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AS.1 Mechanics answers
Page 11
1) The average speed of the horse is 500/40 = 12.5ms
-1
. (2 marks)
2a) The time for the nerve impulse to travel is 1.8/100 = 1.8x10-2s. (2 marks)
2b) The distance travelled = 2 x 1.8x10-2 = 3.6x10-2m. (2 marks)
(2 marks) (2 marks)
Page 13
1a) The distance travelled by the plane is 380km. (1 mark)
1b) The displacement is 34km north. (2 marks)
2) a = (0 – 10)/4 = -2.5ms-2. (3 marks)
Page 15 (7 marks)
1bi) Displacement between A and C = (½ x 20 x 20) + (60 x20) + (½ x 40 x 20) =
200 + 1200 + 400 = 1800m forwards. (2 marks)
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Displacement between C and B = (½ x 10 x 10) + (30 x10) + (½ x 10 x 10) = 50 +
300 + 50 = 400m backwards. (2 marks)
1bii) Displacement between A and B is 1800 – 400 = 1400m forwards. (1 mark)
1biii) Acceleration at point 1 on the graph = 20/20 = 1ms-2. (2 marks)
Acceleration at point 2 on the graph = -20/40 = -0.5 ms-2. (2 marks)
Acceleration at point 3 on the graph = -10/10 = -1ms-2. (2 marks)
Acceleration at point 4 on the graph = 10/10 = 1 ms-2. (2 marks)
Page 17
1) u = 30ms-1 v = 0ms-1 a = -10ms-2
During the reaction time of 0.5 seconds the car travels 0.5 x 30 = 15m
(assuming the driver maintains 30ms-1 whilst thinking about stopping) (2 marks).
To find the braking distance use the equation v2 = u2 + 2as. Braking distance =
(v2 –u2)/2a = (30)2/(2 x 10) = 45m (3 marks).
Total stopping distance = 15 + 45 = 60m. So, unless the pedestrian moves out of
the way, he will be hit by the car. (2 marks)
2) Cheetah’s speed goes from 0 to 20ms-1 in 2 seconds. It’s top speed is 30ms-1
but can only be maintained for a distance of 450m i.e. a time of 15 seconds.
a) Cheetah’s average acceleration = 20/2 = 10ms-2. (2 marks)
bi) Cheetah travels v2/2a = (30)2/(2 x 20) = 45m. (3 marks)
bii) The acceleration takes 3 seconds. You use the formula, s = ut + ½at2. 45 = 0
+ (½ x 10 x t2). (2 marks)
c) 450/30 = 15s for the cheetah to maintain top speed (18s from rest). (2
marks)
d) The antelope is moving for 3 seconds. It reaches its top speed in 2.2 seconds,
using the formula (v-u)/a = t. The antelope’s top speed is 22 ms-1 and its
acceleration is 10ms-2. The distance covered reaching top speed = s = 0 + (½ x 10
x (2.2)2
) = 24.2m. The distance covered in the remaining 0.8s = 22 x 0.8 = 17.6m.
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Constant velocity as the net force on the car is zero. Forward force = backward
force. Downward force = upward force. (4 marks)
(3 marks)
Page 25
(1 mark)
2) The resultant force decreases over time as the drag force increases until it
equals the weight and the resultant force becomes zero to give the terminal
velocity.
When the skydiver opens the parachute the air resistance/drag force increases
hugely so there is a resultant force upwards. The skydiver slows down until the
forces are balanced again and a new, lower terminal velocity is reached. (9
marks)
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Page 27
Correct units; x title + unit; y title + unit; sensible x scale; sensible y scale;
correct plotting; line of best fit (8 marks) and (3 marks)
(9 marks) and (3 marks)
d) The conclusions I have drawn from my graphs are that a is proportional to F
and a proportional to 1/m. (2 marks)
2) 70 tonnes = 70000kg. Acceleration = 1ms-2.
Force F = ma = 7x103N. (3 marks)
3a) acceleration = change in velocity/time = 14/0.5
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Mass = 60kg.
F = ma = 60 x 14/0.15 = 5600N. (3 marks)
3b) Woman’s weight is 60 x 9.81 = 588.6N
5600/588.6 = 9.5 so the force on the driver is about ten times larger than her
weight. (3 marks)
Page 30
1) When the bus moves forward a standing person is thrown backwards. The
reason is that, apart from the feet in contact with the floor, the person is not
part of the bus. The force moving the bus forward doesn’t act on the person
(except for the feet). Their inertia causes them to remain in place (the feet
move forward because of the friction between the feet and the floor). So as
the bus moves forward the passenger “moves” towards the back as they “want”
to stay in the same place. (8 marks)
2) To find the distance to the water you use the formula s + ut + ½at2 where a is
acceleration due to gravity (g) = 9.8ms-2
S = 0 + (½ x 9.81 x (2)
2
) = 19.62m (3 marks)
3) The value of g on the moon = 128N/80kg = 1.6Nkg-1. (3 marks)
4a) To find the time use s + ut + ½at2 and rearrange it. T = (2s/g)½ = (2 x
235/9.81)½ = 6.92 seconds. (3 marks)
4b) time = distance/speed = 235/340 = 0.69s. It would have been of use to
shout a warning. (4 marks)
Extra question:
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Page 31
1a (4 marks), 1b (4marks) and 1c (2 marks)
Page 33
(4 marks)
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Horizontal pull of the bunting acts in the opposite direction to the horizontal
component of the rope, but they are equal in magnitude i.e. 400sin30 = 200N
Page 37
1a) velocity = 15ms-1 angle = 30o
Horizontal component of velocity 15cos30 = 13ms-1. Vertical component =
15sin30 = 7.5ms-1
The vertical component is used in the rest of the calculation to find the time to
reach the maximum height.
v - u = at t = (v – u)/a = 7.5/9.81 = 0.765s
The time of flight = 0.765 x 2 = 1.53s (3 marks)
1b) The range is the distance travelled while the ball is in flight. As the
horizontal component of the velocity is constant distance = horizontal velocity x
time = 13 x 1.53 = 19.89m. (3 marks)
1c) To find the maximum height you use the vertical component of the velocity
in the formula v2 = u2 -2gs = 0 = (7.5)2 – (2 x 9.81 x s)
56.25 = 19.62s, s = 2.87m (3 marks)
Time of the drop is found from
h = ut + ½gt2 36 = 0 + (½ x 9.81 x t2)
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time of drop t = 2.7 seconds
Minimum horizontal velocity is therefore 6.4/2.7 = 2.36ms-1. (5 marks)
3) The easiest way of answering this question is to use the formula on page 36
of the text book where R = u2sin2α/g. R is the range, u is the launch speed, α is
the launch angle from the horizontal and g is the gravitational field strength (or
acceleration due to gravity). In an exam it is more likely that a question will have
lots of parts and you will calculate the range over stages (and not have to
remember the formula). The answer is 93.4ms-1 (3 marks)
h = ut - ½gt2 = 200m = 0 – (½ x 9.8 x t2)
t = 6.39 seconds for the parcel to hit the ground. (3 marks) b) The horizontal distance travelled by the parcel = horizontal velocity x time =
90 x 6.39 = 576m (as the horizontal velocity of the parcel remains the same as
the plane). (3 marks)
(5 marks)
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Page 39
1) When an object slows down when the driving force is removed the drag
forces cause the kinetic energy to be transformed to heat and sound. The total
energy is the same so the law of conservation of energy applies. (4 marks)
2) Energy cannot be created or destroyed so the energy to generate the
electricity can only have the same value as the quantity of energy used to raise
the water upwards (providing the energy isn’t “lost” elsewhere). (4 marks)
Page 41
1a) Heating – energy is transferred from the burning gas to the water in the
kettle. (1 mark)
1b) Working – energy is transferred from the arms to the packing case to make
it move. (1 mark)
2) Work done = force x distance = 6.5kg x 9.81 x 1.5m = 95.65J. (3 marks)
Page 43
1a) KE of the moving lamp = 10J = ½mv2 where it’s mass is 1.2kg. v = (2 x 10/1.2)½
= 4.1ms-1. (3 marks)
1b) To find the height the lamp is raised remember that the maximum k.e. = max
p.e.= 10J.
10J = mgΔh gives Δh as 0.85m. (4 marks)
2a) k.e. = ½mv2 = ½ x 0.16 x (16.8)2 = 22.6J. (4 marks)
2b) 22.6J = mgΔh Δh = 22.6/(0.16 x 9.81) = 14.4m. (4 marks)
2c) 88% of 16.8ms-1 = 14.8ms-1. (4 marks)
Page 45
1) The power of the kettle = 264000J/(2 x 60) = 2200w. (4 marks)
2) Power of the motorboat = Fv = 123000N x 22 = 2.71x106w. (4 marks)
3) The power of the crane = mgΔh/t = 800 x 9.81 x 21/6 = 2.75x104w. (3 marks)
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Page 47
1) m = 0.14kg F = m(v – u)/t = 0.14 x15/0.1 = 21N. (4 marks)
2)
Horizontal component of the speed = 8cos45 = 5.66ms-1. As the horizontal
velocity is constant t = s/v = 6.4/5.66 = 1.13s (4 marks)
Examzone: Topic 1 mechanics
1) u = 1.1ms-1 and v = 0ms-1 S = 1.96m
Deceleration of the trolley is found from
v2 = u2 + 2as (v2-u2)/2s = -(1.1)2/(2 x 1.96) = -0.31ms-2. So the deceleration is
0.31ms-2. (3 marks)
F = ma = 28 x 0.31ms-2 =8.68N is the frictional force overcoming the motion of
the trolley. (2 marks) The power needed to push the trolley at a constant speed = Fv = 8.68 x 1.1 =
9.55w. (2 marks)
Force required to provide the acceleration =
m(v-u)/t = (28 x 1.1)/0.9 = 34N
Steady force required is the force required is the force require to get the
trolley up to 1.1ms-1 plus the friction force = 34 + 8.68 = 42.9N. (3 marks)
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2) Work done = force x distance and in this case is the area under the curve. I
estimated to be 0.05 x 50 x 90 = 22.5J. (4 marks) This is the k.e. To find the
speed use ½mv2. (22.5 x 2 )/0.08 = v2. v = 23.7ms-1. (2 marks)
3) The deceleration of the cars are the gradients of the graphs a = (22 – 30)/1.3 = -6.15ms-2. Deceleration of the cars = 6.15ms-2. (3 marks) The area
under the velocity-time graph is the displacement travelled (1 mark). The
shaded area is the distance travelled required by the question. It is a
parallelogram so use the formula ½(a + b)/h = ½ x 1.8 x 8 = 7.2m. (2 marks)
Collision is more likely if the cars continue to 0ms-1 (1 mark). Car B is still moving
when car A stops. The extra distance covered by B is ½ x 1.8 x 30 = 27m during
the process of going from 30ms-1 to 0ms-1. This is a lot bigger than 7.2m. (3
marks)
4) A body moving at constant speed can be accelerating if its direction is
changing. Therefore there is a changing velocity and therefore an acceleration.
(3 marks)
The Earth provides the force that attracts the Moon and causes it to move in a
circular orbit (1 mark). This force is towards the Earth (1 mark). The Newton’s
third law pair is the Moon attracting the Earth. This force is towards the Moon
(1 mark).
5) Useful work done by the motor increase in gravitational potential energy.
(i) Work = F x d = 3400 x 9.81 x 30 = 1x106J. (2 marks)
(ii) Power = work done/time taken = 1x106/15
= 6.7x104w (2 marks).
Gravitational p.e. k.e. + g.p.e.
Gravitational potential energy at C = mgΔh = 3400 x 9.81 x 12 = 4x105J.
k.e. = 1x106 – 4x105 = 6x105J = ½mv2
v2= 6x105 x 2/3400 v = 18.8ms-1 (4 marks)
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If there are fewer passengers the speed at C would still be the same. The
reason for this is that “m” appears in both formulas and they can be cancelled
out. (2 marks)
Unit 1 topic tests: Mechanics
1) x = ut + ½at2
Unit of x = m; unit of u = ms-1; unit of t = s; unit of a = ms-2; unit of t2 = s2
so m = ms-1s + ms-2s2; s-1s =1 and s-2s2 =1 so m = m + m therefore unit
on the left = unit on the right. (2 marks)
a = the gradient of the velocity-time graph =
(6 -2)/(8 – 5) = 1.33ms-2. (3 marks)
The distance travelled between 6 and 8 seconds is the area under the graph at
those points
(2 x 3.2) + (½ x 2.8) = 9.2m. (2 marks)
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(4 marks)
The contact force is not involved in the energy transfer process as there is no
movement in the direction of this force. (2 marks)
The advantage of keeping x a small movement means the hand force is
increased. This is a force multiplier.
The disadvantage of keeping x a small movement is that the nail only moves a
small distance. (2 marks)
Displacement at the 1.5 seconds is the shaded area. It is about (0.2 x 108) =
21.6m. (2 marks)
Acceleration at 2 seconds is the gradient of the graph at 2.5 seconds. For non-
uniform acceleration this means drawing a tangent to the graph and finding the
gradient. 8/0.5 = 16ms-2. (2 marks)
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The k.e. after 2.5 seconds = ½mv2 =
½ x 420kg x (63)2 = 8.3 x 105J. (2 marks)
Work done on the spring when it is compressed by 4cm = ½ x 22 x 0.04 = 0.44J.
(3 marks)
g.p.e. gained by the frog = mgh = 24x10-3 x 9.81 ms-2 x 0.6 = 0.14J, which is
about 1/3 of the compression energy. (2 marks) The law of conservation energy
says energy cannot be created or destroyed so 2/3 of the compression energy
has been transferred elsewhere, e.g. heating up the spring, surroundings e.t.c.
(2 marks)
Cupboard mass is 10kg. it’s weight is therefore 98.1N. (1 mark)
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Y force is the same as the weight of the cupboard, but acting in the opposite
direction. Magnitude is 98.1N. The forces must be in equilibrium for the
cupboard to stay in place. (1 mark)
Moment of weight about A is 98.1 x 0.15 = 14.7Nm. (2 marks)
Value of x found from clockwise moment – anticlockwise moment
14.7Nm = 0.06x x = 245N (2 marks)
In practice the screws are usually situated as high in the cupboard as possibleso the force at x is as low as possible and the screws are less likely to be pulled
from the wall. (2 marks)