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8/18/2019 As.1 Answers

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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)

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