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VCE Physics. Unit 3 Motion in 1 & 2 Dimensions Revision Questions. Equations of Motion. Motion - Revision Questions Question type:. In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. - PowerPoint PPT Presentation

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  • VCE PhysicsUnit 3Motion in 1 & 2 Dimensions

    Revision Questions

  • Motion - Revision Questions Question type: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 = 0v = ?a = ?s = 400 mt = 19 sChoose the appropriate equation: s = ut + at2400 = 0 + a (19)2 a = 2.22 ms-2Equations of Motion

  • Motion - Revision Questions Question type: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.Q: Which of the graphs, A to F, represents the velocity time graph for the entire journey ?A: Graph BQ: Which of the graphs, A to F, best represents the distance time graph of the car for the entire journey ? A: Graph ESketch Graphs

  • Motion - Revision Questions Question type: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.A: Average Speed = Total Distance Total TimeFor the accelerated part of journey:Distance = 400 m Time = 19 secFor the braking part of journey:Distance = needs to be calculated Time = 5.1 secTo get braking distance, use Eqns of Motionu = ?v = 0a = ?s = ?t = 5.1sList does not contain enough info the calculate sNeed to know u, the initial speed for the braking section which equals the final speed for the accelerating section.For accelerating section:u = 0v = ?a = 2.2 ms-2s = 400 mt = 19 sv = u + at = 0 + (2.2)(19) = 41.8 ms-1Braking list becomesu = 41.8 ms-1 v = 0a = ?s = ?t = 5.1sStill not enough infoNeed to calc accv = u + at0 = 41.8 + a(5.1)a = - 8.2 ms-2Now can calc ss = ut + at2 = (41.8)(5.1) + (-8.2)(5.1)2 = 213.2 - 106.6 = 106.6 mTotal Distance = 400 + 106.6 = 506.6 m Total Time = 19 + 5.1 = 24.1 sSo, Average Speed = 506.6/24.1 = 21 ms-1Average Speed

  • Motion - Revision Questions Question type:In a car the drivers head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the drivers 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 drivers head during this collision.A: Impulse = Change in Momentum Ft = (mv) So F = (mv)/t = (7.0)(8.0)/(1.6 x 10-1) = 350 NMomentum/Impulse

  • Motion - Revision Questions Question type: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 drivers 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 drivers 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 In a car the drivers head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the drivers 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.Air Bags/Crumple Zones

  • Motion - Revision Questions Question type: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.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 collisionMom before = (10 x 103)(6.0) + (5 x 103)(0)Mom after = (15x 103)(v) So v = 60,000/15,000 = 4.0 ms-1Momentum

  • Motion - Revision Questions Question type:Q: Calculate the magnitude of the total impulse that truck Y exerts on truck X A: Impulse = Change in MomentumTruck Xs change in momentum = Final momentum Initial Momentum = (10 x 103)(4.0) (10 x 103)(6.0) = -2.0 x 104 kgms-1The mechanism for this change in momentum is the impulse supplied by Truck YSo, I = 2.0 x 104 NsA 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.Impulse

  • Motion - Revision Questions Question type: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 collisionpBEFORE = (10 x 103)(6.0) + (5 x 103)(0.0) = 6 x 104 kgms-1pAFTER = (15 x 103)(4.0) = 6 x 104 kgms-1

    KEBEFORE = mv2 = (10 x 103)(6.0)2 = 1.8 x 105 JKEAFTER = mv2 = (15 x 103)(4.0)2 = 1.2 x 105 JThus Mom IS conserved but KE is NOT conserved, therefore this is an inelastic collision 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.Inelastic Collisions

  • Motion - Revision Questions Question type:Motion - Revision Questions Question type:A cyclist is towing a small trailer along a level bike track (Figure 1). The cyclist and bike have a mass of90 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 1What driving force is being exerted on the road by the rear tyre of the bicycle?

  • Motion - Revision Questions Question type: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 F = 0Newtons Laws

  • Motion - Revision Questions Question type: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 F = 10,000 - 2000 = 8000 NNow, we know that F = ma So, a = F/m = 8000/2200 = 3.64 ms-2Newtons 2nd LawQ: What is the magnitude of the seaplanes acceleration at the 200 m mark ?

  • Motion - Revision Questions Question type:Q: Estimate the work done by the seaplane against the opposing forces in travelling for a distance of 500 m.

    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 JA: 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 JWork

  • Motion - Revision Questions Question type: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.VectorsA: 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 Adams acceleration at the time shown in Figure C ? Explain your answer.

  • Motion - Revision Questions Question type: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. WNA: The two forces are: (a) Weight Force (W) acting through Adams centre of mass (b) Normal Reaction Force (N) acting from the point of contact b/w Adams feet and the air bag. N.B. N >W to meet requirements that acc be upwards at this timeNOTE: 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.Force Diagrams

  • Motion - Revision Questions Question type:Projectile MotionQ: Which of the directions (A - H), best shows the VELOCITY of the car at X ?A car takes off from a ramp and the path