Microsoft Word - Kinetic and Potential Energy Old Exam Qs.rtf

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Kinetic and Potential Energy Old Exam Qs

Q1. A firework rocket is fired vertically into the air and explodes at its highest point. What are

the changes to the total kinetic energy of the rocket and the total momentum of the rocket

as a result of the explosion?

total kinetic energy

of

rocket

total momentum of

rocket

A

unchanged unchanged

B

unchanged increased

C

increased unchanged

D

increased increased

(Total 1 mark)

Q2. Two ice skaters, initially at rest and in contact, push apart from each other.

Which line, A to D, in the table states correctly the change in the total momentum and the

total kinetic energy of the two skaters?

total momentum total kinetic energy

A

unchanged increases

B

unchanged unchanged

C

increases increases

D

increases unchanged

(Total 1 mark)

Q3. The diagram below shows a child coming down a slide in a playground. The vertical height

of the slide is 3.0 m. The angle between the main slope of the slide and its vertical support

is 50°. Acceleration of free fall g = 9.8 m s

–2

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(a) The child has a mass of 41 kg. Calculate the gain in gravitational potential energy as

the child climbed to the top of the slide.

Gravitational potential energy gained ......................................

(2)

(b) Assume that the slide is frictionless.

(i) Use your answer to part (a) to calculate the speed of the child when reaching

the bottom of the slide.

Speed ...................................

(2)

(ii) Calculate the resultant force acting on the child when in the position shown in

the diagram above.

Resultant force ............................

(2)

(Total 6 marks)

Q4. The speed of an air rifle pellet is measured by firing it into a wooden block suspended

from a rigid support. The wooden block can swing freely at the end of a light inextensible

string as shown in the figure below.

initial position of block

A pellet of mass 8.80 g strikes a stationary wooden block and is completely embedded in

it. The centre of mass of the block rises by 0.63 m. The wooden block has a mass of 450

g.

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(a) Determine the speed of the pellet when it strikes the wooden block.

speed = ........................ m s

–1

(4)

(b) The wooden block is replaced by a steel block of the same mass.

The experiment is repeated with the steel block and an identical pellet. The pellet

rebounds after striking the block.

Discuss how the height the steel block reaches compares with the height of 0.63 m

reached by the wooden block. In your answer compare the energy and momentum

changes that occur in the two experiments.

........................................................................................................................

(4)

(c) Discuss which experiment is likely to give the more accurate value for the velocity of

the pellet.

........................................................................................................................

(2)

(Total 10 marks)

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Q5. The diagram below shows the path of a skier who descends a slope AB.

The skier starts from rest at A and eventually comes to rest again at C on the horizontal

surface BC.

(a) (i) The slope AB has a vertical height of 35 m. The total mass of the skier is 65

kg.

Show that the skier’s loss in gravitational potential energy is about 20 kJ.

(1)

(ii) The kinetic energy of the skier at point B is 11 000 J.

Show that the skier’s speed at point B is about 18 m s

–1

.

(2)

(iii) The average retarding force acting on the skier is 140 N.

Calculate the distance travelled between A and B.

distance travelled .................................................... m

(2)

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(iv) Describe two ways in which the retarding force may arise.

...............................................................................................................

(3)

(b) The skier decelerates uniformly between B and C at 2.8 m s

–2

.

(i) Calculate the time taken to travel from B to C.

time ................................ s

(2)

(ii) Calculate the distance BC.

distance .................................... m

(2)

(Total 12 marks)

Q6. A climber falls 2.3 m before being stopped by his climbing rope that is secured above him.

The weight of the climber is 840 N.

(a) Calculate the loss in gravitational potential energy of the climber.

......................................................................................................................

loss in potential energy .......................................................... J

(2)

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(b) The figure below shows a force-extension graph for the rope being used.

(b) (i) Use the figure above to find the stiffness of the rope when it is being used

with forces up to 350 N. Give the appropriate unit.

.............................................................................................................

stiffness .................................................................................

unit .................................................................................

(4)

(ii) Use the figure above to determine the energy stored in the rope when it is

stretched by 0.25 m.

.............................................................................................................

energy ................................................................................. J

(3)

(Total 9 marks)

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Q7. A type of exercise device is used to provide resistive forces when a person applies

compressive forces to its handles. The stiff spring inside the device compresses as shown

in the figure below.

(a) The force exerted by the spring over a range of compressions was measured.

The results are plotted on the grid below.

(i) State Hooke’s law.

.............................................................................................................

(2)

(ii) State which two features of the graph confirm that the spring obeys Hooke’s

law over the range of values tested.

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.............................................................................................................

(2)

(iii) Use the graph to calculate the spring constant, stating an appropriate unit.

answer = .....................................

(3)

(b) (i) The formula for the energy stored by the spring is

Explain how this formula can be derived from a graph of force against

extension.

.............................................................................................................

(3)

(ii) The person causes a compression of 0.28 m in a time of 1.5 s. Use the graph

in part (a) to calculate the average power developed.

answer = ..................................W

(3)

(Total 13 marks)

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Q8. An ‘E-bike’ is a bicycle that is assisted by an electric motor. The figure below shows an

E-bike and rider with a total mass of 83 kg moving up an incline.

(a) (i) The cyclist begins at rest at A and accelerates uniformly to a speed of 6.7 m

s

–1

at B.

The distance between A and B is 50 m.

Calculate the time taken for the cyclist to travel this distance.

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

(2)

(ii) Calculate the kinetic energy of the E-bike and rider when at B. Give your

answer to an appropriate number of significant figures.

answer = .................................. J

(2)

(iii) Calculate the gravitational potential energy gained by the E-bike and rider

between A and B.

answer = .................................. J

(2)

(b) Between A and B, the work done by the electric motor is 3700 J, and the work done

by the cyclist pedalling is 5300 J.

(i) Calculate the wasted energy as the cyclist travels from A to B.

answer = .................................. J

(2)

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(ii) State two causes of this wasted energy.

Cause 1 ................................................................................................

...............................................................................................................

Cause 2 ................................................................................................

...............................................................................................................

(2)

(Total 10 marks)

Q9. Figure 1 shows car A being towed at a steady speed up a slope which is inclined at 5.0°

to the horizontal. Assume that the resistive forces acting on car A are negligible.

Figure 1

Figure 2 represents a simplified version of the forces acting on car A at the instant shown

in Figure 1.

Figure 2

(a) (i) Car A has a mass of 970 kg. Show that the component of its weight that acts

parallel to the slope is approximately 830 N.

(2)

(ii) Calculate the energy stored in the tow rope as car A is towed up the slope at a

steady speed. The tow rope obeys Hooke’s law and has a stiffness of 2.5 ×

10

4

Nm

–1

.

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energy stored ................................................ J

(4)

(b) The tow rope is attached to a fixing point on car A using a metal hook. During the

ascent of the slope the fixing point snaps and the metal hook becomes detached

from car A. The metal hook gains speed due to the energy stored in the rope. State

and explain how the speed gained by the hook would have changed if the rope used

had a stiffness greater than 2.5 × 10

4

Nm

–1

.

........................................................................................................................

(3)

(Total 9 marks)

Q10. The world record for a high dive into deep water is 54 m.

(a) Calculate the loss in gravitational potential energy (gpe) of a diver of mass 65 kg

falling through 54 m.

loss in gpe = ................................... J

(2)

(b) Calculate the vertical velocity of the diver the instant before he enters the water. Ignore the

effects of air resistance.

velocity = ............................ ms

–1

(2)

(c) Calculate the time taken for the diver to fall 54 m. Ignore the effects of air

resistance.

time = ................................... s

(2)

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(d) Explain, with reference to energy, why the velocity of the diver is independent of his

mass if air resistance is insignificant.

........................................................................................................................

(3)

(Total 9 marks)

Q11.A snowboarder slides down a slope, as shown in the diagram below. Between B and C her

acceleration is uniform.

(a) The snowboarder travels 1.5 m from B to C in a time of 0.43 s and her velocity

down the slope at C is 5.0 ms

–1

.

Calculate her velocity down the slope at B.

velocity = ............................ ms

–1

(3)

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(b) The combined mass of the snowboarder and snowboard is 75 kg and the angle of

the slope is 25°

(i) Calculate the component of the weight of the snowboarder and snowboard

acting down the slope.

weight component = ................................. N

(2)

(ii) At D the snowboarder has reached a constant velocity. She moves a distance

of 2.0 m at constant velocity between D and E.

Calculate the work done against resistive forces as she moves from D to E.

work done = ................................... J

(1)

(c) State and explain what happens to the gravitational potential energy lost between D

and E.

........................................................................................................................

(3)

(Total 9 marks)

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ANSWERS

M1.C

[1]

M2. A

[1]

M3. (a) PE = mgh

C1

= 41 × 9.8 × 3.0 =

1200 or 1210 J

A1

2

(b) (i) mgh = 0.5mv

2

C1

v = 7.7 ms

–1

A1

2

or ecf from (a)

(ii) F = mgcos50

C1

= 258N

A1

2

[6]

M4.(a) Max GPE of block = Mgh = 0.46 × 9.81 × 0.63 = 2.84 J ✓

The first mark is for working out the GPE of the block

1

Initial KE of block = ½ Mv

2

= 2.84 J

Initial speed of block

v

2

= (2 × 2.84) / 0.46

v = 3.51 ms

–1

✓

The second mark is for working out the speed of the block

initially

1

momentum lost by pellet = momentum gained by block

=

Mv = 0.46 × 3.51 = 1.61 kg m s

–1

✓

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The third mark is for working out the momentum of the block

(and therefore pellet)

1

Speed of pellet = 1.58 / m = 1.58 / 8.8 × 10

−3

= 180 ms

–1

(183) ✓

The final mark is for the speed of the pellet

1

At each step the mark is for the method rather than the

calculated answer

Allow one consequential error in the final answer

(b) As pellet rebounds, change in momentum of pellet greater and therefore the

change in momentum of the block is greater ✓

Ignore any discussion of air resistance

1

Initial speed of block is greater ✓

1

(Mass stays the same)

Initial KE of block greater ✓

1

Therefore height reached by steel block is greater than with wooden block ✓

1

(c) Calculation of steel method will need to assume that collision is elastic so that

change of momentum can be calculated ✓

1

This is unlikely due to deformation of bullet, production of sound etc. ✓

1

And therefore steel method unlikely to produce accurate results.

[10]

M5. (a) (i) 65 × 9.8 × 35 seen and evaluated to 22295 or 2231

or 22300J

(ii) correct substitution of 65 kg and either 11000J or

18 m s

–1

in ke formula seen

18.4 (18.397) (m s

–1

) to at least 3 sf

(iii) distance = energy loss/force or work done/force or

numerical equivalent

64-64.3 using E

p

= 20kJ or 79 – 81(m) using 22.3 kJ

(iv) friction

air resistance

further detail eg friction at ski-ice surface

or caused by need to move air when passing through it

8

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(b) (i) time = ∆v/a or numerical equivalent

6.4(3) – 6.6(6.57) (s)

(ii) use of appropriate kinematic equation

(57.8 – 60.4) 58 m or 60 (m) to 2 sf

4

[12]

M6. (a) 840 × 2.3

1900 (J)/1930 (J)

(b) (i) uses gradient

C1

data extraction correct –350 N, 0.3 m

C1

1170

A1

N m

–1

B1

4

(ii) uses area

B1

6.5 to 7.0 squares or 1 square is equivalent to

5 J/area is ½ base × height

B1

32.5 to 35 (J)

B1

3

[9]

M7. (a) (i) F

∆L (1) up to limit of proportionality (1)

accept ‘elastic limit’

F = k∆L with terms defined gets first mark

2

(ii) straight line (1) through origin (1)

2

(iii) working shown and F ≥ 200 N (1) (500/0.385) = 1290 ± 20 (1)

N m

–1

or N/m kg s

–2

(1)

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(b) (i) (∆W = F∆s) so area (beneath line from origin to ∆L)

represents (work done or) energy (to compress/extend) (1)

work done (on or by the spring) linked to energy stored (1)

(area of triangle) =

b × h (therefore E = F∆L) (1)

3

(ii) F = 360 (N) used (1) (1) = 34

(33.6) (W) (1)

ecf from wrong force

3

[13]

M8.(a) (i) (s = ½(u + v) t) t = 2s/v (correct rearrangement, either symbols or values)

(= 100/6.7) = 15

(s) (14.925)

or alternative correct approach

(ii) (KE = 1/2mv

2

= ½ × 83 × 6.7

2

) = 1900 (1862.9 J)

2 sf

2

(iii) GPE = 83 × 9.81 × 3.0 penalise use of 10, allow 9.8

= 2400 (2443 J)

do not allow 2500 (2490) for use of g = 10

2

(b) (i) 5300 + 3700 (or 9000 seen)

or – 2443 – 1863 (or (–) 4306 seen)

= 4700 (J)

(4694) ecf from parts aii & aiii

2

(ii) mention of friction and appropriate location given

mention of air resistance (or drag)

do not allow energy losses or friction within the motor

do not allow energy losses from the cyclist

must give a cause not just eg ‘heat loss in tyres’

2

[10]

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M9. (a) (i) uses trigonometry (mg sin5 or mg cos85 seen)

829.3 / 828.5 (N) at least 3 sf

2

(ii) tension = 830 (N)

E= ½ FL and F = k L identified / or combined to E= ½ (F

2

/k)

correct sub condone power 10 error

13.8 (J) range 13.9 to 13.7

(b) lower speed

less extension

less energy stored (in rope)

[9]

M10.(a) (Ep = mgh)

= 65 × 9.81 × 54

= 3.44 × 10

4

= 3.4 × 10

4

(J) (34433)

max 1 if g =10 used (35100 J)

Correct answer gains both marks

2

(b)

allow 32 (32.3) for the use of 34000

allow 32.6

OR correct use of v

2

= 2 g s

don’t penalise g = 10 (32.863)

2

(c) (s =1 / 2 gt

2

or other kinematics equation)

With use of g= 9.8 or 9.81 or 10 and / or various suvat

equations, expect range 3.2 to 3.4 s.

No penalty for using g= 10 here.

ecf from 1(b) if speed used

2

(d) (all G)PE (lost) is transferred to KE

no (GP)E transferred to 'heat' / 'thermal' / internal energy

OR

Must imply that all GPE is transferred to KE. E.g. accept

‘loss of GPE is gain in KE’ but not: ‘loses GPE and gains

KE’.

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(therefore)

mass cancels.

Accept ‘m’s crossed out

3

[9]

M11.(a)

Correct answer with no working gets 2 out of three.

OR substitution in above equation OR

= 6.9767 – 5.0

= 2.0 (1.98 m s

−1

)

Full credit for use of gsin25 = acceleration down slope. This

yields answer 3.22 m s

1

Allow 1sf answer (2).

3

(b) (i) (F = 75 × 9.81 ×) sin25 (

o

)

= 310 ( 311, 310.94) (N)

use of g = 10 not penalised here

‘sin25’ on its own

Use of g = 10 yields 317

Allow cos65

2

(ii) W = Fs

= 311 × 2.0 = 620 (622 J)

ecf (2bi) × 2.0

1

(c) Idea that GPE is ultimately transferred to: internal (energy) / 'heat'/ 'thermal' (energy

in the surroundings)

Allow transfer of GPE to KE and then to ‘thermal’ etc

Do not allow reference to ‘sound’ on its own

Correct reference to a named resistive force: friction / drag / air resistance

Don’t accept implication that a resistive force is a form of

energy

All GPE becomes 'heat', etc OR no (overall) increase in KE OR reference to

work done against or by a resistive force

Do not allow references to loss of body heat.

Allow: ‘(GPE) not converted to KE’

3

[9]

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