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2 2.1
Mechanics
Kinematics
Assessment statements 2.1.1 Define displacement, velocity, speed, and acceleration. 2.1.2 Explain the difference between instantaneous and average values of speed, velocity and acceleration. 2.1.3 Outline the conditions under which the equations for uniformly accelerated motion may be applied. 2.1.4 Identify the acceleration of a body falling in a vacuum near the Earth’s surface with the acceleration g of free fall. 2.1.5 Solve problems involving the equations of uniformly accelerated motion. 2.1.6 Describe the effects of air resistance on falling objects. 2.1.7 Draw and analyse distance–time graphs, displacement–time graphs, velocity–time graphs and acceleration–time graphs. 2.1.8 Calculate and interpret the slopes of displacement–time graphs and velocity–time graphs, and the areas under velocity–time graphs and acceleration–time graphs. 2.1.9 Determine relative velocity in one and two dimensions.
What is kinematics? Kinematics is the study of motion. As we have already seen, distance is a scalar and displacement is a vector quantity. Distance travelled is a simple measure of length while displacement is defined as the distance travelled in a particular direction. When we want to describe how fast something is moving we say that: distance speed _______ time displacement velocity ___________ time These simple equations can be used for two situations; firstly when the body is moving at a constant speed, and secondly in order to find the average speed. In the second case we need to know the total distance and the total time.
Worked example Calculate your average speed if your journey to school takes half an hour and the distance is 10 km. Solution 10 km speed ______ 0.5 h
distance Examiner’s hint: Speed ________ time
20 kmh1
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This works even though your car or bus has to slow down, speed up and sometimes stop according to the traffic conditions. If your vehicle is changing speed, or direction, it must be changing velocity, which means that it is accelerating. Acceleration is defined as the change in velocity per unit time. Acceleration a can be written as: v a ___ t where v is change in velocity and t is change in time.
Worked example A car starts from rest and accelerates steadily to a speed of 40 ms1 in a time of 5.0 s. Find the magnitude of the acceleration of the car. Solution 1
40 ms a _______ 5.0 s 8.0 ms2
The units of velocity and acceleration are different, and you must always be careful to measure speed and velocity in ms1 but acceleration in ms2. If a ball is rolling down a slope, its speed will be increasing. We can find the average speed of the ball by dividing the total distance by the total time, as seen earlier. Another way to describe the motion is by using the instantaneous speed of the ball; the speed at any instant in time. You can find this by drawing a distance–time graph of the motion and measuring the tangent at a particular instant – we will see how this works later on in this chapter.
The equations for uniformly accelerated motion It is essential that you remember that ‘s ’ in these equations stands for distance travelled, and not for speed.
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If speed or velocity is changing at a steady rate, then the acceleration is constant. In these situations we can use the equations of uniformly accelerated motion, which we will call the suvat equations. These are a set of equations with five variables as follows: s distance (m) u initial velocity (ms1) v final velocity (ms1) a acceleration (ms2) t time (s) UNCORRECTED PROOF COPY
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There are four suvat equations and their power is that if we know any three of the five variables, by using the equations carefully, we can solve for the other two unknown quantities. These are the equations: v u at vt _____ su 2 s ut _12 at 2 v 2 u 2 2as
Worked examples 1 A ball is thrown vertically upwards with an initial velocity of 20 ms1. Taking the acceleration due to gravity as 10 ms2 and neglecting air resistance, find: (a) the time taken to return to the thrower’s hand v � 0 ms�1 (b) the maximum height reached.
Figure 2.1 The maximum speed will be reached just after the ball leaves the girl’s hand.
a � �g � �10 ms�2 u � 20 ms�1
Solution (a) s ? u 20 ms1 v0 a 10 ms2 t?
Examiner’s hint: First list the suvat variables. Then choose the best equation, the one with the relevant variables.
v u at 0 20 10t 10t 20 t2s
Examiner’s hint: Always write down the equation you are using, sometimes you can get a mark for just writing down the correct equation. Examiner’s hint: Next plug in the numbers and solve for the unknown.
Time to return to the thrower’s hand will be twice this 4 s. (b) s ? u 20 ms1 v0 a 10 ms2 t2s
In this case the final velocity, when the ball reaches the maximum height, will be zero. The acceleration will be negative, if we consider the ball travelling upwards, because it will be slowing down. This type of problem involving gravity is always symmetrical; the distance, time and speed going up are the same as those coming back down.
Examiner’s hint: To find the maximum height we again list the variables for the motion upwards.
vt _____ su 2 20 ______ s 02 2 s 20 m
Examiner’s hint: Choose an equation and write it down
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2 A stone is dropped into the sea from a cliff 45 m high. Taking the acceleration due to gravity as 10 ms2 and neglecting air resistance find: (a) the speed of the stone when it hits the sea (b) the time taken to reach the sea. Examiner’s hint: This time the initial speed is zero and the acceleration is positive. Again we list the variables:
(a) s 45 m u0 v? a 10 ms2 t?
u � 0 ms�1 a � g � 10 ms�2
Figure 2.2 Acceleration is positive in this case. The stone accelerates until it hits the water.
v � 30 ms�1
v2 u2 2as v2 0 (2 10 45) 900 v 30 ms1 Examiner’s hint: To find the time in this case we could use any of the other three equations.
(b) s 45 m u0 v 30 ms1 a 10 ms2 t? s ut _12at 2 45 0 (0.5 10 t 2) 45 5t 2 t2 9 t 3s Try this last problem for yourself using the other two equations to make sure you get the same value for the time taken.
To view an animation of motion in 2D, visit heinemann.co.uk/hotlinks, enter the express code 4266S and click on the Weblink 2.1.
In the two examples above, the acceleration of the bodies was g, the acceleration due to gravity. This varies from place to place on the Earth’s surface, but is often taken as 9.81 m s2, or approximated to 10 m s2. The suvat equations work for any situation where the acceleration of the body is constant, and you must learn to spot the type of problem where the suvat equations will provide the solution. Exercises 1 In an experiment a small steel ball is filmed and timed as it falls, from rest, next to a metre ruler. Calculate the time taken for the ball to fall (a) from the zero mark to the 90 cm mark on the ruler (b) from the 90 cm mark to the end of the ruler. Neglect air resistance and take g 9.8 ms2. 2 A car is travelling at 20 ms1 when a dog runs out into the road. If the driver’s reaction time is 0.40 s and the car decelerates steadily at a rate of 20 ms2, calculate the minimum distance travelled before the car can come to rest.
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In the real world, bodies do not fall in a vacuum they experience the effects of air resistance. As long as the body is small and hard, and the height is not too great, then we can ignore these effects. Otherwise air resistance or drag becomes important; obviously the motion of a falling body such as a feather or a leaf will be significantly changed by the air through which it falls. If a feather or a leaf were to fall in a vacuum they would drop just like a stone.
The acceleration due to gravity is slightly lower off the coast of India and slightly higher in the South Pacific, owning to the nature of the rocks in the Earth’s crust.
Imagine what happens to the speed of a skydiver who jumps out of a plane, falls free for a while and finally opens a parachute.
Initially the skydiver’s speed is low, so the air resistance is much smaller than the weight, and the skydiver accelerates. As the speed increases, the air resistance increases until a point is reached when the drag force is equal and opposite to the weight. Then the skydiver will fall at a constant, maximum speed called the terminal velocity. When the parachute opens, the air resistance increases significantly and a new, slower terminal velocity is reached, enabling a safe landing. Figure 2.3 A velocitytime graph of a parachute jump.
B
velocity (ms�1)
C
A D
O
E
Examiner’s hint: When you are asked to sketch a graph, you should draw a pair of labelled axes with units, and a line showing the trend. You do not have to show numbers on a sketch graph but you should draw it clearly with a ruler and pencil.
time (s)
From O to A, the velocity is low and acceleration fairly constant. From A to B, air resistance becomes greater as the velocity increases so the rate of acceleration decreases. From B to C, acceleration is zero, the skydiver is falling at constant terminal velocity. At C, the parachute opens and velocity rapidly decreases. From D to E, a new, lower, terminal velocity is reached. At E, the skydiver lands on the ground. UNCORRECTED PROOF COPY
For a person in free fall, terminal velocity is about 180 kmh1 or 50 ms1, similar to a speeding car.
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In 1944, the plane in which Nicholas Alkemade was flying, was set on fire and his parachute destroyed. Nicholas chose to jump without a parachute rather than stay in the burning plane. Although he fell nearly 6000 m, he was extremely lucky because he blacked out, landing relaxed, through pine trees and into snow. He suffered only minor sprains, cuts and bruises and survived until 1987!
To view an animation of the lunar lander, visit Heinemann.co.uk/hotlinks, enter the express code 4266S and click on Weblink 2.2.
Graphing motion We have already seen some examples of motion graphs, but now we will look at them in more detail. Note that time always goes on the x-axis.
Distance or displacement vs time Figure 2.4 Four distancetime graphs. The car shown in these graphs is: 1. at rest, not moving 2. moving at a constant speed 3. accelerating 4. decelerating
1
d
2
O
t
distance (m)
O
time (s)
d
d
3
O
4
O
t
t
The next graph shows displacement against time; remember that displacement is a vector quantity. The way we show the difference between scalars and vectors on a graph is to draw the scalar above the x-axis only, and the vector both above and below. Figure 2.5 Displacementtime graph. O–A constant velocity forwards A–B decelerating B–C stopped C–D accelerating and changing direction D–E constant velocity back to starting point E–F constant velocity backwards F–G decelerating G–H stopped H–I accelerating and again changing direction I–J constant velocity back to starting point
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B A
C D
displacement O (m)
J
E I
F G
time (s)
H
The slope or gradient of a distance–time graph gives the speed, while for a displacement–time graph the gradient gives the velocity. In Figure 2.5 above, the car is travelling at constant velocity from O to A because the gradient is constant. From A to B, the velocity is decreasing because the gradient is decreasing; in other words the car is decelerating. UNCORRECTED PROOF COPY
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40 s
2s 20 m 40 m
B
A
instantaneous velocity � 402 sm � 20 ms�1
instantaneous velocity � 204 sm � 5 ms�1
Figure 2.6 The instantaneous velocity is found by by measuring the slope of the tangent to the curve. A The tangent line is steep at this point indicating a high instantaneous velocity. B Because the curve has flattened out at this point, the tangent is less steep and the instantaneous velocity is lower.
Velocity vs time It is vitally important that you do not mix up the various different types of motion graph. The next ones we will consider are speed and velocity against time. Try and describe what is happening to a car moving as follows: A
Figure 2.7 Velocity–time graph.
B
velocity (ms�1)
O
C
time (s)
The car starts from rest when the speed is zero, and then from O to A its speed increases at a steady rate. This is known as constant acceleration. Then from A to B the car is not stopped, but is moving at constant speed. From B to C we have constant deceleration and finally the car again comes to rest. As long as the car is travelling in a straight line in the same direction, this graph would be exactly the same with either speed or velocity on the y-axis. The gradient of a velocity–time graph gives the acceleration. So in Figure 2.7 above from O to A the acceleration is constant because the gradient is constant; the graph is a straight line. From A to B the gradient is zero because the rise, or y term is zero; so the acceleration is zero. From B to C the gradient is steeper than from O to A and this tells us that the rate of deceleration was more rapid than the rate of acceleration.
The slope or gradient of a graph y y2 y1 ___ rise ______ ___ run x2 x1 x
Now we will analyse the same graph using numbers. 10
A
Figure 2.8 Finding distance travelled from a velocitytime graph. O–A constant acceleration
B
10 ms 2 ms2 _______ 5s 1
velocity (ms�1)
A–B B–C
O
C 5
15 17 time (s)
constant maximum, velocity of 10 ms1 for 10 s constant deceleration 10 ms 5 ms2 _______ 2s 1
The area under a velocity–time graph gives the displacement. In this case the area is the sum of a triangle, a rectangle and a smaller triangle. Remember that the area of a triangle is half the base times the height. UNCORRECTED PROOF COPY
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Displacement (_12 5 s 10 ms1) (10 s 10 ms1) (_12 2 s 10 ms1) 25 m 100 m 10 m 135 m You need to be able to sketch and label these types of graph for various situations. Exercises 3 A girl throws a ball up in the air and then catches it. Sketch a speed–time graph and then a velocity–time graph for the motion of the ball. 4 A ball is dropped onto the floor and bounces twice. Sketch a velocity–time graph and then a speed–time graph for the motion of the ball.
Acceleration vs time The last type of motion graph we are going to look at is acceleration against time. If the acceleration is uniform or constant, then these graphs will be simply vertical and horizontal straight lines. An acceleration–time graph for the car in Figure 2.8 would look like this: Figure 2.9 Acceleration–time graph O–A constant acceleration of 2 ms2 for 5 s A–B zero acceleration for 10 s B–C constant negative acceleration or deceleration of 5 ms2 for 2 s
acceleration (ms�2)
2
O
A 5
B C 15
17 time (s)
�5
The area under an acceleration–time graph gives the change in velocity. So from O to A, the change in velocity the area of the rectangle (5 s 2 ms2) 10 ms1 From A to B, the area under the graph is zero, so the change in velocity is zero. You can check for yourself that the area from B to C is 10 ms1 in the opposite direction, and that this all fits with the data in Figure 2.8. If the acceleration is not constant, we cannot use the suvat equations to solve the problems. The area under the graph will still tell us the change in velocity, however, as shown in the next example. 22
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Worked example Find the change in velocity of a body that accelerates as shown in the graph below. 10
B
Figure 2.10 Accelerationtime graph. The change in velocity is the area of triangle ABC plus area of rectangle OACD.
acceleration (ms�2)
2
C
A
D 6 time (s)
O
Solution Change in velocity (_12 6 s 8 ms2) (6 s 2 ms2) 24 ms
1
1
12 ms
1
36 ms
To view The moving man simulation, visit Heinemann.co.uk/hotlinks, enter the express code 4266S and click on the Weblink 2.3.
Relative velocity Imagine you are riding in a train travelling at 140 kmh1. Another train overtakes you at a speed of 150 kmh1. It looks like the other train is moving quite slowly. This is because relative to you, the speed of the other train is only 10 kmh1. Think about the following scenario; Diego is standing on the platform and Maria, on a train, passes through. We can talk about the relative motion in two ways; the motion of Maria relative to Diego, or the motion of Diego relative to Maria. The magnitude, or speed, will be the same but the direction will clearly be different. Maria is on the train travelling to the right with a constant velocity 7 ms�1
Figure 2.11 Relative velocity. The velocity of Maria, relative to Diego is 7 ms1 to the right. The velocity of Diego relative to Maria is 7 ms1 to the left.
7 ms�1
Diego is on the platform
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Mechanics
Head-on car collisions are likely to have serious consequences. One reason is that the relative speeds will add up, and so the resulting damage will also increase.
So far we have been looking at relative motion in one dimension. To understand how it works in two dimensions, imagine you are in a car driving down a long straight road. A bird is flying at right angles to the road and you can see it out of the window. motion of car relative to bird
motion of bird relative to car
at 90° to the road Figure 2.12 The motion of the bird relative to the car is different from the motion of the car relative to the bird.
You will learn a lot more about relative motion if you study Option D which includes Einstein’s theory of Special Relativity.
car driving along a straight road at constant velocity
Practical hints on drawing graphs When you are asked to sketch or plot a motion graph, think about the following points: 1 Is this a graph of distance, displacement, speed, velocity or acceleration? They are all usually sketched against time but they are all different graphs! 2 Is this a scalar or a vector graph? In other words do you need to draw it just above the line, or both above and below? 3 Remember to label the axes with both the name and the unit; sometimes you will get a mark for just doing that alone. 4 Does the graph start from the origin (the zero-zero point)? 5 Is the graph going to be a straight line or a smooth curve? 6 Make sure you plot the graph with a sensible scale; use numbers like 2, 4 and 5, not numbers like 3 and 7. 7 Aim for your graph to fill at least half the available space. 8 Measure the gradient or tangent over at least half the space. 9 Link what you need to find to the gradient, intercept or area under the graph. 10 Always remember the equation of a straight line is of the form y mx c and link it whenever possible to any graph you come across. Much of this applies to other types of physics graphs, not just motion graphs.
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2.2
Forces and dynamics
Assessment statements 2.2.1 Calculate the weight of a body using the expression W mg. 2.2.2 Identify the forces acting on an object and draw free-body diagrams representing the forces acting. 2.2.3 Determine the resultant force in different situations. 2.2.4 State Newton’s first law of motion. 2.2.5 Describe examples of Newton’s first law. 2.2.6 State the condition for translational equilibrium. 2.2.7 Solve problems involving translational equilibrium. 2.2.8 State Newton’s second law of motion. 2.2.9 Solve problems involving Newton’s second law. 2.2.10 Define linear momentum and impulse. 2.2.11 Determine the impulse due to a time-varying force by interpreting a force–time graph. 2.2.12 State the law of conservation of linear momentum. 2.2.13 Solve problems involving momentum and impulse. 2.2.14 State Newton’s third law of motion. 2.2.15 Discuss examples of Newton’s third law.
Force, weight and mass In simple terms a force is a pull or a push. Force is a vector quantity and is measured in newtons. There are many different types of force, including • friction • normal or supporting or contact force • tension and compression • air resistance or drag • upthrust or buoyancy • lift • thrust • weight.
Figure 2.13 Force can be a pull or a push.
Weight is the pull of gravity on a body. Assuming that the body is on the surface of the earth, it is directed towards the centre of the earth. Usually we call this direction ‘down’. Strictly speaking, weight and mass are not the same thing. Weight is a vector measured in newtons, and mass is a scalar measured in kilograms. Mass can be described as the amount of matter in a body. In everyday speech, weight and mass are taken to mean the same, but in physics, we need to remember that there is a difference. If you went to the moon, your mass would not change, but your weight would decrease because the pull of gravity is less on the moon.
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Astronauts on the moon were able to take giant steps because their weight was less (although their mass hadn’t changed).
Weight is related to mass in a simple way: W mg W is the weight measured in N m is the mass measured in kg g is the acceleration due to gravity measured in ms2
Worked example What is the weight of a 1.0 kg mass on the surface of the earth where g 9.8 ms2? Solution W mg W 1.0 kg 9.8 ms2 9.8 N
If a body is in translational equilibrium, it means that the forces acting on the body, in all three dimensions, are balanced.
Balanced forces Here, translating means moving from one place to another in a straight line. If a body is in translational equilibrium then it will not be accelerating. This is the basic idea described in Newton’s first law. We have already seen some free-body diagrams in Chapter 1, but here are some more examples. The helium balloon is at rest, so must be in translational equilibrium. upthrust
upthrust weight tension The weight of the balloon will be small but cannot be zero. If the string breaks, then the tension becomes zero and the balloon will accelerate upwards. Figure 2.14 Notice that the direction of the tension in the string must be downwards to balance the forces. 26
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Figure 2.15 If the person pulls harder and just succeeds in lifting the block, then the contact force becomes zero and the pulling force is equal to the weight.
pulling force
contact force
FN Ffr x weight
Figure 2.16 Inclined plane the weight is resolved into x and y components, parallel and perpendicular to the incline. The three forces acting on the block are mg (the weight), FN anf Ffr.
θ y
θ mg
The free-body diagram must show only the three real forces acting on the block. Since the block is accelerating, the forces are not balanced down the incline. x > Ffr y FN where Ffr is the frictional force and FN is the normal (or perpendicular) force. Using geometry, you can see that the angle of the incline is equal to the angle between the weight vector and the y-component. This is because there are 180° in a triangle and the normal force is at 90° to the incline. Using trigonometry, we can then deduce that: x sin ___ mg x mg sin y cos ___ mg y mg cos
Newton’s laws of motion The last recorded outbreak of the Black Death in England was in 1665. In those days people thought that the plague would remain local, so they returned to their villages, often spreading the infection. In 1665 Isaac Newton was a young Cambridge professor, working on some problems regarding the physics of motion. The plague forced him into isolation at his family home in Woolsthorpe and he worked alone there for eighteen months. During this time he made several completely original contributions to mathematics and science, effectively inventing calculus and revolutionizing our ideas about gravitation and light.
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Constant velocity means two things, constant speed and constant direction; in other words constant speed in a straight line.
Aristotle (384322 BC) was one of the most important of the ancient Greek philosophers and he tutored Alexander the Great. Aristotle thought that a heavy object would fall more quickly than a light one, and that a force was necessary for an object to move. Galileo (15641642) was born in Pisa, Italy and in many ways was the founder of modern physics. Galileo thought that a heavy object and a light object should fall at the same rate and that a body could move forever if no force were acting. Who was right, Aristotle or Galileo?
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Newton’s first law states that a body will either remain at rest, or move with constant velocity, unless acted on by an unbalanced force. We have already seen how this law works, and it is simply common sense that something does not start to move unless there is resultant force acting on it. More difficult to understand is the idea that a body can be in equilibrium, even though it is moving. It can be difficult to accept that a body can move without a force being exerted, as this goes against both common sense and the teaching of Aristotle. Newton’s second law tells us what happens when an unbalanced force acts and is expressed in the most important equation of basic mechanics: F m a
F ma is the resultant or unbalanced force measured in N is the mass measured in kg is the acceleration measured in ms2
Note that the earlier definition of weight is a special case of this general law. For a constant mass, this means that force is directly proportional to acceleration. If we take a constant out of a simple equation, then we get a relationship that is directly proportional. For example if: y mx and m is a constant then
If two variables are directly proportional to each other it means that if one increases, the other will increase in proportion, or at the same rate. A graph of the two variables will be a straight line through the origin.
yx
Conversely if two things are directly proportional to each other and we put in a constant then we get an equality or equation. For example if: Fx and k is a constant then
F kx
Worked example Find the acceleration of a block of mass 500 g pulled across a table with a force of 12 N, if there is a frictional force of 6 N. Figure 2.17 The resultant force 6 N to the right.
force of friction
500 g
pulling force 12 N
6N
Solution The force in Newton’s second law is always the resultant or unbalanced force.
F ma F a __ m 6N a _____ 0.5 kg a 12 ms2 to the right.
Practical hints on calculations The basic method for numerical physics problems is not complicated. 1 Choose the correct equation 2 Usually you can find the equations in the Data Booklet but sometimes you simply have to memorise them. 28
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3 Make sure you know what the symbols (letters) stand for. 4 When solving the problem, always write down the equation you are using sometimes you will get a mark for just writing the equation. 5 Plug in the numbers. 6 Make sure the units are consistent; if not, change them all to SI units. 7 Give your final answer to the correct number of significant figures and make sure you have included the correct unit. 8 Before you move on, ask yourself whether the answer you have given is sensible and realistic. Exercises 5 A train of mass 1.5 105 kg is travelling at 40 ms1 when the brakes are applied and it decelerates steadily. The train travels a distance of 250 m before coming to a halt. (a) Calculate the deceleration of the train. (b) Find the average braking force.
To view a simulation of forces in one dimension – The ramp, visit Heinemann.co.uk/hotlinks, enter the express code 4266S and click on the Weblink 2.4.
6 A large helium balloon is attached to the ground by two fixing ropes. Each rope makes an angle of 50° with the ground. There is a force F vertically upwards of 2.15 103 N. The total mass of the balloon and its basket is 1.95 102 kg. (a) State the magnitude of the resultant force when it is attached to the ground. (b) Calculate the tension in either of the fixing ropes. (c) The fixing ropes are released and the balloon accelerates upwards. Calculate the magnitude of this initial acceleration. (d) The balloon reaches a terminal velocity 10 s after take-off. The upward force F remains constant. Describe how the magnitude of the air resistance on the balloon varies during the first 10 s of its flight.
Momentum and impulse The momentum of a body is a property that depends on its mass and its velocity. Linear momentum is defined as the product of mass and velocity. Linear simply means in a straight line. Note that the definition refers to velocity, not speed. momentum mass velocity p mv p m v
Momentum is a vector quantity.
is the momentum measured in kg ms1 is the mass measured in kg is the velocity measured in ms1
Given that mass usually remains constant, this is often written: p mv
Worked example Find the momentum of a car of mass 600 kg travelling at a constant velocity of 72 kmh1. Solution
72 1000 ms1 72 kmh1 _________ (60 60) 20 ms1 p mv p 600 kg 20 ms1 p 12 000 kg ms1
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Examiner’s hint: The velocity is not given in SI units so we must convert.
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Conservation of momentum Momentum is important for solving problems involving collisions and explosions. We say that momentum is conserved in these interactions. This means that the total momentum before an event is equal to the total momentum after the event.
Worked example A ball of mass 1.0 kg is dropped and hits the floor with a velocity of 8 ms1. It bounces back with a velocity of 6 ms1. Find the change of momentum. Figure 2.18
8 ms�1 6 ms�1
For a vector problem like this, we must use a convention for direction. Usually we take the directions up or to the right as positive and down or to the left as negative. Figure 2.19 Vector quadrant.
Solution The change in momentum final momentum initial momentum �
�
�
�
p mv mu p (1.0 kg 6 ms1) (1.0 kg 8ms1) p 6 kg ms1 8 kg ms1 p 14 kg ms1
This idea can be expressed as a problem-solving equation in different ways: m1v1 m2v2 The law of conservation of linear momentum states that the total momentum remains constant in any interaction, providing there is no external force.
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m1v1 m2v2 m3v3 m1v1 m2v2 m3v3 The version we choose depends on the problem, whether it is a collision or explosion and whether the two bodies stick together after the collision. UNCORRECTED PROOF COPY
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Worked example A car of mass 600 kg travelling at 20 ms1 collides head on with a truck of mass 2400 kg. If both vehicles come to rest immediately after the crash what was the velocity of the truck? Solution m1v1 m2v2 m3v3 (600 kg 20 ms1) (2400 kg v2) (3000 kg 0 ms1) 12 000 kg ms1 2400v2 0 12 000 v2 ______ 2400 v2 5 ms1
Examiner’s hint: Since the bodies involved stick together after the collision, use this equation Examiner’s hint: The final velocity, v3, is zero, and we will take the direction of motion of the car as positive.
Impulse The change in momentum of a body is also known as the impulse. Impulse is defined as the product of a force and the time during which it acts. Impulse Ft mv The unit for impulse is Ns and this can also be used as a unit of momentum. Imagine what happens if a girl kicks a football and a boy kicks a brick of the same mass, say 500 g. Both the ball and the brick are accelerated from rest to a velocity of 10 ms1. The difference is that the brick is harder than the ball so the girl’s foot is in contact with the ball for a longer time interval. If the girl’s foot is in contact with the ball for 0.2 s then we can calculate the force exerted like this: Ft mv F 0.2 s 0.5 kg 10 ms1 5 F ___ 0.2 F 25 N If the boy’s foot is in contact with the brick for 0.01 s then the force would be much greater: Ft mv F 0.01 s 0.5 kg 10 ms1 5 F ____ 0.01 F 500 N Figure 2.20 It is not a good idea to kick a brick. For the same change in momentum, if the time of contact is smaller, the force will be greater. We say that the time is inversely proportional to the force.
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Remember that Newton’s second law is: F ma As we saw earlier in the chapter: v a ___ t Substituting for a in the second law gives: ____ F mv t Knowing that: mv p We can re-write Newton’s second law as: p F ___ t Notice this can be rearranged to give the impulse equation: Ft p mv In words we say that impulse is the change in momentum while force is the rate of change of momentum.
Newton’s third law of motion Newton’s third law states that if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. Clearly there must always be two different bodies involved in the third law. An easy way to demonstrate this is with two spring balances.
N 0 5 10 15 20 25 30
A
Education is not the same in every country and depends on differences in language, politics, culture and religion, among many other things. In all educational systems there is however, agreement in the validity of the ideas behind Newton’s three laws of motion. 32
15 N N 0 5 10 15 20 25 30
15 N
Figure 2.21 Two spring balances. The reading on A will always be the same as the reading on B.
B
There are examples of third law pairs of forces all around us; here are just a few examples. • When you kick a football (or a brick), your foot exerts a force on the ball and the ball exerts an equal and opposite force on your foot. The ball moves, hopefully towards the goal, and you can feel the force on your foot. • When a cannon is fired, the cannon ball moves forward but the ball exerts an equal and opposite force on the cannon, so it recoils. • If you play tennis, your racket exerts a force on the ball and the ball exerts an equal and opposite force on your racket. • The Earth exerts a force on the Moon and the Moon exerts an equal and opposite force on the Earth. We know the Earth pulls the Moon because it remains in orbit around us and we know the Moon pulls on the Earth because it influences the tides in the oceans. UNCORRECTED PROOF COPY
The photo shows the racket strings being stretched and deformed.
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2.3
Work, energy and power
Assessment statements 2.3.1 Outline what is meant by work. 2.3.2 Determine the work done by a non-constant force by interpreting a force-displacement graph. 2.3.3 Solve problems involving the work done by a force. 2.3.4 Outline what is meant by kinetic energy. 2.3.5 Outline what is meant by change in gravitational potential energy. 2.3.6 State the principle of conservation of energy. 2.3.7 List different forms of energy and describe examples of the transformation of energy from one form to another. 2.3.8 Distinguish between elastic and inelastic collisions. 2.3.9 Define power. 2.3.10 Define and apply the concept of efficiency. 2.3.11 Solve problems involving work, energy and power.
Physical work Sometimes in science we use words in a different sense to the way they are normally used in the English language. Although a word often looks, sounds and is spelt the same, it sometimes has a different meaning in physics. One example is the definition of work in the study of physics. Work is defined as the force times the distance moved in the direction of the force. This means that if there is no movement in the direction of the force, then no work is done. If your car is stuck in the mud or the sand, you can push it as hard as you like, and get completely exhausted, but if you do not move it then you have done no work (on the car)! Work is a scalar quantity and is measured in joules. It is independent of the mass of the body being moved and the path taken. It only depends on the magnitude of the force and the distance moved in the direction of the force. The equation for work is: W Fscos W F s
is the work measured in J is the force measured in N is the distance moved measured in m is the angle between the applied force and the direction of motion
Worked example applied force 40 N
A force of 40 N acting at 60° to the horizontal pulls a block of mass 10 kg a distance of 2.0 m across a smooth surface. How much work is done? 1 kg
60°
Figure 2.22 Work done pulling a block.
x 2m
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Solution The size of the mass makes no difference to the work done. If the surface is smooth, there is no friction, so the resultant force in the xdirection is the x-component of the 40 N force. Since the x-component is adjacent to 60° angle: x cos 60° ___ 40 x 40 cos 60° N W Fscos W 40 cos 60° N 2 m W 40 J F
When a series of weights are hung on a spring causing it to stretch, work is done. The force is the weight and the distance moved is the extension of the spring. As long as the spring is not stretched beyond a certain point, called the elastic limit, then the force is directly proportional to the extension. A graph of force against extension is a straight line through the origin.
force (N)
O
extension (cm)
x
Figure 2.23 A graph of force against extension for a stretched spring.
The work done by this nonconstant force is found from the area under the graph. work done _12 x F W _12 x kx W _12 kx 2
The gradient k _Fx F kx k is the spring constant.
In physics, energy and work are very closely linked; in some senses they are the same thing. Both are scalar quantities and both are measured in joules. If you have energy, you can do work. In general, we say that: work done energy converted
Energy is defined as the ability to do work.
W E When work is done, energy is converted or changed to different types. In the case of the spring being stretched, the work done increases the elastic potential energy of the spring. elastic potential energy _12 kx 2
Kinetic energy Examiner’s hint: Kinetic energy is often abbreviated to KE or Ek.
Kinetic energy (Ek) is the energy a body has because it is moving. The word kinetic refers to motion. Imagine a block of mass m accelerated from rest by a resultant force, F. After travelling a distance s the mass has an acceleration a.
Examiner’s hint: Sometimes in an exam you will be asked to derive an equation, such as the kinetic energy derivation here. Many students find this difficult and there is really only one way to improve your skills. You simply have to practise and repeat them with the book closed and then practise some more! 34
The work done, W Fs but from Newton’s second law F ma so that W mas (substituting for F) From the suvat equations, v u 2 2as since u 0 then v 2 2as as _12 v 2 Since W mas E and the energy converted is kinetic energy kinetic energy mas m _12 v 2 Ek _12mv 2 UNCORRECTED PROOF COPY
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Worked example Find the kinetic energy of a truck of mass 4000 kg travelling at a speed of 30 ms1 Solution Ek _12mv2 Ek _12 4000 kg (30 ms1)2 Ek 1.8 106 J
Potential energy Gravitational potential energy (Ep) is the energy a body has because of its position, in particular, its vertical height relative to a given point.
Examiner’s hint: Potential energy is often abbreviated to PE or Ep.
Now imagine a mass m lifted a vertical height h. Here work is being done against the pull of gravity. the work done, W Fs here the force is the weight mg and the distance is the height h so the work W mgh The work is increasing the gravitational potential energy of the mass. W Ep Ep mg h
Worked example The Leaning Tower of Pisa is 56 m high. (a) Find the gravitational potential energy of a 10 kg cannon ball at the top of the tower. (b) If the cannon ball is dropped from the top of the tower, what will be its speed when it hits the ground? (Neglect air resistance and take g 10 ms2) Solution (a)
Ep mg h Ep 10 kg 10 ms2 56 m Ep 5600 J
(b)
Ep Ek 5600 J _12 mv 2 5600 J _12 10 kg v 2 5600 v 2 ____ 5 v 33 ms1
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Examiner’s hint: We could solve this using the suvat equations, but if air resistance is neglected, all the gravitational potential energy will be changed to kinetic energy. There is a famous story about Galileo dropping cannon balls of different weights from the top of the Leaning Tower of Pisa to demonstrate that heavy objects and lighter objects fall at the same rate. This story may or may not be true.
To view a more recent version of Galileo’s experiment, visit Heinemann.co.uk/hotlinks, enter the express code 4266S and click on Weblink 2.5. 35
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Conservation of energy The idea of one form of energy changing into another is summarised in a very important principle or law of physics, the principle conservation of energy. The principle of conservation of energy states that energy cannot be created or destroyed; only changed from one form to another.
We usually classify energy as the following forms: 1 Kinetic energy 2 Gravitational potential energy 3 Elastic potential energy
All of the energy on Earth has come originally from the Sun. Is this statement true? How do you know?
4 Thermal or heat energy 5 Light energy 6 Sound energy 7 Chemical energy 8 Electrical energy 9 Magnetic energy 10 Nuclear energy
Examiner’s hint: In an exam do not say that energy is ‘lost’ but instead say it has been changed or converted to other forms.
In any given transformation, some of the energy is almost inevitably changed to heat. There are energy changes happening all around us all of the time. Here are a few examples to think about. • A bouncing ball The gravitational potential energy is changed to kinetic energy and back again with each bounce. While the ball is in contact with the ground, some energy is stored as elastic potential energy. Eventually the ball comes to rest and all the energy has become low grade heat. • An aeroplane taking off As fuel is burned in the engines, chemical energy is converted to heat, light and sound. The plane accelerates down the runway and its kinetic energy increases. There will be heat energy due to friction between the tyres and the runway. As the plane takes off and climbs into the sky, its gravitational potential energy increases. • A nuclear power station Nuclear energy from the uranium fuel changes to thermal energy that is used to boil water. The kinetic energy of the steam molecules drives turbines, and as the kinetic energy of the turbines increases, it interacts with magnetic energy to give electrical energy. • A laptop computer The electrical energy form the mains or chemical energy from the battery is changed to light on the screen and sound through the speakers. There is kinetic energy in the fan, magnetic energy in the motors, and plenty of heat is generated. • A human body Most of the chemical energy from our food is changed to heat to keep us alive. Some is changed to kinetic energy as we move, or gravitational potential energy if we climb stairs. There is also elastic potential energy in our muscles, sound energy when we talk and electrical energy in our nerves and brain.
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There is a useful relationship between kinetic energy and momentum which will help you to solve numerical problems. mv 2 Ek _12 mv 2 ____ 2 2 2 m _____ Multiplying top and bottom by m gives Ek v 2m Momentum p mv Squaring gives Finally substituting in the above equation
p2 m 2v 2 p2 Ek ___ 2m
Collisions When we looked at momentum earlier, we mentioned that it was important for collisions. Now we must distinguish between two different types of collisions. When elastic collisions occur, momentum is conserved and kinetic energy is also conserved. This means that no energy is changed to heat in an elastic collision. In fact there is really no such thing as a perfectly elastic collision, although, of course, some collisions are more elastic than others. After an elastic collision the bodies always move separately. Billiard ball collisions are highly elastic because the balls are so hard. In a perfectly elastic collision, one ball could stop, and the other move off with the same speed as the first. In practice, this cannot happen because some of the kinetic energy changes to heat and sound. Particle physicists believe some collisions between sub-atomic particles are perfectly elastic.
During inelastic collisions, momentum is conserved, but kinetic energy is not conserved. In other words, some of the kinetic energy is converted to heat and other forms of energy. After an inelastic collision the bodies may stick together.
Examiner’s hint: In exam questions about collisions and explosions, especially multiple-choice, assume that momentum is always conserved.
Kinetic energy can sometimes increase, for example, if there is an input of chemical energy from a fuel or explosive material.
Power Power is a measure of how fast work is done or how quickly energy is converted. Power is a scalar quantity. energy or work power _____________ time If the energy or work is measured in joules and the time in seconds, then the power will be measured in watts (W).
Power is defined as the rate of doing work.
1 W 1 Js1 A 100 W light bulb converts electrical energy to heat and light at the rate of 100 J every second. Sometimes you will see power given in kW or even MW. UNCORRECTED PROOF COPY
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Worked example Calculate the power of a worker in a supermarket who stacks shelves 1.5 m high with cartons of orange juice, each of mass 6.0 kg, at the rate of 30 cartons per minute. Solution work power _____ time Fs power __ t (30 60 N) 1.5 m power _________________ 60 s power 45 W
This engraving shows a steam engine designed by James Watt.
James Watt (17361819) was a Scotsman whose work on the steam engine was a major influence on the Industrial Revolution.
From the days of Aristotle, through the time of Galileo, Newton, Watt and Joule right up until Einstein in the early 20th century, scientists worked mostly alone. Nowadays, scientists often work in multinational teams. These teams may make important discoveries in fields such as particle physics, or work on solving the great global problems like HIV/Aids and global warming.
There is another way to calculate power. Fs F _s Fv power __ t t In other words, power is equal to force times velocity, providing the velocity is constant.
Efficiency
One of the most efficient devices is the electric transformer. They can be over 99% efficient because, since there are no moving parts, there are no energy losses due to friction or air resistance.
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Efficiency is a ratio of how much work, energy or power we get out of a system compared to how much is put in. useful output efficiency ___________ total input Since it is a ratio, there is no unit, and we can express efficiency as a percentage by multiplying by 100%. No real machine or system can ever be 100% efficient, because there will always be some energy changed to heat due to friction, air resistance or other causes. UNCORRECTED PROOF COPY
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Worked example A car engine has an efficiency of 20% and produces an average of 25 kJ of useful work per second. How much energy is converted to heat per second? Solution
useful output efficiency ___________ total input 25 000 J 0.2 _________ total input
Examiner’s hint: The efficiency is 20% which equals 0.2. Examiner’s hint: 25 kJ 25 000 J.
25 000 125 000 J total input ______ 0.2 If the total input is 125 kJ, and 25 kJ of useful work are obtained then the heat produced must be: 125 kJ 25 kJ 100 kJ
Exercises 7 This question is about driving a metal bar into the ground. Large metal bars can be driven into the ground using a heavy falling object.
object mass � 2.0 � 103 kg
bar mass � 400 kg
In the situation shown, the object has a mass 2.0 × 103 kg and the metal bar has a mass of 400 kg. The object strikes the bar at a speed of 6.0 m s1. It comes to rest on the bar without bouncing. As a result of the collision, the bar is driven into the ground to a depth of 0.75 m. (a) Determine the speed of the bar immediately after the object strikes it. (b) Determine the average frictional force exerted by the ground on the bar. 8. This question is about estimating energy changes for an escalator (moving staircase). The diagram below represents an escalator. People B step on to it at point A and step off at point B. (a) The escalator is 30 m long and makes an angle of 40° with the horizontal. At full capacity, 30 m 48 people step on at point A and step off at point B every minute. (i) Calculate the potential energy gained by a person of weight 7.0 × 102 N in 40° moving from A to B. A (ii) Estimate the energy supplied by the escalator motor to the people every minute when the escalator is working at full capacity. (iii) State one assumption that you have made to obtain your answer to (ii). The escalator is driven by an electric motor that has an efficiency of 70 %. (b) Using your answer to (a) (ii), calculate the minimum input power required by the motor to drive the escalator.
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2.4
Uniform circular motion
Assessment statements 2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the centre of the circle. 2.4.2 Apply the expression for centripetal acceleration. 2.4.3 Identify the force producing circular motion in various situations. 2.4.4 Solve problems involving circular motion.
Motion in a circle If a body is travelling in a circular path, then it is constantly changing direction; at no point is it travelling in a straight line. Even if it is travelling at constant or uniform speed, a body that is moving round a circle must be changing velocity. This is because velocity is a vector and a change in velocity can be a change of speed or direction. If the velocity of the body is changing, then, by definition the body must be accelerating. This type of acceleration is called centripetal acceleration. The word centripetal means centre-seeking and the direction of the acceleration is always towards the centre of the circle. v
Figure 2.25 A particle moving in a circle. The instantaneous velocity, v, is the tangent to the circle. The direction of the centripetal force will be in the same direction as the centripetal acceleration.
a
The equation for centripetal acceleration is: v2 a __ r a centripetal acceleration ms2 v speed ms1 r radius of the circle m Since the body moves with constant speed we can use the simple equation: distance speed _______ time If the body goes round the entire circle once, then this can be written: r v 2____ T 2 r is the circumference of a circle. T is the period (or time period) and is the time taken for one complete revolution. If we square this equation it becomes: 2 2
4 r v 2 _____ T2 40
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Substituting for v 2 in the centripetal acceleration and cancelling one of the r terms gives: 2r a 4____ T2 If a body is accelerating, we know from Newton’s second law that it must be being acted on by an unbalanced force. The force causing the centripetal acceleration is called the centripetal force. The direction of the centripetal force is also towards the centre of the circle; this is logical because bodies accelerate in the direction of the resultant force. To understand this, think about a car driving round a circular track. If the friction is too low, for example, if the car skids on a patch of ice, then the car will fly off at a tangent.
The centripetal force is simply an extra name we give to a force that is already there; it is not a new or different force.
Figure 2.26 There must be friction between the car tyres and the road to keep the car going round.
v
Fc
The friction is the force that turns the car into the bends and it is directed towards the center of the circle. The friction here is the centripetal force. In the same way, when the Moon orbits the Earth there is a centripetal force from the Moon towards the centre of the Earth. In this case the force is gravitational. If you whirl a rubber bung or a stone tied to a piece of string around your head, then the centripetal force that keeps the stone turning is the tension in the string. If the string breaks and the tension becomes zero, the bung or stone flies off in a straight line.
Examiner’s hint: If you are asked to draw the force on a body moving in a circle, always draw it pointing towards the centre of the circle. Do not be tempted to add another arrow pointing outwards to balance the forces; they are not balanced because the body is accelerating.
If the equation for centripetal acceleration is: v2 a __ r and we know that
F ma
we can substitute for a to give: 2
mv F ____ r This is the equation for centripetal force.
Worked example A ball of mass 50 g tied to the end of a piece of string is whirled by a student in a horizontal circle of radius 1.2 m at constant speed. The ball makes 1.5 revolutions per second. (a) Find the acceleration of the ball. (b) Find the force that the student must exert on the string. UNCORRECTED PROOF COPY
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Solution distance speed _______ time 1.5 2 1.2 m v ______________ 1.0 s v 11.3 ms1
(a) Examiner’s hint: Since the ball is moving in a circle, the acceleration will be centripetal.
2
v a __ r (11.3 ms1)2 a __________ 1.2 m 107 ms2 2
mv F ____ r
(b)
0.050 kg (11.3 ms1)2 ____________________ 1.2 m 5.3 N
Exercises 9
An aircraft accelerates from rest along a horizontal straight runway and then takes off. The mass of the aircraft is 8.0 × 103 kg. (a) The average resultant force on the aircraft while travelling along the runway is 70 kN. The speed of the aircraft just as it lifts off is 75 m s1. Estimate the distance travelled along the runway. (b) The aircraft climbs to a height of 1250 m. Calculate the potential energy gained during the climb. (c) When approaching its destination, the pilot puts the aircraft into a holding pattern. This means the aircraft flies at a constant speed of 90 m s1 in a horizontal circle, and the radius of the circle is 500 m. For the aircraft in the holding pattern, (i) calculate the magnitude of the resultant force on the aircraft (ii) state the direction of the resultant force.
10 A geostationary satellite, for example a communications satellite, orbits the Earth with a time period of 24 hours. If the distance from the satellite to the centre of the Earth is 4.2 107 m, calculate the acceleration of the satellite.
Practice questions
1 A ball, initially at rest, takes time t to fall through a vertical distance h. If air resistance is ignored, the time taken for the ball to fall from rest through a vertical distance 9h is A 3t B 5t C 9t D 10t 2 A raindrop falling through air reaches a terminal velocity before hitting the ground. At terminal velocity, the frictional force on the raindrop is A zero. B less than the weight of the raindrop. C greater than the weight of the raindrop. D equal to the weight of the raindrop. 42
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3 An athlete runs round a circular track at constant speed. Which one of the following graphs best represents the variation with time t of the magnitude d of the displacement of the athlete from the starting position during one lap of the track? A d
0 0
C
B d
0 0
t
D d
d
0 0
t
0
t
4 A ball is dropped from rest at time t 0 on to a horizontal surface from which it rebounds. The graph shows the variation of time t with speed v of the ball.
v
0
t
A C
0
B
D
t
Which one of the following best represents the point at which the ball just loses contact with the surface after the first bounce? A B C D 5 Juan is standing on the platform at a railway station. A train passes through the station with speed 20 m s1 in the direction shown measured relative to the platform. Carmen is walking along one of the carriages of the train with a speed of 2.0 m s1 measured relative to the carriage in the direction shown. Velocity is measured as positive in the direction shown on the diagram. The velocity of Carmen relative to Juan is A 22 m s1 20 ms�1 B 18 m s1 C 18 m s1 D 22 m s1
Carmen 2.0 ms�1
Juan platform positive direction
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6 The graph below shows the variation with time of the distance moved by a car along a straight road. During which time interval does the car have its greatest acceleration? distance moved
A
B
C
time
D
7 The variation with time t of the speed v of a car moving along a straight road is shown below. v
S1 0
0
S2
S3 t
Which area, S1, S2 or S3, or combination of areas, represents the total distance moved by the car during the time that its speed is reducing? A S1 B S3 C S1 S3 D S1 S2 S3 8 When a body is accelerating, the resultant force acting on it is equal to its A change of momentum. B rate of change of momentum. C acceleration per unit of mass. D rate of change of kinetic energy. 9 A ball of mass m, travelling in a direction at right angles to a vertical wall, strikes the wall with a speed v1. It rebounds at right angles to the wall with a speed v2. The ball is in contact with the wall for a time t. The magnitude of the force that the ball exerts on the wall is
m (v1 v2) A _________ t D m (v1 v2)t m (v1 v2) B _________ t C m (v1 v2)t 44
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10 A sphere of mass m strikes a vertical wall and bounces off it, as shown below. momentum pB momentum pA
The magnitude of the momentum of the sphere just before impact is pB and just after impact is pA. The sphere is in contact with the wall for time t. The magnitude of the average force exerted by the wall on the sphere is (pB pA) A ________ t pB pA) (________ B t (pB pA) ________ C mt pB pA) (________ D mt 11 A truck collides head on with a less massive car moving in the opposite direction to the truck. During the collision, the average force exerted by the truck on the car is F T and the average force exerted by the car on the truck is FC. Which one of the following statements is correct? A F T will always be greater in magnitude than FC. B F T will always be equal in magnitude to FC. C F T will be greater in magnitude than FC only when the speed of the car is less than the speed of the truck. D F T will be equal in magnitude to FC only when the speed of the truck is equal to the speed of the car. 12 A small boat in still water is given an initial horizontal push to get it moving. The boat gradually slows down. Which of the following statements is true for the forces acting on the boat as it slows down? A There is a forward force that diminishes with time. B There is a backward force that diminishes with time. C There is a forward force and a backward force both of which diminish with time. D There is a forward force and a backward force that are always equal and opposite. 13 The graph below shows the variation with displacement 5 d of the force F applied by a spring on a cart. 4 The work done by the force in moving the cart through a distance of 2 cm is A 10 102 J 3 B 7 102 J F/N C 5 102 J 2 D 2.5 102 J 1 0
0
1 2 d/10�2 m
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14 Which one of the following is a true statement about energy? A Energy is destroyed due to frictional forces. B Energy is a measure of the ability to do work. C More energy is available when there is a larger power. D Energy and power both measure the same quantity. 15 An electric train develops a power of 1.0 MW when travelling at a constant speed of 50 ms1. The net resistive force acting on the train is A 50 MN B 200 kN C 20 kN D 200 N 16 Which of the following quantities is/are conserved in an inelastic collision in an isolated system of two objects? Linear momentum of system Kinetic energy of system A
Yes
Yes
B
Yes
No
C
No
Yes
D
No
No
17 The diagram below represents energy transfers in an engine. input energy EIN
engine
useful output energy EOUT
wasted energy EW
The efficiency of the engine is given by the expression
EW A ___ EIN EW B ____ EOUT EOUT C ____ EIN EOUT D ____ EW 18 A machine lifts an object of weight 1.5 103 N to a height of 10 m. The machine has an overall efficiency of 20%. The work done by the machine in raising the object is A 3.0 × 103 J B 1.2 × 104 J C 1.8 × 104 J D 7.5 × 104 J. 46
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19 An electric motor, with an input power of 250 W, produces 200 W of mechanical power. The efficiency of the motor is A 20%. B 25%. C 55%. D 80%. 20 A stone of mass m is attached to a string and moves round in a horizontal circle of radius R at constant speed V. The work done by the pull of the string on the stone in one complete revolution is A zero B 2mV 2 2mV 2 C ______ R 2mV 2 D ______ R 21 Two objects X and Y are moving away from the point P. The diagram below shows the velocity vectors of the two objects.
velocity vector for object Y
P
velocity vector for object X
Which of the following velocity vectors best represents the velocity of object X relative to object Y? A
B
C
22 This question is about throwing a stone from a cliff. Antonia stands at the edge of a vertical cliff and throws a stone vertically upwards.
D
v � 8.0 ms�1
The stone leaves Antonia’s hand with speed v 8.0 ms1. The acceleration of free fall g is 10 m s2 and all distance measurements are taken from the sea point where the stone leaves Antonia’s hand. (a) Ignoring air resistance, calculate (i) the maximum height reached by the stone. (ii) the time taken by the stone to reach its maximum height. The time between the stone leaving Antonia’s hand and hitting the sea is 3.0 s. (b) Determine the height of the cliff.
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23 This question is about the collision between two railway trucks (carts). (a) Define linear momentum. In the diagram on the right, truck A is moving along a horizontal track. It collides with a stationary truck B, and on collision, the two join together. Immediately before the collision, truck A is moving with speed 5.0 ms1. Immediately after collision, the speed of the trucks is v.
5.0 ms�1 A
B
immediately before collision v A
B
immediately after collision
The mass of truck A is 800 kg and the mass of truck B is 1200 kg.
(b) (i) Calculate the speed v immediately after the collision. (ii) Calculate the total kinetic energy lost during the collision. 24 An elevator (lift) starts from rest on the ground floor and comes to rest at a higher floor. Its motion is controlled by an electric motor. A simplified graph of the variation of the elevator’s velocity with time is shown below.
0.80 0.70 0.60 0.50 velocity 0.40 ( ms�1) 0.30 0.20 0.10 0.00 0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0 11.0 12.0 time (s)
The mass of the elevator is 250 kg. Use this information to calculate (a) the acceleration of the elevator during the first 0.50 s (b) the total distance travelled by the elevator (c) the minimum work required to raise the elevator to the higher floor (d) the minimum average power required to raise the elevator to the higher floor (e) the efficiency of the electric motor that lifts the elevator, given that the input power to the motor is 5.0 kW.
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25 This question is about momentum and the kinematics of a proposed journey to Jupiter. (a) State the law of conservation of momentum. A solar propulsion engine uses solar power to ionize atoms of xenon and to accelerate them. As a result of the acceleration process, the ions are ejected from the spaceship with a speed of 3.0 104 ms1. spaceship mass � 5.4 � 102 kg
xenon ions speed � 3.0 � 104 ms�1
(b) The mass (nucleon) number of the xenon used is 131. Deduce that the mass of one ion of xenon is 2.2 1025 kg. (c) The original mass of the fuel is 81 kg. Deduce that, if the engine ejects 7.7 1018 xenon ions every second, the fuel will last for 1.5 years. (1 year 3.2 107 s) (d) The mass of the spaceship is 5.4 × 102 kg. Deduce that the initial acceleration of the spaceship is 8.2 × 105 m s2. The graph below shows the variation with time t of the acceleration a of the spaceship. The solar propulsion engine is switched on at time t 0 when the speed of the spaceship is 1.2 103 m s1. 10.0
9.5 a/�10�5 ms�1 9.0
8.5
8.0 0.0
1.0
2.0
3.0 t/�107 s
4.0
5.0
6.0
(e) Explain why the acceleration of the spaceship is increasing with time. (f) Using data from the graph, calculate the speed of the spaceship at the time when the xenon fuel has all been used. (g) The distance of the spaceship from Earth when the solar propulsion engine is switched on is very small compared to the distance from Earth to Jupiter. The fuel runs out when the spaceship is a distance of 4.7 1011 m from Jupiter. Estimate the total time that it would take the spaceship to travel from Earth to Jupiter.
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