What is the uncertainty in the volume of the cube? The percentage uncertainty of which quantity has the largest effect on the L percentage uncertainty in Q? Which of the following graphs will result in a straight-line graph? He measures the height. Using a stopwatch, he measures the time for the ball to drop to the floor. The speed of the motorboat in still water is 4. The river is 30 m wide. There is a current in the river whose speed with respect to the shore is 3.
Determine the distance from P at which he will reach the shore. Determine the direction in which she has to turn her boat to do this. She graphed the following results. Explain to the student how the results show she is not correct. This section is an introduction to the basic concepts used in describing motion. We will begin with motion in a straight line with constant velocity and then constant acceleration.
Knowledge of uniformly accelerated motion allows analysis of more complicated motions, such as the motion of projectiles.
We will begin our discussion of motion with straight line motion in one dimension. This means that the particle that moves is constrained to move along a straight line. The position of the particle is then described by its coordinate on the straight line Figure 2.
If the line is horizontal, we may use the symbol x to represent the coordinate and hence the position. If the line is vertical, the symbol y is more convenient. In general, for an arbitrary line we may use a generic name, s, for position. So in Figure 2. Understand the difference between distance and displacement. Understand the difference between speed and velocity.
Understand the concept of acceleration. Analyse graphs describing motion. Solve motion problems using the equations for constant acceleration. Discuss the motion of a projectile. Show a qualitative understanding of the effects of a fluid resistance force on motion. Understand the concept of terminal speed.
As the particle moves on the straight line its position changes. In uniform motion the graph of position against time is a straight line Figure 2. In equal intervals of time, the position changes by the same amount. This means that the slope of the position—time graph is constant. The average velocity during an interval of time t is the ratio of the change in position s during that time interval to t.
The velocity is the same at all times. Positive velocity means that the coordinate s that gives the position is increasing. Negative velocity means that s is decreasing. Worked example 2. Determine the time and position at which they will meet. The position of A is given by the formula:. They will meet when they are the same position, i.
Consider two motions shown in Figure 2. The distance travelled is the actual length of the path followed and in this case is also 20 m. The second motion is an example of motion with changing direction. The change in the position of this particle, i. But the distance travelled by the particle the length of the path is 8.
So we must be careful to distinguish distance from displacement. Distance is a scalar but displacement is a vector. Numerically, they are different if there is a change of direction, as in this example. For constant velocity, the graph of velocity versus time gives a horizontal straight line Figure 2. An example of this type of motion is coasting in a straight line on a bicycle on level ground Figure 2.
Figure 2. Uniformly accelerated motion In the last section we discussed uniform motion. This means motion in a straight line with constant velocity. In such motion the graph of position versus time is a straight line. In most motions velocity is not constant. In uniformly accelerated motion the graph of velocity versus time is a non-horizontal straight line Figure 2.
In equal intervals of time the velocity changes by the same amount. The slope of the velocity—time graph is constant. Acceleration is the rate of change of velocity. When the acceleration is positive, the velocity is increasing Figure 2. Negative acceleration means that v is decreasing. The plane reaches a take-off speed of km h—1 about 72 m s—1 in about 2 seconds, implying an average acceleration of about 36 m s—2. The distance travelled until take-off is about 72 m. Determine the time at which the particle stops instantaneously.
The particle is getting slower. At some point it will stop instantaneously, i. Defining velocity in non-uniform motion But how is velocity defined now that it is not constant?
We would like to have a concept of the velocity at an instant of time, the instantaneous velocity. We need to make the time interval t very small. In other words, instantaneous velocity is the average velocity obtained during an interval of time that is very, very small. As there is uniform acceleration, the graph is a curve. Choose a point on this curve. Draw the tangent line to the curve s and at the point. In Figure 2.
Instantaneous velocity can be positive or negative. The magnitude of the instantaneous velocity is known as the instantaneous speed. We define the average speed to be the total distance travelled divided by the total time taken. The average velocity is defined as the change in position i. Consider the graph of velocity versus time in Figure 2. Imagine approximating the straight line with a staircase. The area under the staircase is the change in position since at each step the velocity is constant.
If we make the steps of the staircase smaller and smaller, the area under the line and the area under the staircase will be indistinguishable and so we have the general result that:.
Graphs of position versus time for uniformly accelerated motion are parabolas Figure 2. If not, the acceleration is negative. Exam tip The table summarises the meaning of the slope and area for the different motion graphs. Graph of …. Worked examples 2. Find its displacement after Displacement is the change of position, i. After a displacement of 20 m, its velocity becomes 7.
Find the acceleration of the car. After 6. Determine the displacement of the body in the 6. We know u, v and t. The slow ball starts first and the other 4. Determine the position of the balls when they meet. Let the two balls meet t s after the first ball starts moving. The position of the slow ball is:. For this motion find: a the change in position, i. The problem is best solved using the velocity—time graph, which is shown in Figure 2.
Thus, after 5. The change in position, i. The object moved toward the right, stopped and returned to its starting position we know this because the displacement was 0. The distance travelled is 50 m in moving to the right and 50 m coming back, giving a total distance travelled of m. Sketch the position—time graph for this motion for the first 20 s.
This is the function we must graph. The result is shown in Figure 2. At 5 s the object reaches the origin and overshoots it. The furthest it gets from the origin is 25 m. At 10 s the velocity is zero. A special acceleration Assuming that we can neglect air resistance and other frictional forces, an object thrown into the air will experience the acceleration of free fall while in the air. This is an acceleration caused by the attraction between the Earth and the body.
The magnitude of this acceleration is denoted by g. The direction of this acceleration is always vertically downward. We have motion on a vertical line so we will use the symbol y for position Figure 2. We make the vertical line point upwards. One is to use the displacement arrow shown in blue in Figure 2. Another way of looking at this is shown in Figure 2. Here we start at the highest point and make the line along which the ball moves point downwards.
Now, the initial velocity is zero because we take our initial point to be at the top. This is the time to fall to the sea. It took 2. If you preferred the diagram in Figure 2. Projectile motion Figure 2. We see that in the vertical direction, both objects fall the same distance in the same time. How do we understand this fact? Consider Figure 2. A blue ball is allowed to drop vertically from the same height. But suppose there is an observer Y, who moves to the right with velocity 2v with respect to the ground.
What does Y see? So this observer will determine that the two bodies reach the ground at the same time. Since time is absolute in Newtonian physics, the two bodies must reach the ground at the same time as far as any other observer is concerned as well.
Observer Y is moving to the right with v velocity with respect to the ground. The discussion shows that the motion of a ball that is projected at some angle can be analysed by separately looking at the horizontal and the vertical directions.
All we have to do is consider two motions, one in the horizontal direction in which there is no acceleration, and another in the vertical direction in which we have an acceleration, g. At some later time t the components of velocity are vx and vy.
In the x-direction we do not have any acceleration and so:. The red arrows in Figure 2. We would like to know the x- and y-components of the position vector. We now use the formula for position. Let us collect what we have derived so far.
We have four equations with which we can solve any problem with projectiles, as we will soon see:. Exam tip Always choose your x- and y-axes so that the origin is the point where the launch takes place.
Find the x- and y-components of the initial velocity in each case. The graphs are shown in Figure 2. Determine: a the time at which it will hit the ground b the horizontal distance travelled c the speed with which it hits the ground. Exam tip This is a basic problem — you must know how to do this! Determine: a the vertical component of velocity after 4.
The position of the object is shown every 0. Note how the dots get closer together as the object rises the speed is decreasing and how they move apart on the way down the speed is increasing. It reaches a maximum height of 5. The photo in Figure 2.
At what point in time does the vertical velocity component become zero? The time when the vertical velocity becomes zero is, of course, the time when the object attains its maximum height. What is this height? Exam tip You should not remember these formulas by heart. You should be able to derive them quickly.
What about the maximum displacement in the horizontal direction sometimes called the range? At this point the vertical component of displacement y is zero. The second time is what we want — the time in which the range is covered.
This equation also says that there are two different angles of launch that give the same range for the same initial speed. These two angles add up to a right angle can you see why? Determine the maximum height reached.
Find the initial velocity of launch. Fluid resistance The discussion of the previous sections has neglected air resistance forces. In general, whenever a body moves through a fluid gas or liquid it experiences a fluid resistance force that is directed opposite to the velocity.
The magnitude of this force increases with increasing speed. Imagine dropping a body of mass m from some height. Initially, the only force on the body is its weight, which accelerates it downward. As the speed increases, the force of air resistance also increases.
Eventually, this force will become equal to the weight and so the acceleration will become zero: the body will then move at constant speed, called terminal speed, vT. The speed eventually becomes the terminal speed and the acceleration becomes zero. The initial acceleration is g. The effect of air resistance forces on projectiles is very pronounced. With air resistance forces the range and maximum height are smaller and the shape is no longer symmetrical.
The projectile hits the ground with a steeper angle. Determine the mass of the projectile. Nature of science The simple and the complex Careful observation of motion in the natural world led to the equations for motion with uniform acceleration along a straight line that we have used in this section. Thinking about what causes an object to move links to the idea of forces. The falling leaf is complicated because it is acted upon by several forces: its weight, but also by air resistance forces that constantly vary as the orientation and speed of the leaf change.
In addition, there is wind to consider as well as the fact that turbulence in air greatly affects the motion of the leaf. So the physics of the falling leaf is far away from the physics of motion along a straight line at constant acceleration.
But learning the principles of physics in a simpler context allows its application in more involved situations.
Uniform motion 1 A car must be driven a distance of km in 2. During the first 1. Calculate the average speed for the remainder of the journey. The initial position is zero. You do not have to put any numbers on the axes. As soon as the fly reaches B it immediately turns around and flies towards A, and so on until A and B meet. Accelerated motion 5 The initial velocity of a car moving on a straight road is 2. It becomes 8. Find the acceleration. Find the distance it travels.
Find the acceleration of the particle. Determine when its position will become 16 m. Find the take-off velocity. The driver sees an emergency ahead and 0. The deceleration of the car is 4.
Without performing any calculations, state whether the answer to c would now be less than, equal to or larger than before. Explain your answer. One of the balls is dropped 1. Initially the velocity of the object is 2. Find the acceleration at 2. Draw the graph showing the variation of the velocity of the object with time.
Draw the graph showing the variation of the position of the object with time. Draw the graph showing the variation of the acceleration of the object with time. Draw the graph showing the variation of the position of the object with time assuming a zero initial position. You do not have time to get the keys, so you begin to push the car towards the garage.
The maximum acceleration you can give the car is 2. Find the least time it takes to put the car in the garage. Assume that the car, as well as the garage, are point objects. Four points on this graph have been selected. The speed just before each impact with the floor is the same as the speed just after impact. Assume that the time of contact with the floor is negligibly small. When the cart hits the ends of the air track it reverses direction with the same speed it had right before impact.
Assume the time of contact of the cart and the ends of the air track is negligibly small. After a few seconds she opens a parachute. Eventually she will reach a terminal speed and will then land. The initial velocity of the stone is 8. Determine: a the maximum height of the stone b the time when it hits the sea c the velocity just before hitting the sea d the distance the stone covers e the average speed and the average velocity for this motion.
It reaches the bottom of the cliff 6. Projectile motion 25 A ball rolls off a table with a horizontal speed of 2. The table is 1. Calculate how far from the table the ball will land. They are thrown horizontally with the same speed, 4. Calculate u. Draw graphs to show the variation with time of: a the horizontal position b the vertical position. The chimp lets go of the branch as soon as the hunter pulls the trigger. Treating the chimp and the bullet as point particles, determine if the bullet will hit the chimp.
Air resistance and other frictional forces are neglected. The graph shows the position of the ball every 0. Draw the path of the ball on your graph. Explain why the ball will reach terminal speed. Classical physics is based to a great extent on these laws. It was once thought that knowledge of the present state of a system and all forces acting on it would enable the complete prediction of the state of that system in the future. This classical version of determinism has been modified partly due to quantum theory and partly due to chaos theory.
Forces and their direction A force is a vector quantity. It is important that we are able to correctly identify the direction of forces. In this section we will deal with the following forces. Treat bodies as point particles. Construct and interpret freebody force diagrams. Solve problems involving solid friction. Weight This force is the result of the gravitational attraction between the mass m of a body and the mass of the planet on which the body is placed.
X 26 Outline how the exchange of gluons by quarks results in the strong nuclear force between Y nucleons? The 2 2 assumptions made are the usual ones see question 19 i no losses of energy due to frictional forces as the turbines turn ii no turbulence in the air, and iii that all the air stops at the turbines so that the speed of the air behind the turbines is zero which is impossible. Information regarding prices, travel timetables, state and in those 10 s?
Find the uncertainty in: 0 0 0. As the book falls. R 47 Describe the energy transformations taking place f F when a body of mass 5. In which of the following calculated quantities is the percentage uncertainty the greatest.
The potential increases as we move to the right. MYP Core. Published on Apr Remaining distance is 15 km and must be covered in 15 km 1. Predict how your answers to b and c would change if at all. Determine the strings as shown in the diagram.
The weak interaction being weaker than the strong would naturally lead to a long lifetime decay. A black body appears black when its temperature is very low. It absorbs all the radiation incident on it and reflects none. The power P radiated by the source is distributed over the P P surface area A of this imaginary sphere. The power per unit area i. The radiated energy is in the infrared region of the electromagnetic spectrum. Gases in the atmosphere absorb part of this radiated energy and reradiate it in all directions.
Some of this radiation returns to the earth surface warming it further. See page for sources. I i. You must be sure as to whether the question wants you to assume a black body or not. Strictly speaking, in a model without an atmosphere the earth surface cannot be taken to be a black body — if it were no radiation would be reflected!
In addition there is conduction to the atmosphere, as well as convection. Reducing the albedo means that less radiation is reflected and more is absorbed and so an increase in temperature might be expected.
The increase in temperature might involve additional evaporation and so more rain. These are both higher in the case of the tropical ocean water and evaporation will be more significant in that case. The fact that region b is warmer is further supported by the somewhat greater conduction which would be expected if the difference in temperature between the atmosphere and the land were larger.
If however the amplitude of oscillations is small the force does become approximately proportional to the displacement and the oscillations are then approximately simple harmonic. Therefore, 4 5 6 7 6. The displacement is zero all the time when 6. Thus the 0. The frequency is 5. The g 9. Hence the net force on the mass will still be 2kx so the period will not change. The forces on the woman at the position in a are her weight vertically downwards and the tension in the spring upwards.
The acceleration is not proportional to the displacement so it looks we do not have SHM. But we must measure displacement from an equilibrium position.
Hence we do have the condition for SHM. The b 0. In ii there will be no change since both wavelength and width halve. Notice however that if we were to pay attention to the vertical axis scale, with a smaller slit width less light would go through so in both cases the intensity would be less.
This means that the phase difference between them keeps changing with time very fast, on a time scale of nanoseconds. Thus, whatever interference pattern is produced at any moment in time, a different pattern will be produced a nanosecond later. Therefore all we can observe is an average of the rapidly changing patterns on the screen, i. Monochromatic light means light of the same wavelength. Intensity 5 4 3 2 1 —0. When this is about equal to the angular 4.
This means d 8 3. However the intensity of the light decreases with increasing m and so it is preferable to increase N instead.
The wave will then be reflected with frequency f 1. The car is now acting as an approaching source. The frequency received back at the source f 2 will be higher than that emitted from the car. This is the case of a double Doppler effect.
This 45 a The frequency emitted is f. The stars are neither approaching or moving away from the observer at that time. In this way the function can be plotted on a calculator to give the graph shown here. From then on the moon will pull it 1 in.
At B it is opposite to the velocity and so the speed decreases. The spherical equipotential surfaces of the right mass are much less distorted and so this is the larger mass.
The equipotential surfaces of a single point mass are spherical. At a maximum value the derivative is zero. F2 F1 F3 They have magnitudes: 8. So the potential at the center is kQ1 kQ2 kQ3 kQ4 8.
The wire has to be long so that the charge of one sphere will not affect the charge distribution on the other so that both are uniformly charged. Hence the potential 1. It is clear that the net field will point in the negative y — direction. It has magnitude 2kQ 2kQ d 2kQd. This is the energy that must be supplied. But v T and so we deduce that. This happens only for orbit 2. More simply, both spacecraft and astronaut are in free fall with the same acceleration.
GMm GMm. The firing of the rockets when the satellite is in the lower orbit makes the satellite move on an elliptical orbit. After half a revolution the satellite will be at A and further from the earth than in the original position at P. As the satellite gets to A its kinetic energy is reduced and the potential energy increases.
At A the speed is too low for the new circular orbit and the engines must again be fired to increase the speed to that appropriate to the new orbit. If the engines are not fired at A then the satellite will remain in the elliptical orbit and will return to P.
Therefore since the total energy is conserved, the kinetic energy and hence the speed are greatest at P. R 34 a We have done this before. This implies that the distance R will decrease. From the period formula in b the period will decrease as well. In the first orbit this evaluates to 2r 8. The change is 1. It equals the slope which is 2. In the next 2 s the slope and hence the emf is constant at 2 V. In the last 4 s the slope is 1 V.
This gives the graph here. We have a straight line graph with slope 6. In the next 2 s the emf is zero which means that the flux is constant. Similarly, in the last 4 the emf is constant so the flux is increasing at a constant rate, i. A possible graph is shown here. Since the current is increasing the flux is increasing. This can be done by having the induced current create a magnetic field directed out of the page. The current must then be counter-clockwise.
So the induced current must produce a magnetic field directed away from us. By the right hand rule for the magnetic field direction, the current must be clockwise. As the ring moves away from the magnet we see magnetic field coming towards us and the flux is decreasing. So we must produce a current whose magnetic field comes towards us and so the current must be counter-clockwise. When the ring is half way down the length of the magnet the current must be zero.
So the induced current must create a magnetic field coming towards us and so the current is counter-clockwise.
As the ring leaves, the flux is decreasing, the field is going away from us and the current must produce its own magnetic field away from us, i. Half way it must be zero. The forces on two diametrically opposite points on the ring are as shown. The net force of these two forces is upwards. The same is true for any other two diametrically opposite points and so the net magnetic force on the ring is vertically upwards, making the ring fall slower than in free fall.
Hence the left end will be negatively charged. The electrons that have moved to the left have left the right end of the rod positively charged. Hence the flux is increasing. To oppose this increase the induced current must produce a magnetic field into the page and so the current must be clockwise. Hence the flux is decreasing. To oppose this decrease the induced current must produce a magnetic field out of the page and so the current must be counter-clockwise.
The force on the movable rod is thus directed to the right. Note: this could have been guessed since by moving the rod to the right we decrease the are and hence the flux. This is what must happen to oppose the change in flux which is an increase since the field is increasing. The oscillations are simple harmonic. As the magnet moves downward the flux in the coil is increasing. Assume the lower pole is a north pole.
The induced current will produce a magnetic field upward and so there will be a retarding force on the magnet that will make the oscillations die out. A similar argument applies when the magnet moves upwards. Since it is conducting a current will be established as well.
The current must produce a field out of the page so as to oppose the increase in flux which means that the current is clockwise. On the lower part of the ring in the region of the field there will therefore be a magnetic force directed upward. Similarly there will be an upward force as the ring leaves the region of the magnetic field.
In both cases the upward force delays the fsll of the ring so this ring land last. But since we want the dependence on angle and not time the period of the graph will not change.
This gives the graphs shown here. I rms c The period is 1. This is because at double the rotation speed the induced emf doubles and so the power that depends on the square of the induced emf increases by a factor of 4. This gives the graph shown here. The frequency is unchanged so it stays 50 Hz. The vertical axis is flux axis not power. The power lost in the cables is then V 1. The average power is 24 2 The charge is then 8. The average current is Q 2. After connecting the charged capacitor to the uncharged charge will move form to the other until the voltage across each is the same.
The total charge on both capacitors will be pC by charge conservation. This is, approximately, the time for which the current is appreciable enough to light the lamp. This is represented by the area of the thin strip in the graph. Hence the total area is the total energy stored. Hence VQ0 6. The E max 1. When light is incident on the metal an electron from within the metal may absorb one photon and so its energy will increase by an amount equal to the photon energy.
If this energy is sufficient to overcome the potential well the electron finds itself in, the electron may be free itself from the metal. In the photon theory of light the energy is supplied to an electron by a single photon in an instantaneous interaction.
The energy of the photon depends on frequency so if the photon energy is less than the work function the electron cannot be emitted. The intensity of light depends on the number of photons present and so this will affect the number of electrons emitted not their energy. To accumulate the energy equal to the work function we need 4.
Thus an electron with energy Of course the electron may just lose no energy to the atom in which case it will have an elastic collision moving away with the same energy as the original, i. P nhf where n is the number of photons incident per second. This fraction is called the quantum efficiency, which in general does depend on wavelength in a complex way.
This can be understood if we accept that the energy in atoms has specific values so that differences in levels also have specific values. Hence hc hc 6. Then h h p2. Hence 2m h 6. The difference with the previous answer is a question of significant figures.
Chapter 19 contains questions that mix together different parts of the course — a favourite trick Cambridge University Press, We use cookies to give you the best possible experience. By using our website you agree to our use of cookies. Dispatched from the UK in 2 business days When will my order arrive?
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