck that is stationary at a set of traffic lights. After the collision they are stuck together and move off with a speed of 2.0 ms–1 Q: If the collision took place over a period of 0.3 s, what was the average force exerted by the car on the truck? For the truck Impulse = Change in momentum Ft = mv F(0.3) = 3000 F = 10,000 N

Relative Motion A train is travelling at a constant velocity on a level track. Lee is standing in the train, facing the front, and throws a ball vertically up in the air, and observes its motion. Q: Describe the motion of the ball as seen by Lee. Lee sees the ball move straight up and down. Sam, who is standing at a level crossing, sees Lee throw the ball into the air. Q: Describe and explain the motion of the ball as seen by Sam. From Sam’s point of view, the ball follows a parabolic path made up of the vertical motion imparted by Lee and the horizontal motion due to the train.

Newtons 2nd Law In Figure 2, a car of mass 1000 kg is being towed on a level road by a van of mass 2000 kg. There is a constant retarding force, due to air resistance and friction, of 500 N on the van, and 300 N on the car. The vehicles are travelling at a constant speed. Figure 2 Q: What is the magnitude of the force driving the van? As the vehicles are travelling at constant speed, a = 0 and thus ΣF = 0 Thus driving force = total retarding force Thus driving Force = 800 N Q: What is the value of the tension, T, in the towbar? Looking at the car alone the forces acting are the tension in the towbar and the retarding force. Since the car is travelling at constant velocity, ΣF = 0, so tension = retarding force = 300 N

Equations of Motion When travelling at a speed of 15.0 ms–1 the van driver stops the engine, and the van and car slow down at a constant rate due to the constant retarding forces acting on the vehicles. Q: How far will the van and car travel before coming to rest? Determine deceleration ΣF = ma so a = ΣF/m = 800/3000 = ms-2 2. Use equations of motion to find distance u = 15.0 ms-1 v = 0 a = ms-2 s = ? t = ? v2 = u2 + 2as 0 = (15)2 +2(-0.266)s s = 422 m Another approach: Change in Kinetic Energy = work done mathematically ½ mv2 = Fd so d = 0.5(3000)(15)2 800 = 422 m

Work and Energy In a storeroom a small box of mass 30.0 kg is loaded onto a slide from the second floor, and slides from rest to the ground floor below, as shown in Figure 4. The slide has a linear length of 6.0 m, and is designed to provide a constant friction force of 50 N on the box. The box reaches the end of the slide with a speed of 8.0 m s–1 Figure 4 Q: What is the height, h, between the floors? At the second floor the box has Potential Energy = mgh On reaching the ground floor this has been converted to Kinetic Energy (½mv2) plus the work done against friction in moving down the slope thus, mgh = ½ mv2 + Fd (30.0)(10)h = ½ (30)(8.0)2 + 50(6.0) h = 4.2 m

Elastic Potential Energy Figure 4 The box then slides along the frictionless floor, and is momentarily stopped by a spring of stiffness N m–1 Q: How far has the spring compressed when the box has come to rest? KE of box is converted to Elastic Potential Energy in the spring Thus ½ mv2 = ½ kx2 ½ (30.0)(8.0)2 = ½ (30,000)(x)2 x = 0.25 m

Work and Energy A model rocket of mass 0.20 kg is launched by means of a spring, as shown in Figure 1. The spring is initially compressed by 20 cm, and the rocket leaves the spring as it reaches its natural length. The force-compression characteristic of the spring is shown in Figure 2. Q: How much energy is stored in the spring when it is compressed? A: Compressing the spring requires work to be done on it. (this work is stored as elastic potential energy in the spring) Work done = area under graph up to a compression of 0.2 m = ½ (0.2)(1000) = 100 J

Energy Conversion Q: What is the speed of the rocket as it leaves the spring? A: All the elastic potential energy stored at max compression (100 J) will be converted to Kinetic Energy at release. So 100 = ½ mv2 v = √1000 = 31.6 ms-1

Equations of Motion Q: What is the maximum height, above the spring, reached by the rocket? You should ignore air resistance on the way up since the rocket is very narrow. v2 = u2 +2as 0 = (31.6)2 + 2(-10)s s = 1000/20 = 50 m u = 31.6 ms-1 v = 0 a = -10 ms-2 s = ? t = ? +ve

Newtons 2nd Law Retarding Force (R) (Newtons) time (s) When the rocket reaches its maximum height, the parachute opens and the system begins to fall. In the following questions you should still ignore the effects of air resistance on the rocket, but of course it is critical to the force on the parachute. This retarding force due to the parachute is shown as R in Figure 3, and its variation as a function of time after the parachute opened is shown in Figure 4. Q: What is the acceleration of the rocket at a time 5 s after the parachute opens? R +ve A: At t = 5.0 s, R = 1.8 N Weight of Rocket W = mg = (0.2)(10) = 2.0 N W ΣF = ma a = ΣF/m = (2.0 – 1.8)/ 0.2 = 1.0ms-2

Centripetal Force The safe speed for a train taking a curve on level ground is determined by the force that the rails can take before they move sideways relative to the ground. From time to time trains derail because they take curves at speeds greater than that recommended for safe travel. Figure 5 shows a train at position P taking a curve on horizontal ground, at a constant speed, in the direction shown by the arrow. Fc Q: At point P shown on the figure, draw an arrow that shows the direction of the force exerted by the rails on the wheels of the train.

Centripetal Force The radius of curvature of a track that is safe at 60 km/h is approximately 200 m. Q: What is the radius of curvature of a track that would be safe at a speed of 120 km/h, assuming that the track is constructed to the same strength as for a 60 km/h curve? Fc = mv2 R A: Centripetal Force This can be solved as a ratio question m(60) = m(120)2 x x = 800 m A: B

Vector Addition Q: At point Q the driver applies the brakes to slow down the train on the curve. Which of the arrows (A to D) indicates the direction of the net force exerted on the wheels by the rails? A: B

Momentum A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0 tonne. They remain locked together as they move off. The speed immediately after the collision was known to be 7.0 ms-1 from the jammed reading on the car speedometer. Robin, one of the police investigating the crash, uses conservation of momentum to estimate the speed of the truck before the collision. Q: What value did Robin obtain? A: PBEFORE = PAFTER PBEFORE = (3000)(x) PAFTER = ( )(7.0) So, 3000x = 28,000 x = 9.3 ms-1 The calculated value is questioned by the other investigator, Chris, who believes that conservation of momentum only applies in elastic collisions. Q: Explain why Chris’s comment is wrong. A: Momentum is conserved in ALL types of collisions whether they be elastic or inelastic. KE is not conserved in this type (inelastic) collision.

Elastic/Inelastic Collisions Q: Use a calculation to show whether the collision was elastic or inelastic. A: Total KE before collision = ½ mv2 = ½ (3000)(9.3)2 = J Total KE after collision = ½ mv2 = ½ (4000)(7.0)2 = 98,000 J KE is NOT conserved So collision is INELASTIC

Projectile Motion Fred is playing tennis on the deck of a moving ship. He serves the ball so that it leaves the racket 3.0 m above the deck and travels perpendicular to the direction of motion of the ship. The ball leaves the racket at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 ms-1. You may ignore air resistance in the following questions. Q: With what speed, relative to the deck, did the ball leave Fred’s racket? Give your answer to three significant figures. 30 ms-1 x 8o x = initial speed 30 ms-1 A: Assuming projectile behaviour: VHORIZONTAL = 30 ms-1 throughout the flight thus, Cos 8 = 30/x x = 30/Cos 8 = 30.3 ms-1

Projectile Motion Q: At its highest point, how far was the ball above the deck? 30 ms-1 x 8o VVERT A: Using the equations of motion for the balls flight from racquet to point of max height, where VVERT = 0 v2 = u2 + 2as 0 = (30.3 Sin 8)2 + 2(-10)s s = 0.89 m u = VVERT = 30.3 Sin 8 v = 0 a = -10 ms-2 s = ? t = ? +ve So Height above the DECK = = 3.89 m

Relative Motion The ship is travelling straight ahead at a velocity of 10 ms-1 Q: When the ball is at its highest point at what speed is it moving relative to the ocean? 30 ms-1 10 ms-1 A: V BALL REL OCEAN = VSHIP + VBALL VBALL + VSHIP = Q: at what angle is the ball travelling relative to the direction of the ship’s travel? 30 10 VBALL REL OCEAN θ Tan θ = 30/10 = 3.0 θ = 71.60 VBALL REL OCEAN = √(30)2 + (10)2 = 31.6 ms-1

Universal Gravitation Newton was the first person to quantify the gravitational force between two masses M and m, with their centres of mass separated by a distance R as F= GMm R2 where G is the universal gravitational constant, and has a value of 6.67 × N m2 kg2. For a mass m on the surface of Earth (mass M) this becomes F = gm, where g = GM Q: Which one of the expressions (A to D) does not describe the term g? A. g is the gravitational field at the surface of Earth. B. g is the force that a mass m feels at the surface of Earth. C. g is the force experienced by a mass of 1 kg at the surface of Earth. D. g is the acceleration of a free body at the surface of Earth. A: B

Gravitational Force Q: What is the magnitude of the force exerted by Earth on a water molecule of mass 3.0 × kg at the surface of Earth? A: F = GMm R2 = (6.67 x 10-11)( 5.98 x 1024)(3.0 x 10-26) (6.37 x 106)2 = 2.95 x N

Kepler’s 3rd Law A satellite in a circular orbit of radius 3.8 × 108 m around Earth has a period of 2.36 × 106 s. Q: Calculate the mass of Earth. You must show your working. A: R3 = GMe T π2 Me = (4π2) (3.8 x 108)3 (6.67 x 10-11) (2.36 x 106)2 = 5.83 x 1024 kg

Universal Gravitation The radius of the orbit of Earth in its circular motion around the Sun is 1.5 × 1011 m Q: Indicate on the diagram, with an arrow, the direction of the acceleration of Earth. (Figure 3). Q: Calculate the mass of the Sun. Take the value of the gravitational constant G = 6.67 × 10–11 N m2 kg–2. Question assumes you know the period of rotation of the earth around the sun ie. 365 days = 365 x 24 x 60 x 60 = s (3.15 x 107 s) FC = Fg me4π2 R/T2 = Gmems/R2 ms = (3.14)2(1.5 x 1011)3 6.67 x 10-11(3.15 x 107)2 ms = 4π2R3 GT2 ms = 2 x 1030 kg

Gravitational Field Strength
Motion - Revision Questions Question type: Gravitational Field Strength Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a constant speed at a radius R = 4.22 x 107 m from the centre of Earth. Q: Calculate the magnitude of the Earth’s gravitational field at the orbit radius, R = 4.22 x 107 m, of Nato III. Give your answer to 3 sig figs. You MUST show your working. G = 6.67 x Nm2kg-2 Me = 5.98 x 1024 kg. A: g = GMe/R2 = (6.67 x 10-11)(5.98 x 1024) (4.22 x 107)2 = Nkg-1

Universal Gravitation
Motion - Revision Questions Question type: Universal Gravitation Q: What is the speed of Nato III in its orbit ? A: The centripetal force (FC) required by Nato III to complete its circular orbit is supplied by the force of gravitational attraction (Fg) between Earth and Nato III. Thus Fg = FC Thus GMem/R2 = mv2/R So v = √GMe/R = √(6.67 x 10-11)(5.98 x 1024) (4.22 x 107) = 3074 ms-1 = 3.1 x 103 ms-1

Circular Motion Motion - Revision Questions Question type:
Q: Which ONE of the following statements (A - D) about Nato III is correct ? A: The net force acting on Nato III is zero and therefore it does not accelerate. B: The speed is constant and therefore the net force acting on Nato III is zero. C: The is a net force acting on Nato III and therefore it is accelerating. D: There is a net force acting on Nato III, but it has zero acceleration. A: Alt C is correct

Weight & Weightlessness
Motion - Revision Questions Question type: Weight & Weightlessness The Russian space station MIR (Russian meaning - peace) is in a circular orbit around the Earth at a height where the Gravitational Field Strength is 8.7 Nkg-1 Q: Calculate the magnitude of the gravitational force exerted by Earth on the astronaut of mass 68 kg on MIR A: W = mg = (68)(8.7) = 592 N When the astronaut wishes to rest he has to lie down and strap himself into bed. Q: What is the magnitude of the force that the bed exerts on the astronaut before he begins to fasten the strap ? A: 0 N

Weightlessness Motion - Revision Questions Question type:
Newspaper articles about astronauts in orbit sometimes speak about zero gravity when describing weightlessness. Q: Explain why the astronaut in the orbiting MIR is not really weightless. A: Weight is the action of a gravitational field acting on a mass. For true weightlessness to exist the gravitational field strength needs to be zero. There is no place in the universe where the gravitational field strength is zero, so nothing is truly “weightless”. The astronaut actually does have weight, as calculated earlier. The astronaut’s “apparent weight” is zero because he is not subject to a reaction force inside the spacecraft since both he and the spacecraft are in a state of constant free fall.

Unit 3 Motion in 1 & 2 Dimensions Revision Questions

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