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1. Two perpendicular forces of magnitudes F N and 8N act at a
point O (see diagram). Their resultant has magnitude 17 N. a.
Calculate F and find the angle which the resultant makes with the 8
N force. (4) A third force of magnitude E N, acting in the same
plane as the two original forces, is now applied at the point O.
The three forces of magnitudes E N, F N and 8 N are in equilibrium.
b. State the value of E and the angle between the directions of the
E N and 8 N forces. (2)
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A-Level Starter Activity
Topic: Resultant Forces Chapter Reference: Mechanics 1, Chapter
10
7 minutes
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Solutions
1a. F2 = 172 – 82 M1 F = 15 M1 cos x = 8
17 M1
x = 61.9o M1 1b.
E = 17 M1 Angle = 118.1o or 241.9o M1
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1. A particle P of mass 2 kg is attached to one end of a light
string, the other end of which is attached to a fixed point O. The
particle is held in equilibrium, with OP at 30º to the downward
vertical, by a force of magnitude F newtons. The force acts in the
same vertical plane as the string and acts at an angle of 30º to
the horizontal, as shown in the figure below. Find, the value of F
and the tension in the string. (7)
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A-Level Starter Activity
Topic: Motion in 2 Directions Chapter Reference: Mechanics 1,
Chapter 10
8 minutes
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Solutions 1. Resolving Vertically: M1 T cos 30 + F cos 60 = 2g
M1 Resolving Horizontally: M1 T cos 60 – F cos 30 = 0 M1 Solving
simultaneously/elimination M1 F = g = 9.8N M1 T = √3𝑔𝑔 = 17 N
M1
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1.A particle of mass m kg is attached at C to two light
inextensible strings AC and BC. The other ends of the strings are
attached to fixed points A and B on a horizontal ceiling. The
particle hangs in equilibrium with AC and BC inclined to the
horizontal at 30° and 60° respectively, as shown in the figure.
Given that the tension in AC is 20 N, find, a. The tension in BC
(4) b. The value of m (4)
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A-Level Starter Activity
Topic: Motion in 2 Directions Chapter Reference: Mechanics 1,
Chapter 10
8 minutes
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Solutions
1a. Resolving horizontally M1 20 cos 30 = M1 20 cos 30 = T cos
60 M1 T = 20√3 N M1
1b.
Resolving vertically, M1 mg = M1 mg = 20 sin 30 + T sin 60 M1 m
= 40
𝑔𝑔= 4.08 𝑘𝑘𝑘𝑘 M1
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1. Two forces P and Q act on a particle O. The angle between the
lines of action of P and Q is 120o as shown in the figure below.
The force P has magnitude 20 N and the force Q has magnitude X
newtons. The results of P and Q is the for R. Given that the
magnitude of R is 3X newtons, find, giving your answers to 3
significant figures. a. The value of X (4) b. The magnitude of (P –
Q) (3)
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A-Level Starter Activity
Topic: Resultant Forces (D) Chapter Reference: Mechanics 1,
Chapter 10
8 minutes
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Solutions
1a. Use of the cosine rule: (3X)2 = 202 + X2 – 2.20Xcos 60
M1
8X2 + 20X – 400 = 0 M1 X = −5± √25+800
4 M1
X = 5.93 M1 1b.
Use of the cosine rule M1 |𝑃𝑃 − 𝑄𝑄|2 = 202 + X2 – 2X x 20 x cos
120 M1 |𝑃𝑃 − 𝑄𝑄| = 23.5 N M1
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1. A particle of weight W newtons is attached at C to two light
inextensible strings AC and BC. The other ends of the strings are
attached to fixed points A and B on a horizontal ceiling. The
particle hangs in equilibrium with AC and BC inclined to the
horizontal at 30° and 50° respectively, as shown in the figure.
Given that the tension in BC is 6N, find, a. The tension in AC (3)
b. The value of W (3)
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A-Level Starter Activity
Topic: Motion in 2 Directions Chapter Reference: Mechanics 1,
Chapter 10
7 minutes
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Solutions
1a. Resolving horizontally M1 T cos 30 = 6 cos 50 M1 T = 4.45 N
(No units, no mark) M1
1b.
Resolving vertically M1 W = 6 cos 40 + T cos 60 M1 W = 6.82 N
(No units, no mark) M1
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1. Two forces F1 and F2 act on a particle P. The force F1 is
given by F1 = (-i + 2j)N and F2 acts in the direction of the vector
(i + j). Given that the resultant of F1 and F2 actions in the
direction of the vector (i + 3j). Find F2. (6) 2. Three forces (15i
+ j)N, (5qi – pj)N and (-3pi – qj)N, where p and q are constants,
act on a particle. Given that the particle is in equilibrium, find
the value of p and the value of q. (5)
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A-Level Starter Activity
Topic: Forces and Vectors Chapter Reference: Mechanics 1,
Chapter 10
8 minutes
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Solutions
1. F1 + F2 = F3 M1 (-1 + k)i + (2 + k)j = ai + 3aj M1 Equating i
and j values -1 + k = a 2 + k = 3a
M1
Solving simultaneously: -3 + 3k = 2 + k 2k = 5
M1
k = 2.5 M1 F2 = 2.5i + 2.5j M1
2.
15 + 5q – 3p = 0 M1 Equating vertical forces, 1 – p – q = 0 p =
1 – q
M1
Solving by substitution, 15 + 5q – 3(1 – q) = 0 M1
12 + 8q = 0 q = -3
2 M1
p = 52 M1
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1. A vertical rope AB has its end B attached to the top of the
scale plan. The scale pan has mass 0.5kg and carries a brick of
mass 1.5kg, as shown in the figure. The scale pan is raised
vertically upwards with constant acceleration 0.5 ms-2 using the
rope AB. The rope is modelled as a light inextensible string. a.
Find the tension in the rope AB. (4) b. Find the magnitude of the
force exerted on the scale pan by the brick (4)
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A-Level Starter Activity
Topic: Forces and Acceleration (scale-pan) Chapter Reference:
Mechanics 1, Chapter 10
6 minutes
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Solutions
1a. Using F = ma in the vertical direction M1 T – W = (0.5 +
1.5) x 0.5 T – 2g = 2 x 0.5 M1
T = 2g + 1 M1 T = 20.6 N M1
1b.
Using F = ma in the vertical direction M1 R – W = ma R = ma + W
M1
R = 1.5 x 0.5 + 1.5 x 9.8 M1 R = 15.45 N M1
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1. A particle P of mass 0.5 kg is moving under the action of a
single force (3i – 2j)N. Show that the magnitude of the
acceleration of P is 2√13 ms-2. (5) 2. Two cars P and Q are moving
on a straight horizontal roads with constant velocities. The
velocity of P is (15i + 20j) ms-1 and the velocity of Q is (20i –
5j)ms-1. Find the direction of motion of Q, giving your answer as a
bearing to the nearest degree. (3)
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A-Level Starter Activity
Topic: Forces, Acceleration and Vectors Chapter Reference:
Mechanics 1, Chapter 10
5 minutes
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Solutions
1. Use of F = ma M1 (3i – 2j) = 0.5 x a M1 a = 6i - 4j M1 a =
√62 + 42 M1 a = 2√13 ms-2 M1
2.
Bearing of Q = 90 + x M1 Q = 90 + tan-1( 5
20) M1
Q = 90 + 14.0 Q = 104o M1
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1. Two particles A and B have masses 2m and 3m respectively. The
particles are attached to the ends of a light inextensible string.
Particle A is held at rest on a smooth horizontal table. The string
passes over a small smooth pulley which is fixed at the edge of the
table. Particle B hangs at rest vertically below the pulley with
the string taut, as shown in the figure. Particle A is released
from rest. Assuming that A has not reached the pulley, find, a. the
acceleration of B (6) b. the tension in the string (1) c. the
magnitude and direction of the force exerted on the pulley by the
string (2)
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A-Level Starter Activity
Topic: Forces and Acceleration (Connected Particles) Chapter
Reference: Mechanics 1, Chapter 10
8 minutes
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Solutions
1a. Using F = ma on A M1 T = 2ma M1 Using F = ma on B M1 3mg – T
= 3ma M1 Solving simultaneously, 3mg = 5ma M1
a = 3𝑔𝑔5
(or 5.88) ms-2 M1 1b.
Tension = 6𝑚𝑚𝑔𝑔5
(11.76) N M1 1c.
F = √𝑇𝑇2 + 𝑇𝑇2 = 6𝑚𝑚𝑔𝑔
5 √2 N M1 tan 𝜃𝜃 = 1 𝜃𝜃 = 45o M1
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1. Two particles A and B have masses 2m and 3m respectively. The
particles are connected by a light inextensible string which passes
over a smooth light fixed pulley. The system is held at rest with
the string taut. The hanging parts of the string are vertical and A
and B are above a horizontal plane, as shown in Figure 2. The
system is released from rest. Show that the tension in the string
immediately after the particles are released is 12
5𝑚𝑚𝑚𝑚. (8)
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A-Level Starter Activity
Topic: Forces and Acceleration (Connected Particles) Chapter
Reference: Mechanics 1, Chapter 10
6 minutes
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Solutions
1. F = ma for particle B M1 3mg - T = 3ma M1 F = ma for particle
A M1 T – 2mg = 2ma M1 Solving simultaneously, 2mg – T + T – 2mg =
3ma + 2ma M1
mg = 5ma a = 𝑔𝑔
5 M1
T – 2mg = 2m(𝑔𝑔5
) T = 2m(𝑔𝑔
5+ 𝑚𝑚)
M1
T = 125𝑚𝑚𝑚𝑚 M1
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1. A vertical light rod PQ has a particle of mass 0.5kg attached
to it at P and a particle of mass 0.75kg attached to it at Q, to
form a system, as shown in the figure below. The system is
accelerated vertically upwards by a vertical force of magnitude 15N
applied to the particle at Q. Find the thrust in the rod. (8)
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A-Level Starter Activity
Topic: Forces and Acceleration (Connected Particles with Thrust)
Chapter Reference: Mechanics 1, Chapter 10
8 minutes
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Solutions
1. Using F = ma at P M1 T – 0.5g = 0.5a M1 Using F = ma at Q M1
15 – T – 0.75g = 0.75a M1 Solving simultaneously, 15 – 1.25g =
1.25a
M1
a = 2.2 ms-2 M1 Hence, T = 0.5 x 2.2 + 0.5g M1 T = 6N M1
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1. A woman travels in a lift. The mass of the woman is 50 kg and
the mass of the lift is 950 kg. The lift is being raised vertically
by a vertical cable which is attached to the top of the lift. The
lift is moving upwards and has constant deceleration of 2 ms-2. By
modelling the cable as being light and inextensible find. a. The
tension in the cable (3) b. The magnitude of the force exerted on
the woman by the floor of the lift (3)
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A-Level Starter Activity
Topic: Force and Acceleration (Lifts) Chapter Reference:
Mechanics 1, Chapter 10
6 minutes
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Solutions
1a.
F = ma M1 T – 1000g = 1000 x -2 M1 T = 7810 N M1
1b.
Resolving vertically, R – 50g M1
Using F = ma R – 50g = 50 x – 2 M1
R = 390 N M1
Mech 1 C10 Resultant Forces 1Mech 1 C10 Resultant Forces 2Mech 1
C10 Resultant Forces 3Mech 1 C10 Resultant Forces (Difficult)Mech 1
C10 Motion in 2 DirectionsMech 1 C10 Forces and VectorsMech 1 C10
Forces and Acceleration (Scale Pan)Mech 1 C10 Forces, Acceleration
and VectorsMech 1 C10 Connected ParticlesMech 1 C10 Connected
Particles (Pulleys)Mech 1 C10 Connected Particles with ThrustMech 1
C10 Forces and Acceleration (Lifts)