Download AS Unit G481: Mechanics

Sample Student Answers
AS Unit G481: Mechanics
Module 2: Forces in action
Question 1
Total marks: 17
Figure 3 shows a crate resting on the back of the flatbed of a moving lorry.
(a)
The lorry brakes and decelerates to rest.
Describe and explain what happens to the crate if the flatbed of the lorry is smooth.
Marks available: 2
Student answer:
(a) The crate carries on going with the same velocity because there is no force to slow it down.
Examiner comments:
(a) Good answer, giving both what happens and an explanation.
(b) A rough flatbed allows the crate to stay in position on the lorry when the lorry brakes. In what
direction must a force act on the crate to allow this?
Marks available: 1
Student answer:
(b) To the left.
Examiner comments:
(b) Correct.
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(c) Explain how seat belts worn by rear-seat passengers can reduce injuries when a car is involved
in a head-on crash.
Marks available: 3
Student answer:
(c) The seat belt produces a force on the passenger in the opposite direction to the motion of
the car. This force will stop the passenger hitting the front seats.
Examiner comments:
(c) This is a good start to the answer, but you could have gone on to explain how seat belts
stretch, increasing the time over which the passenger is stopped, and how this in turn will
reduce the force acting on the passenger. You could also mention that seat belts have a
large area, which reduces the pressure on the passenger.
(d) A skydiver jumps from an aircraft at altitude 4000 m and delays opening his parachute until he
reaches altitude 1000 m.
In this question, consider only vertical velocity.
During the first 3.0 s after leaving the aircraft, air resistance to his fall can be neglected.
Describe qualitatively how his velocity will change during the rest of the time he is falling but
before the parachute is opened. Explain why his velocity changes in the way you have
described.
Marks available: 5
Student answer:
(d) At first he falls under gravity. This is called free fall, with an acceleration of about 10 m s−2.
After the first three seconds his velocity continues to increase and therefore the air
resistance acting upwards will also increase. This means his acceleration will decrease.
When the air resistance upwards equals his weight downwards then his acceleration will
become zero and he will reach a terminal velocity.
Examiner comments:
(d) This is a perfect answer in which you have explained everything that has happened. You
wasted time writing the first sentence because the question only asked about what
happened after the third second; you would not be awarded any marks for this sentence.
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(e) When he opens his parachute at 1000 m, he is travelling downwards with a velocity of 60 m s−1.
Draw two labelled sketch graphs, on the same time-axis, to show qualitatively the variation
with time of
(i)
the parachutist’s velocity as it is reduced to 5 m s−1 for landing
(ii) the force which the parachute exerts on the parachutist.
Marks available:
(i) 3 (ii) 3
Student answer:
Examiner comments:
(e) Your velocity graph is absolutely correct but your force diagram needed a little more care. It
should have started at the origin and then shot upwards so that its maximum occurred at the
same time as the maximum deceleration (steepest gradient).
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Module 2: Forces in action
Question 2
Total marks: 11
(a) (i) Define the moment of a force. Illustrate your answer with a diagram.
(ii) What is a couple? What effect does a couple have on the motion of a body?
(iii) Define torque of a couple. Again illustrate your answer with a diagram.
Marks available:
(i) 2 (ii) 2 (iii) 3
Student answer:
(a) (i) The moment of a force about a point is the product of the force multiplied by the
perpendicular distance of the force from the point.
(ii) A couple is two equal and parallel forces which are acting in opposite directions and not in
the same straight line. A couple makes a body rotate.
(iii) The torque of a couple is its turning effect.
Torque = one of the forces × perpendicular distance between them
Examiner comments:
(a) These are really good definitions. Remember that definitions should always be exact and
that equations will often help.
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(b) An electricity cable is attached to a pole at a height of 6.0 m above the ground as shown in
Figure 5.
The cable exerts a force of 280 N on the pole at an angle of 90o to the pole. So that there is zero
turning moment on the pole itself, a wire under tension is attached to the pole at a height of
4.0 m, and it makes an angle of 35o to the pole.
(i) Calculate the moment which the cable exerts about P, a point in the pole level with the
ground.
(ii) Calculate the tension necessary in the wire.
Marks available:
(i) 2 (ii) 2
Student answer:
(b) (i) Moment = 280 × 6.0
= 1680 N m
(ii) Perpendicular distance = 4 sin 35o
=2.29 m
Using the Principle of Moments
1680 = T × 2.29
T = 732 N
Examiner comments:
(b) Good answers. Always remember to try to use words in calculations to help explain what
you are doing. This helps avoid answers which are just a jumble of figures followed by an
answer.
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