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G481
MECHANICS MODULE 2:
FORCES IN ACTION
Answer Booklet
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RAB Plymstock School
Lesson 16 questions – Newton’s 2nd Law
1a)
Define a Newton.
……The force that causes 1 kg to accelerate with 1ms-2
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (1)
b)
Fig 1.1 shows a car of mass 1200kg, pulling a caravan of mass 400kg along a horizontal
road.
Fig 1.2
Calculate, for the first 20s of the journey
i)
The acceleration of the car
Gradient = 13/20
acceleration = …0.65…..ms-2 (2)
ii)
the resultant force acting on the car
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Force = ma
1200x0.65
resultant force on the car = 780…N (2)
iii)
the resultant force acting on the caravan.
Force = ma
400x0.65
resultant force acting on the caravan = ……260…. N (2)
Use the graph to describe how the acceleration changes after the first 20s of the
c
i)
journey.
……gradient is less hence acceleration is less
……reaches terminal velocity…………………………(1)
ii)
Suggest the possible reason for this change.
……resultant force is less / resistive forces increasing / driver eases off accelerator / climbing a
hill
……………………………………………………………………………………………… (1)
Total [9]
2
a)
Define acceleration.
……rate of change of velocity ………………………………………………….
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (2)
b)
An aircraft of total mass 1.5 x 105 kg accelerates, at maximum thrust from the engines,
from rest along a runway for 25s before reaching the required speed for take-off of 65ms-1.
Assume that the acceleration of the aircraft is constant.
Calculate
i)
the acceleration of the aircraft
acceleration = (v-u)/t
=(65-0)/25
ii)
F=ma
acceleration…2.6……………. ms-2 (3)
the force acting on the aircraft to produce this acceleration
1.5 x 105 x 2.6
iii)
force = ……390000 (3.9x105)………….. N (2)
the distance travelled by the aircraft in this time.
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distance = (u+v)t/2
(0+65)x25/2
distance = …813…(allow 810)…………….. m (2)
c)
The length of runways at some airports is less than the required distance for take-off by
this aircraft in (b)(iii).
State and explain one method that could be adopted for this aircraft so that it could reach the
required take-off speed on shorter runways.
……Reduce the mass…………………………………………………………………………….
……remove baggage / people / fuel /
since F=ma, less mass needs less Force for same acceleration so greater acceleration can be
attained……………………………………………………………………………………………
….
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (2)
Total [11]
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RAB Plymstock School
Lesson 18 questions – Forces in equilibrium
1)
Fig 1.1 shows two strings supporting an object of weight 12N. The tension in one of the
strings is 6.0 N. The tension in the other string is T.
i)
Calculate the magnitude of the vertical component of the tension T in order for the
object to be in equilibrium.
Tv + 6sin30 = 12
Tv + 3 = 12
Tv = 12 – 3
Tv = 9N
vertical component………9……………N (2)
ii)
Hence calculate the magnitude of T.
cos30=9/T
T=9/cos30
T=10.39N
magnitude of T…………10…………N (2)
Total [4]
2)
A girl travels down a pulley-rope system that is set up in an adventure playground. Fig
2.1 shows the girl at a point on her run where she has come to rest.
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a)
Explain why the vector sum of the three forces must be zero.
…………The girl is at rest………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
b)i)
Sketch a labelled vector triangle of the forces acting on the pulley wheel.
T2
60˚
500N
40˚
80˚
T1
(3)
ii)
Determine by scale diagram or calculation the forces T1 and T2 the rope exerts on the
pulley wheel.
Using Sine Rule:
a
sinA
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=
b
sinB
=
c
sinC
RAB Plymstock School
500
sin40
=
T2
sin80
=
T1
sin60
T2=766N (766.04N)
T1=674N (673.65N)
(3)
Total [7]
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RAB Plymstock School
Lesson 20 questions – moments and torque
1a)
Define centre of gravity
…Point where all the mass/weight of the body
seems/appears to act………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
b)
Define moment of a force
……Force x perpendicular distance ……………………………………………
……from pivot (to the line of action of the force)…………………………………………
………………………………………………………………………………………… (2)
c)
Fig 1.1 shows a student being weighed. The student, of mass 50kg, stands 0.30 m from
end A of a uniform plank of mass 8.0 kg and length 2.0 m. The plank is pivoted 0.50 m from
end A, and an object of mass 5.0 kg is moved from and B until the plank balances.
normal from
pivot
s
d
50N
weight 80N
fig 1.1
i)
500N
On fig 1.1, draw and label the forces acting on the plank.
ii)
Show by using the principle of moments that the 5.0 kg mass must be placed 0.30m
from end B in order for the plank to balance.
Moments about P: clockwise moments = anticlockwise moments
500 x 0.2
= 80 x 0.5 + 50s
s
= 1.2m
d
= 0.3m
s + d = 1.5
(4)
iii)
The arrangement shown in Fig 1.1 has a maximum mass that it can determine. Suggest
two changes that can be made to the arrangement to increase this maximum possible mass.
Explain your answers.
1.
……Move 5.0kg mass towards B
this will make the clockwise moment larger meaning a larger anticlockwise moment could be
achieved. ……………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
2.
……Move the pivot towards A,
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this will make the clockwise moment larger meaning a larger anticlockwise moment could be
achieved………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (3)
Total [13]
2a)
Using a sketch define the moment of a force.
s
moment = Fs
F
……moment (Nm) = force(N) x perpendicular distance (m)
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
b)
Fig 2.1 shows a computer resting on a tabletop that is hinged at A.
W
50N
fig 2.1
The tabletop has a mass of 5.0 kg and its centre of gravity is 0.40m from the axis of the hinge
at A. The computer has a weight of 200N acting through a point 0.25 m from the hinge at A.
The tabletop is supported to maintain it in a horizontal position by a force F acting vertically at
B. The distance AB is 0.80m.
i)
Calculate the weight of the tabletop.
5.0 x g = 5.0 x 10
weight = ……50……..N (1)
ii)
On fig 2.1 draw and label an arrow to represent the weight W of the tabletop. (1)
iii)
Apply the principle of moments about the hinge at A to determine the vertical force F
applied at B that is required to maintain the tabletop in equilibrium.
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clockwise moments = anticlockwise moments; about A.
200 x 0.25 + 0.4 x 50 = F x 0.8
70/0.8 = F
force F = …87(.5)………..N (3)
iv)
The tabletop must experience a resultant force of zero in order to be in equilibrium.
Explain how the forces acting on the tabletop fulfil this condition.
…………Upward force acts at hinge
So F and the force at hinge equals weight of table and computer
(the upward force = downward force)
……………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
Total [9]
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RAB Plymstock School
Lesson 21 questions – Density and pressure
1
a)
Define density.
…………Density = mass (kg) / volume (m3)…………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………… (1)
b)
Fig 1.1 shows a swimming pool with a base of surface area 80m2. The pool contains
water of density 1000kgm-3 to a uniform depth of 1.2m.
i)
fig 1.1
Show that the weight of water in the pool is 9.4 x 105N.
Volume = 1.2 x 80 = 96 m3
density x volume = mass
1000 x 96 = 96000kg
w=mg = 96000 x 9.81 = 941760N ≈ 9.4 x 105N
(2)
ii)
Calculate the pressure exerted on the base of the pool by the water.
pressure = force / area
= 9.4 x 105 / 80
=
pressure = ……11750……………….. unit …Pa……… (3)
c)
A second pool, of base area 40m2, has water of the same density and depth as that of the
pool in part (b). Determine the pressure of the water on the base of this pool.
pressure = 9.4 x 105 / 40
pressure = …23500 Pa…………………..
comment on your answer.
…………half the surface area so double the pressure
………………………………………………………………………………………….
…………………………………………………………………………………………………(2)
Total [8]
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RAB Plymstock School
Lesson 22 questions – Forces on vehicles
1. Use the internet to help you answer the following questions. A good start would be:
http://www.howstuffworks.com
a)
Label the following diagram
b)
What happens when you hit the brake pedal?
………the piston pushes the brake shoes against the drum
………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
c)
Label the following diagram:
d)
How are disc brakes different to normal bike brakes?
……… Bicycle brakes have a calliper, which squeezes the brake pads against the wheel. In a disc brake, the
brake pads squeeze the rotor instead of the wheel, and the force is transmitted hydraulically instead of through a
cable. …………………………………………………………………………………
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…………………………………………………………………………………………
…………………………………………………………………………………… (2)
e)
What force between the pads and the disc slows the disc down?
………………Friction……………………………………………… (1)
Total [12]
2
a)i)
Define pressure
…………pressure is the force per unit area………………………………………..
…………………………………………………………………………………………..
ii)
State the SI unit of pressure.
……………Pascals……or Nm-2…………………………………………………… (3)
b)
Fig 2.1 shows the arrangement of the master cylinder and the slave cylinder for
part of a braking system. The master cylinder piston, of surface area 8.0cm2, exerts a force of
28N on the brake fluid when the driver presses the brake pedal.
Fig 2.1
i)
Calculate the pressure on the brake fluid produced by the master piston.
P=F/A
=28/8x10-4
pressure = ……35…….. unit …kPa…… (2)
This pressure is transmitted to the slave cylinder piston that has a surface area of 40cm2.
ii)
Calculate the force exerted on this piston.
ecf
F=PxA
=35000x4x10-3
force = ……140………….. N (2)
iii)
State two methods of increasing the force produced by the slave cylinder
piston.………decreasing area of master piston ………………………………..
………………increasing force on master……………………………………..
………………increasing area of slave……………………………………… (2)
Total [9]
3)
a)
Explain the following terms when associated with driving a car.
i)
thinking distance
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………distance travelled by car between seeing obstacle to be stopped before and pressing
brakes………………………………………………………………..
ii)
braking distance.
………distance travelled by car after pressing brakes to coming to a complete stop
safely……………………………………………………………………………… (2)
b)
car.
Fig 3.1 shows the variation with speed v of the thinking distance d for the driver of a
fig 3.1
i)
Explain why the graph is a straight line through the origin.
………the speed of the car is directly proportional to its thinking distance…… (1)
ii)
Use the graph to determine the time taken for the driver to react when he is travelling at
-1
16ms .
speed = distance/time
time = distance/speed
=8.8m/16ms-1
time taken = ………0.55……(0.6)………. s (2)
iii)
The driver is travelling at 30ms-1 in a car which brakes with an acceleration of -6.5ms-2.
Calculate
1.
the thinking distance
ecf
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distance = speed x reaction time
=30x0.55
thinking distance = ………16.5…(18 if 0.6 used)………… m
2.
the overall stopping distance.
braking distance
v2=u2+2as
0=302+2(-6.5)s
s=900/13
=69.23
Total distance = thinking + braking
= 16.5 + 69.23 (ecf)
stopping distance =………85.73…(87.2 if 18 used)………….. m
(5)
c)
Explain the effect of the road conditions and tyre tread on the stopping distance.
………Wet or icy roads can decrease friction and so increase braking distance therefore
increasing total stopping distance.………………………………………..
………If the tyre tread is worn or bald it cannot increase surface area between tyre and road as
the water will get in way and could lead to aquaplaning which reduces friction between road
and tyre and leads to increase in braking distance and therefore stopping
distance.………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………… (3)
Total [13]
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RAB Plymstock School
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RAB Plymstock School