4 Forces in action
Exam-style mark scheme
OCR Physics A
Question
Answer
Marks
Guidance
1ai
Tension T1  3  9.81  29.4 N
A1
A is at rest so ΣF  0
1 a ii 1
3  9.81  T2  3a
C1
Applying F  ma to A
T2  1.8  9.81  1.8a
C1
and to B.
adding gives 3  9.81 – 1.8  9.81  (3  1.8) a
a  2.45 m s  2
1 a ii 2
s  ut 
Note tension is less than T1
A1
A is initially at rest so u  0
1
a t²
2
1
 2.45 t 2
2
t  1.07 s
C1
1.4 
A1
1bi
Force  19  9.81  186 N
1 b ii
From 0 to t1, the lift, and hence the suitcase, are
accelerating upwards. So newtonmeter reading is greater
than 186 N.
B1
From t1 to t2, the lift has constant speed so newtonmeter
reading  weight  186 N
B1
From t2 to t3, the lift has constant deceleration so
newtonmeter reading is less than weight.
At constant speed there is no
resultant force on suitcase.
B1
The newtonmeter reading
will actually decrease
(suitcase appears to be
lighter) by the same amount
as it increased in first stage
since the graph is symmetric.
Force which produces an acceleration of 1 m s–2 on a
mass of 1 kg.
B1
F  m a gives: (170 – 90)  950 a
C1
a  8.4  10–2 m s–2
A1
The car is accelerating so its speed is increasing. This
increases the air resistance.
B1
Resultant force decreases so acceleration decreases.
B1
v  u  a t only applies if a is constant.
B1
2a
2bi
2 b ii
3a
The torque of a couple is the product of one of the forces
and the perpendicular distance between the forces.
B1
B1
Newtonmeter records the
weight of the suitcase (m g)
Upward force from
newtonmeter >weight, giving
resultant upward force to
provide acceleration.
Remember to use the
resultant force not just the
driving force.
Students often try to use the
SUVAT equations in
situations where acceleration
is not constant. This is
incorrect physics and would
lead to loss of marks.
It is usually easier to give a
word equation for definitions
i.e.
Torque  (One force) 
(perpendicular distance
between forces)
3b
Torque is measured in newton metre so
[Torque]  [kg m s–2] [m]  [kg m2 s–2]
© Oxford University Press 2015
B1
SI base units are kilogram,
metre and second in
mechanics.
www.oxfordsecondary.co.uk/acknowledgements
This resource sheet may have been changed from the original
1
4 Forces in action
Exam-style mark scheme
OCR Physics A
Question
3c
3d
4ai
4 a ii
Answer
Marks
Hang plate and a plumb line freely from hole A.
Mark a vertical line from A, indicated by plumb line, onto
the plate.
B1
Repeat for other holes.
B1
Centre of mass is the point where the lines intersect.
B1
Mass plate  8850  (0.11  (5.0  10–3))  4.87 kg
C1
(4.87  9.81)  0.15  0.22 T
C1
T  32.6 N
A1
G is the point where vertical line from hook meets the
centre of the beam.
B1
Horizontally:
T1 sin 55  T2 sin 20
hence
T1  T2
sin 20
sin 55
C1
Vertically T1 cos55 T2 cos 20  12  103
Substituting
T2
sin 20
cos 55  T2 cos 20  12  10 3
sin 55
T2  10 .2  10 3 N
4 b ii
4 b iii
5a
5b
Horizontal and vertical
components of the three
forces must equate to zero
since the hook is in
equilibrium. The equations
must be solved
simultaneously.
C1
A1
Mass  (7.9  10³)  (7  10–2)  (5  10–2)  (3  10–2)
C1
Mass  0.83 kg
A1
Moments about P gives:
10  MC  (13  2.0)  (36  0.83)
C1
MC  5.59 kg
A1
When immersed the block experiences an upthrust which
reduces the clockwise moment about P.
M1
This means the anticlockwise moment must be reduced
by reducing mass at C.
A1
Tangent at t  0 (s)
M1
Gradient  2.0 m s–2
A1
Two forces are acting on sphere at t  0 s, the weight and
the upthrust.
B1
Resultant force on sphere is therefore less than the
weight.
B1
© Oxford University Press 2015
Note conversion of mm to m
and use of acceleration of
free fall to convert mass to
weight.
A1
T1  4.25  10 3 N
4bi
Guidance
The density formula must be
rearranged and cm
converted to m correctly.
Since all lengths are in cm
there is no need to convert
them to m in this equation.
Likewise it is possible to
work consistently with
masses here rather than
convert each one to weight.
The reason must be given in
order for the second mark to
be scored. This prevents
anyone getting a mark by
guessing the change in
mass.
www.oxfordsecondary.co.uk/acknowledgements
This resource sheet may have been changed from the original
2
4 Forces in action
Exam-style mark scheme
OCR Physics A
Question
5ci
Answer
Marks
F  ma
Using Newton’s 2nd law
mg – U  ma
5 c ii
5 c iii
12  10–3  9.81 – U  12  10–3  2.0
C1
U  9.37  10–2 N
A0
Upthrust on a body in fluid is equal to the weight of fluid
displaced by the body
Density, ρ 
V
V
5d
6a
6b
Guidance
Upthrust, U  weight of liquid
displaced
B1
U / g 
V
U /g

C1
2
(9.37  10 / 9.81)
1600
V  5.97  10–6 m3
A1
As velocity increases drag increases (and acceleration
decreases).
B1
From 0 s to 0.35 s, the gradient of graph becomes less
steep.
B1
After 0.35 s, drag  weight, and velocity remains
constant.
B1
Diagram and/or description of apparatus.
B1
Measurement of diameter of sphere (and hence
calculation of R) using micrometer or Vernier callipers.
B1
Allow: sphere to fall in tube of liquid until terminal velocity
is reached.
B1
Measure time to fall a set distance at terminal velocity
(allow: strobe lighting and camera or light gates and
timer/stopwatch or video).
B1
Plot graph of v against 1/R (or 1/diameter)
B1
Straight line through origin suggests theory is true.
B1
© Oxford University Press 2015
Upthrust is constant
throughout but increase in
drag causes resultant force
to decrease, which reduces
gradient (acceleration).
Terminal velocity is reached
at t  0.35 s
www.oxfordsecondary.co.uk/acknowledgements
This resource sheet may have been changed from the original
3