Module 1 - HACH A-Level Physics



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|Unit 4 |Momentum and Collisions |

|Lesson 1 | |

|Learning Outcomes |To be able to calculate momentum and know the units |

| |To be able to explain the difference between elastic and inelastic collisions |

| |To be able to find the velocity of an object after a collision or explosion |MR. C - SJP |

Momentum (Also seen in GCSE Physics 2)

The momentum of an object is given by the equation: momentum = mass x velocity [pic]

Since it depends on the velocity and not speed, momentum is a vector quantity. If we assign a direction to be positive for example if ( was positive, an object with negative velocity would be moving (. It would also have a negative momentum.

Momentum is measured in kilogram metres per second, kg m/s or kg m s-1

Conservation (Also seen in GCSE Physics 2)

In an isolated system (if no external forces are acting) the linear momentum is conserved.

We can say that: the total momentum before = the total momentum after

The total momentum before and after what? A collision or an explosion.

Collisions (Also seen in GCSE Physics 2)

There are two types of collisions; in both cases the momentum is conserved.

Elastic – kinetic energy in conserved, no energy is transferred to the surroundings

If a ball is dropped, hits the floor and bounces back to the same height it would be an elastic collision with the floor. The kinetic energy before the collision is the same as the kinetic energy after the collision.

Inelastic – kinetic energy is not conserved, energy is transferred to the surroundings

If a ball is dropped, hits the floor and bounces back to a lower height than released it would be an inelastic collision. The kinetic energy before the collision would be greater than the kinetic energy after the collision.

[pic] [pic]

Before After

In the situation above, car 1 and car 2 travel to the right with initial velocities u1 and u2 respectively. Car 1 catches up to car 2 and they collide. After the collision the cars continue to move to the right but car 1 now travels at velocity v1 and car 2 travels a velocity v2. [( is positive]

Since momentum is conserved the total momentum before the crash = the total momentum after the crash.

The total momentum before is the momentum of A + the momentum of B

The total momentum after is the new momentum of A + the new momentum of B

We can represent this with the equation: [pic]

Explosions (Also seen in GCSE Physics 2)

We look at explosions in the same way as we look at collisions, the total momentum before is equal to the total momentum after. In explosions the total momentum before is zero. [( is positive]

[pic] [pic]

Before After

If we look at the example above we can see that the whole system is not moving, so the momentum before is zero. After the explosion the shell travels right with velocity v2 and the cannon recoils with a velocity v1.

The momentum of the system is given as: [pic]

So the equation for this diagram would be: [pic]

But remember, v1 is negative so: [pic] ( [pic]

|Unit 4 |Force and Impulse |

|Lesson 2 | |

|Learning Outcomes |To be able to calculate force from change in momentum |

| |To be able to explain and calculate impulse |

| |To know the significance of the area under a force-time graph |MR. C - SJP |

Force (Also seen in GCSE Physics 2)

If we start at F = ma we can derive an equation that links force and momentum.

[pic] we can replace a in this equation with [pic] from Unit 2

[pic] multiplying out makes the equation

[pic] or [pic] where ∆ means ‘the change in’

This states that the force is a measure of change of momentum with respect to time. This is Newton’s Second Law of Motion:

The rate of change of an object’s linear momentum is directly proportional to the resultant external force. The change in the momentum takes place in the direction of the force.

If we have a trolley and we increase its velocity from rest to 3m/s in 10 seconds, we know that it takes a bigger force to do the same with a trolley that’s full of shopping. Ever tried turning a trolley around a corner when empty and then when full?

Force is measured in Newtons, N

Car Safety (Also seen in GCSE Physics 2)

Many of the safety features of a car rely on the above equation. Airbags, seatbelts and the crumple zone increase the time taken for the car and the people inside to stop moving. Increasing the time taken to change the momentum to zero reduces the force experienced.

Catching

An Egg: If we held our hand out and didn’t move it the egg would smash. The change in momentum happens in a short time, making the force large. If we cup the egg and move our hands down as we catch it we make it take longer to come to a complete stop. Increasing the time taken decreases the force and the egg remains intact.

Cricket Ball: If we didn’t move our hands it would hurt when the ball stopped in our hands. If we make it take longer to stop we reduce the force on our hands from the ball.

Impulse

[pic] multiply both sides by t ( [pic]

[pic] multiply both sides by t ( [pic]

We now have an equation for impulse. Impulse is the product of the force and the time it is applied for.

An impulse causes a change in momentum.

Impulse is measured in Newton seconds, Ns

Since [pic], the same impulse (same force applied for the same amount of time) can be applied to a small mass to cause a large velocity or to a large mass to cause a small velocity

Ft = mv = mv

Force-Time Graphs

The impulse can be calculated from a force-time graph, it is the same as the area under the graph.

The area of the first graph is given by:

height x length = Force x time = Impulse

|Unit 4 |Circular Motion |

|Lesson 3 | |

|Learning Outcomes |To be able to calculate the angular displacement of an object moving in a circle |

| |To be able to calculate the angular speed of an object moving in a circle |

| |To be able to calculate the speed of an object moving in a circle |MR. C - SJP |

To the right is the path a car is taking as it moves in a circle of radius r.

Angular Displacement, θ

As the car travels from X to Y it has travelled a distance of s and has covered a section of the complete circle it will make. It has covered and angle of θ which is called the angular displacement.

[pic] [pic]

Angular Displacement is measured in radians, rad

Radians

1 radian is the angle made when the arc of a circle is equal to the radius.

For a complete circle [pic] ( [pic] ( [pic] ( [pic]

A complete circle is 360° so 360° = 2π rad

1° = 0.017 rad 57.3° = 1 rad

Angular Speed, ω

Angular speed is the rate of change of angular displacement, or the angle that is covered every second.

[pic]

Angular Speed is measured in radians per second, rad/s or rad s-1

Frequency, f

Frequency is the number of complete circles that occur every second.

For one circle; [pic], if we substitute this into the equation above we get [pic]

This equation says that the angular speed (angle made per second) is equal to one circle divided by the time taken to do it. Very similar to speed = distance/time

Since [pic] the above equation can be written as [pic]

Frequency is measures in Hertz, Hz

Speed, v

The velocity of the car is constantly changing because the direction is constantly changing. The speed however, is constant and can be calculated.

[pic] If we rearrange the top equation we can get [pic], the speed then becomes

[pic] Now if we rearrange the second equation we get [pic], the equation becomes

[pic] Cancel the t’s and we finally arrive at our equation for the speed.

[pic]

Speed is measured in metres per second, m/s or m s-1

|Unit 4 |Centripetal Force and Acceleration |

|Lesson 4 | |

|Learning Outcomes |To be able to calculate the centripetal acceleration of an object moving in a circle |

| |To be able to calculate the centripetal force that keeps an object moving in a circle |

| |To be able to explain why the centrifugal force does not exist |MR. C - SJP |

Moving in a Circle (Also seen in GCSE Physics 3)

For an object to continue to move in a circle a force is needed that acts on the object towards the centre of the circle. This is called the centripetal force and is provided by a number of things:

For a satellite orbiting the Earth it is provided by gravitational attraction.

For a car driving around a roundabout it is provided by the friction between the wheels and the road.

For a ball on a string being swung in a circle it is provided by the tension in the string.

Centripetal force acts from the body to the centre of a circle

Since F=ma the object must accelerate in the same direction as the resultant force. The object is constantly changing its direction towards the centre of the circle.

Centripetal acceleration has direction from the body to the centre of the circle

Centrifugal Force

Some people thought that an object moving in a circle would experience the centripetal force acting from the object towards the centre of the circle and the centrifugal force acting from the object away from the centre of the circle.

They thought this because if you sit on a roundabout as it spins it feels like you are being thrown off backwards.

If someone was watching from the side they would see you try and move in a straight line but be pulled in a circle by the roundabout.

The centrifugal force does not exist in these situations.

Centripetal Acceleration

The centripetal acceleration of an object can be derived if we look at the situation to the right. An object of speed v makes an angular displacement of ∆θ in time ∆t.

[pic]

If we look at the triangle at the far right we can use [pic] when θ is small. This becomes: [pic]

We can rearrange this to give: [pic]

Acceleration is given by[pic] substitute the above equation into this one

[pic] this is the same as [pic]

In lesson 3 (Circular Motion) we established that[pic], substitute this into the equation above

[pic]

If we use [pic]we can derive two more equations for acceleration

[pic] [pic] [pic]

Centripetal Acceleration is measured in metres per second squared, m/s2 or m s-2

Centripetal Force

We can derive three equations for the centripetal force by using [pic]and the three equations of acceleration from above.

[pic] [pic] [pic]

Centripetal Force is measured in Newtons, N

|Unit 4 |Simple Harmonic Motion |

|Lesson 5 | |

|Learning Outcomes |To know what simple harmonic motion is |

| |To be able to describe the acceleration of an SHM system |

| |To be able to calculate the displacement, velocity and acceleration of an SHM system |MR. C - SJP |

Oscillations

In each of the cases below there is something that is oscillating, it vibrates back and forth or up and down.

Each of these systems is demonstrating Simple Harmonic Motion (SHM).

[pic]

SHM Characteristics

The equilibrium point is where the object comes to rest, in the simple pendulum it at its lowest point.

If we displace the object by a displacement of x there will be a force that brings the object back to the equilibrium point. We call this the restoring force and it always acts in the opposite direction to the displacement.

We can represent this as: [pic]

Since [pic] we can also write: [pic]

For an object to be moving with simple harmonic motion, its acceleration must satisfy two conditions:

*The acceleration is proportional to the displacement

*The acceleration is in the opposite direction to the displacement (towards the equilibrium point)

Equations

The following equations are true for all SHM systems but let us use the simple pendulum when thinking about them.

The pendulum bob is displaced in the negative direction when at point 1, it is released and swings through point 2 at its maximum speed until it reaches point 3 where it comes to a complete stop. It then swings to the negative direction, reaches a maximum speed at 4 and completes a full cycle when it stops at 5.

Displacement, x

The displacement of the bob after a time t is given by the equation: [pic] (CALCS IN RAD)

Since [pic] the equation can become: [pic] ( [pic]

(where t is the time into the cycle and T is the time for one complete cycle)

The maximum displacement is called the amplitude, A. [pic] ( MAXIMUM

Velocity, v

The velocity of the bob at a displacement of x is given by the equation: [pic]

The maximum velocity occurs in the middle of the swing (2 and 4) when displacement is zero (x = 0)

[pic] ( [pic] ( [pic] ( [pic] ( MAXIMUM

Acceleration, a

The acceleration of the bob at a displacement of x is given by the equation: [pic]

As discussed before the acceleration acts in the opposite direction to the displacement.

The maximum acceleration occurs at the ends of the swing (1, 3 and 5) when the displacement is equal to the amplitude (x = A).

[pic] ( [pic] ( MAXIMUM

|Unit 4 |SHM Graphs |

|Lesson 6 | |

|Learning Outcomes |To be able to sketch the graphs of displacement, velocity and acceleration for a simple pendulum |

| |To be know what the gradients represent |

| |To be able to explain the energy in a full cycle and sketch the graph |MR. C - SJP |

Pendulum

Consider the simple pendulum drawn below. When released from A the bob accelerates and moves to the centre point. When it reached B it has reached a maximum velocity in the positive direction and then begins to slow down. At C it has stopped completely so the velocity is zero, it is at a maximum displacement in the positive and accelerates in the negative direction. At D it is back to the centre point and moves at maximum velocity in the negative direction. By E the velocity has dropped to zero, maximum negative displacement and a massive acceleration as it changes direction.

This repeats as the pendulum swings through F, G, H and back to I.

Below are the graphs that represent this:

Gradients

Since [pic]the gradient of the displacement graph gives us velocity. At C the gradient is zero and we can see that the velocity is zero.

Also since [pic]the gradient of the velocity graph gives us acceleration. At C the gradient is a maximum in the negative direction and we can see that the acceleration is a maximum in the negative direction.

Energy

In all simple harmonic motion systems there is a conversion between kinetic energy and potential energy. The total energy of the system remains constant. (This is only true for isolated systems)

For a simple pendulum there is a transformation between kinetic energy and gravitational potential energy.

At its lowest point it has minimum gravitational and maximum kinetic, at its highest point (when displacement is a maximum) it has no kinetic but a maximum gravitational. This is shown in the graph.

For a mass on a spring there is a transformation between kinetic energy, gravitational potential energy and the energy stored in the spring (elastic potential). At the top there is maximum elastic and gravitational but minimum kinetic. In the middle there is maximum kinetic, minimum elastic but it still has some gravitational. At its lowest point it has no kinetic, minimum gravitational but maximum elastic.

|Unit 4 |SHM Time Periods |

|Lesson 7 | |

|Learning Outcomes |To be able to calculate the time period of a simple pendulum |

| |To be able to calculate the time period of a mass on a spring |

| |To be able to describe the experiment to find g |MR. C - SJP |

The Simple Pendulum

In the diagram we can see that the restoring force of the pendulum is: [pic]

When ( is less than 10° (in radians) [pic]so the equation can become: [pic]

Since both [pic]and [pic](for SHM) the equation now becomes: [pic]

This simplifies to: [pic]

Rearranging for f gives us [pic]

And since [pic] then: [pic]

Time is measured in seconds, s

Mass on a Spring

When a spring with spring constant k and length l has a mass m attached to the bottom it extends by an extension e, this is called the static extension and is the new equilibrium point. The tension in the spring is balanced by the weight. We can represent this as: [pic]

If the mass is pulled down by a displacement x and released it will undergo SHM.

The net upwards force will be: [pic]

This can be multiplied out to become: [pic]

Since [pic] this can become: [pic]

It simplify to: [pic]

Since both [pic]and [pic](for SHM) the equation now becomes: [pic]

This simplifies to: [pic]

Rearranging for f gives us: [pic]

And since [pic] the equation becomes: [pic]

Time is measured in seconds, s

Finding g

We can find the value of the gravitational field strength, g, on Earth by carrying out the following experiment.

Set up a simple pendulum of length l and measure the time for one oscillation.

If we measure the time taken for 20 oscillations and divide it by 20 we reduce the percentage human error of the reading and make our experiment more accurate.

If we look at the equation [pic] and rearrange it to become: [pic], by plotting a graph of T2 against l we can find the value of g from the gradient which will be = [pic].

|Unit 4 |Resonance and Damping |

|Lesson 8 | |

|Learning Outcomes |To know what free and forced vibration are and the phase difference between the driver and driven |

| |To know what resonance is and how it is reached |

| |To know what light, heavy and critical damping are and their affects on resonance |MR. C - SJP |

Free Vibration

Free vibration is where a system is given an initial displacement and then allowed to vibrate/oscillate freely. The system will oscillate at a set frequency called the natural frequency, f0. We have seen from the last lesson that the time period for a pendulum only depends on the length and gravitational field strength whilst the time period of a mass and spring only depends on the mass and the spring constant. These factors govern the natural frequency of a system.

Forced Vibration

Forced vibration is where a driving force is continuously applied to make the system vibrate/oscillate. The thing that provides the driving force will be moving at a certain frequency. We call this the driving frequency.

Resonance

If I hold one end of a slinky and let the other oscillate freely we have a free vibration system. If I move my hand up and down I force the slinky to vibrate. The frequency of my hand is the driving frequency.

When the driving frequency is lower than the natural frequency the oscillations have a low amplitude

When the driving frequency is the same as the natural frequency the amplitude increases massively, maybe even exponentially.

When the driving frequency is higher than the natural frequency the amplitude of the oscillations decreases again.

Phase Difference between driver and driven

When the driving force begins to oscillate the driven object the phase difference is 0.

When resonance is achieved the phase difference between them is π.

When the driving frequency increases beyond the natural frequency the phase difference increases to π/2.

Damping

Damping forces oppose the motion of the oscillating body, they slow or stop simple harmonic motion from occurring. Damping forces act in the opposite direction to the velocity.

Galileo made an important observation while watching lamps swing in Pisa cathedral. He noticed that the swinging gradually died away but the time taken for each swing stayed roughly the same. The swing of the lamp was being damped by air resistance.

[pic]

Light damping slowly reduces the amplitude of the oscillations, but keeps the time period almost constant.

Heavy damping allows the body to oscillate but brings it quickly to rest.

Critical damping brings the body back to the equilibrium point very quickly with out oscillation.

Over damping also prevent oscillation but makes the body take a longer time to reach equilibrium.

Damping and Resonance

Damping reduces the size of the oscillations at resonance. There is still a maximum amplitude reached but it is much lower than when the system is undamped. We say that damping reduces the sharpness of resonance. This becomes clearer if we look at the graph on the right.

It shows the amplitude of oscillation against frequency for different levels of damping.

Q1.(a)    Figure 1 shows how the kinetic energy, Ek, of an oscillating mass varies with time when it moves with simple harmonic motion.

Figure 1

[pic]

(i)      Determine the frequency of the oscillations of the mass.

 

 

 

 

frequency of oscillation .............................................. Hz

(2)

(ii)     Sketch, on Figure 1, a graph showing how the potential energy of the mass varies with time during the first second.

(2)

(b)     Figure 2 shows a ride called a ‘jungle swing’.

Figure 2

[pic]

The harness in which three riders are strapped is supported by 4 steel cables. An advert for the ride states that the riders will be released from a height of 45 m above the ground and will then swing with a period of 14.0 s. It states that they will be 1.0 m above the ground at the lowest point and that they will travel at speeds of ‘up to 120 km per hour’.

(i)      Treating the ride as a simple pendulum, show that the distance between the pivot and the centre of mass of the riders is about 49 m.

 

 

 

 

 

 

 

(2)

(ii)     The riders and their harness have a total mass of 280 kg.

Calculate the tension in each cable at the lowest point of the ride, assuming that the riders pass through this point at a speed of 120 km h–1. Assume that the cables have negligible mass and are vertical at this point in the ride.

 

 

 

 

tension in each cable ................................................ N

(4)

(iii)    Show that the maximum speed stated in the advert is an exaggerated claim.

Assume that the riders are released from rest and neglect any effects of air resistance.

 

 

 

 

 

(4)

(iv)    The riders lose 50% of the energy of the oscillation during each half oscillation. After one swing, the speed of the riders as they pass the lowest point is 20 m s–1.

Calculate the speed of the riders when they pass the lowest point, travelling in the same direction after two further complete oscillations.

 

 

 

 

 

speed of riders .......................................... ms–1

(3)

(Total 17 marks)

Q2.Figure 1 shows a fairground ride called a Rotor. Riders stand on a wooden floor and lean against the cylindrical wall.

Figure 1

[pic]

The fairground ride is then rotated. When the ride is rotating sufficiently quickly the wooden floor is lowered. The riders remain pinned to the wall by the effects of the motion. When the speed of rotation is reduced, the riders slide down the wall and land on the floor.

(a)     (i)      At the instant shown in Figure 2 the ride is rotating quickly enough to hold a rider at a constant height when the floor has been lowered.

Figure 2

[pic]

Draw onto Figure 2 arrows representing all the forces on the rider when held in this position relative to the wall.

Label the arrows clearly to identify all of the forces.

(3)

(ii)     Explain why the riders slide down the wall as the ride slows down.

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(2)

A Rotor has a diameter of 4.5 m. It accelerates uniformly from rest to maximum angular velocity in 20 s.

The total moment of inertia of the Rotor and the riders is 2.1 × 105 kg m2.

(b)     (i)      At the maximum speed the centripetal acceleration is 29 m s–2.

Show that the maximum angular velocity of a rider is 3.6 rad s–1.

 

 

 

 

(2)

(ii)     Calculate the torque exerted on the Rotor so that it accelerates uniformly to its maximum angular velocity in 20 s.

State the appropriate SI unit for torque.

 

 

 

torque ................................................. SI unit for torque ........................

(3)

(iii)     Calculate the centripetal force acting on a rider of mass 75 kg when the ride is moving at its maximum angular velocity.

Give your answer to an appropriate number of significant figures.

 

 

 

centripetal force .................................................. N

(1)

(c)     Figure 3 shows the final section of a roller coaster which ends in a vertical loop. The roller coaster is designed to give the occupants a maximum acceleration of 3g. Cars on the roller coaster descend to the start of the loop and then travel around it, as shown.

Figure 3

[pic]

(i)      At which one of the positions marked A, B and C on Figure 3 would the passengers experience the maximum reaction force exerted by their seat?

Circle your answer below.

 

A                      B                         C

 

(1)

(iii)     Explain why the maximum acceleration is experienced at the position you have chosen.

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(2)

(Total 14 marks)

Q3.( a)     (i)      Name the two types of potential energy involved when a mass–spring system performs vertical simple harmonic oscillations.

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(1)

(ii)     Describe the energy changes which take place during one complete oscillation of a vertical mass-spring system, starting when the mass is at its lowest point.

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(2)

(b)     Figure 1 shows how the total potential energy due to the simple harmonic motion varies with time when a mass-spring system oscillates vertically.

Figure 1

 [pic]

time / s

(i)      State the time period of the simple harmonic oscillations that produces the energy–time graph shown in Figure 1, explaining how you arrive at your answer.

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(2)

(ii)     Sketch a graph on Figure 2 to show how the acceleration of the mass varies with time over a period of 1.2 s, starting with the mass at the highest point of its oscillations. On your graph, upwards acceleration should be shown as positive and downwards acceleration as negative. Values are not required on the acceleration axis.

Figure 2

 [pic]

(2)

(c)     (i)      The mass of the object suspended from the spring in part (b) is 0.35 kg. Calculate the spring constant of the spring used to obtain Figure 1. State an appropriate unit for your answer.

 

 

 

spring constant ...................................... unit ....................................

(3)

(ii)     The maximum kinetic energy of the oscillating object is 2.0 × 10–2 J. Show that the amplitude of the oscillations of the object is about 40 mm.

 

 

 

(4)

(Total 14 marks)

Q4.(a)    A simple pendulum is given a small displacement from its equilibrium position and performs simple harmonic motion.

State what is meant by simple harmonic motion.

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(2)

(b)    (i)       Calculate the frequency of the oscillations of a simple pendulum of length 984 mm. Give your answer to an appropriate number of significant figures.

 

 

frequency ....................................... Hz

(3)

(ii)     Calculate the acceleration of the bob of the simple pendulum when the displacement from the equilibrium position is 42 mm.

 

 

 

 

acceleration .................................... ms–2

(2)

(c)     A simple pendulum of time period 1.90 s is set up alongside another pendulum of time period 2.00 s. The pendulums are displaced in the same direction and released at the same time.

Calculate the time interval until they next move in phase. Explain how you arrive at your answer.

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time interval .......................................... s

(3)

(Total 10 marks)

Q5.          The two diagrams in the figure below show a student before and after she makes a bungee jump from a high bridge above a river. One end of the bungee cord, which is of unstretched length 25 m, is fixed to the top of a railing on the bridge. The other end of the cord is attached to the waist of the student, whose mass is 58 kg. After she jumps, the bungee cord goes into tension at point P. She comes to rest momentarily at point R and then oscillates about point Q, which is a distance d below P.

BEFORE AFTER

[pic] 

(a)     (i)      Assuming that the centre of mass of the student has fallen through a vertical distance of 25 m when she reaches point P, calculate her speed at P.

You may assume that air resistance is negligible.

 

                                                          answer = ........................... ms–1

(2)

(ii)     The bungee cord behaves like a spring of spring constant 54 Nm–1.

Calculate the distance d, from P to Q, assuming the cord obeys Hooke’s law.

 

 

                                                           answer = ................................ m

(2)

(b)     As the student moves below P, she begins to move with simple harmonic motion for part of an oscillation.

(i)      If the arrangement can be assumed to act as a mass-spring system, calculate the time taken for one half of an oscillation.

 

 

                                                           answer = ................................ s

(2)

(ii)     Use your answers from parts (a) and (b)(i) to show that the amplitude A, which is the distance from Q to R, is about 25 m.

 

 

 

(3)

(c)     Explain why, when the student rises above point P, her motion is no longer simple harmonic.

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(2)

(d)     (i)      Where is the student when the stress in the bungee cord is a maximum?

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(1)

(ii)     The bungee cord has a significant mass. Whereabouts along the bungee cord is the stress a maximum? Explain your answer.

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(2)

(Total 14 marks)

 

Q6.(a)     A vibrating system which is experiencing forced vibrations may show resonance.

Explain what is meant by

forced vibrations ............................................................................................

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resonance.......................................................................................................

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(3)

(b)     (i)      Explain what is meant by damping.

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(ii)     What effect does damping have on resonance?

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(2)

(Total 5 marks)

M1.(a)     (i)      correct period read from graph or use of f=1/T 0.84±0.01

C1

2.4 Hz gets C1

correct frequency 1.2 (1.18 − 1.25 to 3 sf)

A1

(ii)     correct shape (inverse)

B1

Crossover PE = KE

B1

(b)    (i)       [pic]

C1

48.7 (49) m

A1

(ii)     v = 120 000 / 3600 = 33(.3) m s−1

B1

Use of F = m v2/r (allow v in km h−1 )

B1

Total tension = 6337 + (280 × 9.81) = 9.083 × 103 N

Allow their central force

B1

Divide by 4      2.27 × 103 N

Allow their central force

B1

(iii)    mgh = ½ mv2

B1

Condone: Use of v = 2πfA (max2)

9.8×44 = 0.5 v2      Allow 45 in substitution

B1

Condone      22 m s−1

29.4 m s−1          (Use of 45 gives 29.7)

B1

106 km h−1 (their m s−1 correctly converted)

Or compares with 33 m s−1

B1

(iv)    1/16th(0.625) % of KE left if correct

M1

Allow 1/8 (0.125)or 1/32(0.313)

KE at start = 5.6 × 104 J or states energy ∝ speed2 so speed is ¼

M1

Allow for correct subn E =½ 280 × 202 x factor from incorrect number of swings calculated correctly

Final speed calculated = 5 m s−1

A1

Must be from correct working

[17]

M2.(a)     (i)      Weight / W / mg − vertically downwards from some point

on the body

B1

Friction − vertically upwards and touching both the

wall and the body

B1

Centripetal force / normal reaction / R − horizontally

to the left from the body

Each must be correct and correctly labelled

Minus one for each additional inappropriate force

B1

3

(ii)     Centripetal force / reaction / R is smaller

B1

Frictional force reduces

Frictional force is less than weight

Resultant force is downward

Friction is proportional to (normal) reaction

B1

2

(b)     (i)       rω2 = 29 or

v2 / r = 29

B1

Use of correct radius leading to 3.590 (rad s−1 ) to at least

3 sig figs

2.54 using wrong r = 1 mark

B1

2

(ii)     Angular acceleration, α = 3.6 / 20 OR 3.59 / 20 or 0.18

or 0.1795

C1

3.8 (3.77, 3.78) × 104 cao

A1

N m or kg m2 s−2

B1

3

(iii)    2200 N cao

B1

1

(c)     (i)      C

B1

1

(ii)     Speed greatest (as all PE turned to KE)

B1

Total reaction force = mrω2 + mg or v2 / r + mg or R is

largest or

R = ma + mg

OR

Acceleration = v2 / r

B1

2

[14]

M3.(a)     (i)      elastic potential energy and gravitational potential energy ✓

For elastic pe allow “pe due to tension”, or “strain energy” etc.

1

(ii)     elastic pe → kinetic energy → gravitational pe

                                                        → kinetic energy → elastic pe ✓✓

[or pe→ke→pe→ke→pe is ✓ only]

[or elastic pe → kinetic energy → gravitational pe is ✓ only]

If kinetic energy is not mentioned, no marks.

Types of potential energy must be identified for full credit.

2

(b)     (i)      period = 0.80 s ✓

during one oscillation there are two energy transfer cycles

(or elastic pe→ke→gravitational pe→ke→elastic pe in 1 cycle)

or there are two potential energy maxima per complete oscillation ✓

Mark sequentially.

2

(ii)     sinusoidal curve of period 0.80 s ✓

– cosine curve starting at t = 0 continuing to t = 1.2s ✓

For 1st mark allow ECF from T value given in (i).

2

(c)    (i)      use of T = [pic] gives 0.80 = [pic] ✓

∴ [pic] = 22 (21.6) ✓    N m–1 ✓

 

Unit mark is independent: insist on N m–1.

Allow ECF from wrong T value from (i): use of 0.40s gives 86.4 (N m–1).

3

(ii)     maximum ke = ( ½ mvmax2) = 2.0 × 10–2 gives

vmax2 = [pic] ✓ (= 0.114 m2s–2)  and  vmax = 0.338 (m s–1) ✓

         vmax = 2πfA gives A = [pic] ✓

                               and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm

[or maximum ke = (½ mvmax2) = ½ m (2πfA)2 ✓

        ½ × 0.35 × 4π2 × 1.252 × A2 = 2.0 × 10–2 ✓

∴ A2 = [pic] ✓ ( = 1.85 × 10–3 )

    

    and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]

[or maximum ke = maximum pe = 2.0 × 10–2 (J)

    maximum pe = ½ k A2 ✓

    ∴ 2.0 × 10–2 = ½ × 21.6 × A2 ✓

from which A2 = [pic] ✓ ( = 1.85 × 10–3 )

    

and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]

First two schemes include recognition that f = 1 / T i.e. f = 1 / 0.80 = 1.25 (Hz).

Allow ECF from wrong T value from (i) – 0.40sgives A = 2.15 × 10–2m but mark to max 3.

Allow ECF from wrong k value from (i) –86.4Nm–1 gives A = 2.15 × 10–2m but mark to max 3.

4

[14]

M4.(a)    acceleration is proportional to displacement (from equilibrium)  [pic]

Acceleration proportional to negative displacement is 1st mark only.

acceleration is in opposite direction to displacement

or towards a fixed point / equilibrium

Don’t accept “restoring force” for accln.

position  [pic]

2

(b)    (i)       [pic]

3SF is an independent mark.

 [pic]

When g = 9.81 is used, allow either 0.502 or 0.503 for 2nd and 3rd marks.

 [pic]  [pic] ]

Use of g = 9.8 gives 0.502 Hz: award only 1 of first 2 marks if quoted as 0.502, 0.503 0.50 or 0.5 Hz.

answer to 3SF  [pic]

3

(ii)      [pic]

Allow ECF from any incorrect f from (b)(i).

= 0.42 (0.419) (m s−2) [pic]

2

(c)     recognition of 20 oscillations of (shorter) pendulum

and / or 19 oscillations of (longer) pendulum  [pic]

Explanation: difference of 1 oscillation or phase change of 2[pic]

or Δt = 0.1 so n = 2 / 0.1 =20, or other acceptable point  [pic]

time to next in phase condition = 38 (s)  [pic]

Allow “back in phase (for the first time)” as a valid explanation.

[ or (T = 1.90 s so) (n + 1) × 1.90 = n × 2.00  [pic]

gives n = 19 (oscillations of longer pendulum)  [pic]

minimum time between in phase condition = 19 × 2.00 = 38 (s)  [pic] ]

3

[10]

M5.(a)      (i)     speed at P, v (=[pic]) = [pic]  [pic]

= 22(.1) (m s–1)  [pic]

2

(ii)     use of F = k∆L gives [pic]  [pic]

= 11 (10.5) (m)  [pic]

2

(b)      (i)      period T = 2[pic]= 2π[pic] [pic] (= 6.51 s)

time for one half oscillation = 3.3 (3.26) (s)  [pic]

2

(ii)     frequency [pic]  [pic] (= 0.154 (Hz))

use of v = ±2πf [pic] when x = 10.5 m and v = 22.1 m s–1 gives 22.12

= 4π2 × > 0.1542 (A2 – 10.52)  [pic]

from which A = 25.1 (m)  [pic]

        [alternatively, using energy approach gives ½ mvP2 + mg∆L = ½ k(∆L)2  [pic]

        ∴ (29 × 22.12) + (58 × 9.81 × ∆L) = 27 (∆L)2

        solution of this quadratic equation gives ∆L = 35.7 (m)  [pic]

        from which A = 25.2 (m)  [pic]]

3

(c)     bungee cord becomes slack  [pic]

student’s motion is under gravity (until she returns to P)  [pic]

has constant downwards acceleration or acceleration is not ∝ displacement  [pic]

2

(d)     (i)      when student is at R or at bottom of oscillation  [pic]

1

(ii)     at uppermost point or where it is attached to the railing  [pic]

because stress = F/A and force at this point includes weight of whole cord  [pic]

[accept alternative answers referring to mid-point of cord because cordwill show thinning there as it stretches or near knots at top or bottomof cord where A will be smaller with a reference to stress = F/A]

2

[14]

M6.(a)     vibrations are forced when periodic force is applied (1)

frequency determined by frequency of driving force (1)

resonance when frequency of applied force = natural frequency (1)

when vibrations of large amplitude produced

[or maximum energy transferred at resonance] (1)

(max 3)

(b)     (i)      damping when force opposes motion [or damping removes energy] (1)

(ii)     damping reduces sharpness of resonance

[or reduces amplitude at resonant frequency] (1)

(2)

[5]

[pic]

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