Edexcel International A Level Maths · M1 · Mechanics 1
Modelling and vectors in mechanics
28 free, high-quality past-paper-style questions on Modelling and vectors in mechanics for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 260 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-M1-1-B-01M1 · ASEasy6 marks
A window cleaner raises a bucket of water of mass 4.5 kg to a balcony. A rope is attached to the bucket, passes over a fixed pulley above the balcony, and the window cleaner pulls vertically downwards on the free end. Both parts of the rope are vertical. In a model, the bucket is a particle, the rope is light and inextensible, and the pulley is small and smooth. Take g = 9.8 m s−2.
(a)
State what is meant by modelling the bucket as a particle.
(1)
(b)
Explain what the assumption that the rope is light tells you about the tension in the rope.
(1)
(c)
Explain why the pulley is modelled as smooth.
(1)
(d)
The bucket rises at a constant speed. Find the size of the force with which the window cleaner pulls on the rope.
(2)
(e)
Suggest one way in which the model could be refined.
(1)
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Answer
Mark
Notes
Part (a)
the mass of the bucket is concentrated at a single point / its size (and any rotation or swinging) is ignored
B1
Part (b)
the mass of the rope is ignored, so the tension is the same at every point along each part of the rope
B1
Part (c)
there is no friction at the pulley, so the tension is the same on both sides of it
B1
Part (d)
constant velocity, so the resultant force on the bucket is zero: tension = weight
M1
4.5 × 9.8 = 44.1 N, and by the smooth pulley this is the pull on the free end
A1
Part (e)
any one: include the mass of the rope / include friction at the pulley / include air resistance / treat the bucket as an object that can swing
B1
Exam tip: In a “state what is meant by” modelling question, name the object from the question and say what is ignored — “the bucket's size is ignored” scores where a bare definition may not.
[In this question i and j are unit vectors due east and due north respectively.] A robotic lawnmower crosses a flat field in two straight stages. Its displacement in the first stage is a = (7i − 4j) m and in the second stage is b = (−2i + 9j) m.
(a)
Find the magnitude of a.
(2)
(b)
Find the angle between a and the unit vector i.
(2)
(c)
Find the total displacement of the lawnmower.
(1)
(d)
Write down the bearing of the lawnmower's final position from its starting point.
(1)
(e)
Find a unit vector in the direction of the total displacement.
(1)
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Answer
Mark
Notes
Part (a)
√(72 + (−4)2) = √65
M1
= 8.06 m
A1
Part (b)
tan−1(4 ÷ 7)
M1
= 29.7° (below i, i.e. south of east)
A1
Part (c)
a + b = (5i + 5j) m
B1
Part (d)
045°
B1
Part (e)
(5i + 5j) ÷ 5√2 = (1/√2)(i + j) or equivalent, e.g. (√2/2)i + (√2/2)j
B1
Common error: Using tan−1(7 ÷ 4) and giving 60.3°. That is the angle with j, not with i. A sketch shows which side of the triangle is opposite the angle you want.
[In this question i and j are unit vectors due east and due north respectively.] A hiker walks 3.0 km on a bearing of 040° and then 5.0 km on a bearing of 130°.
Figure: the two stages of the walk, with bearings measured clockwise from north; not to scale.
(a)
Write the displacement for the first part of the walk in the form (pi + qj) km, giving p and q to 3 significant figures.
(2)
(b)
Write the displacement for the second part of the walk in the same form.
(1)
(c)
Find the distance of the hiker from the starting point at the end of the walk.
(2)
(d)
Find the bearing of the hiker's final position from the starting point.
(2)
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Answer
Mark
Notes
Part (a)
3.0 sin 40° i + 3.0 cos 40° j
M1
= (1.93i + 2.30j) km
A1
Part (b)
5.0 sin 130° i + 5.0 cos 130° j = (3.83i − 3.21j) km
B1
Part (c)
resultant = 5.76i − 0.916j, then Pythagoras
M1
5.83 km (√34; the two stages are at right angles)
A1
Part (d)
angle below east = tan−1(0.916 ÷ 5.76) = 9.04°, or 40° + tan−1(5 ÷ 3)
M1
bearing = 099° (99.0°)
A1
Exam tip: With i east and j north, a bearing θ gives the components (sin θ, cos θ) — sine goes with east. The signs of sin 130° and cos 130° then look after the direction for you.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O on the bank of a pond.] A model boat is at the point with position vector (−2i + 5j) m when t = 0, and moves with constant velocity (1.5i − 0.8j) m s−1.
(a)
Find the speed of the boat.
(2)
(b)
Find the position vector of the boat at time t seconds.
(1)
(c)
Find the time at which the boat is due east of O, and its distance from O at that time.
(3)
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Answer
Mark
Notes
Part (a)
√(1.52 + 0.82) = √2.89
M1
= 1.7 m s−1
A1
Part (b)
r = ((−2 + 1.5t)i + (5 − 0.8t)j) m
B1
Part (c)
due east of O ⇒ j-component zero: 5 − 0.8t = 0
M1
t = 6.25 s
A1
i-component = −2 + 1.5 × 6.25 = 7.38 m (7.375, positive so east ✓)
A1
Exam tip: “Due east of O” means the j-component of the position vector is zero. Then check that the i-component is positive — otherwise the boat would be due west.
Figure 1 shows a child sitting on a sledge on horizontal snow. A parent pulls the sledge with a rope. The tension in the rope is 60 N and the rope makes the angle shown in Figure 1 with the horizontal. The child and sledge have a total mass of 18 kg and are modelled as a single particle; the rope is modelled as light and the snow as smooth.
(a)
State what is meant by modelling the child and sledge as a particle.
(1)
(b)
State what is meant by modelling the snow as smooth.
(1)
(c)
Find the horizontal and vertical components of the tension in the rope.
(2)
(d)
Find the acceleration of the sledge.
(2)
(e)
In practice the sledge accelerates more slowly than this. Give one reason, in terms of the model, why.
(1)
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Answer
Mark
Notes
Part (a)
all the mass is taken to act at a single point / the size and shape of the child and sledge are ignored
B1
Part (b)
there is no friction between the sledge and the snow
B1
Part (c)
horizontal: 60 cos 25° = 54.4 N
B1
vertical: 60 sin 25° = 25.4 N
B1
Part (d)
only the horizontal component accelerates the sledge: 54.38 = 18a
M1
a = 3.02 m s−2
A1
Part (e)
any one: the snow is not smooth, so friction acts / air resistance has been ignored
B1
Common error: Using the full 60 N in F = ma. Only the component along the direction of motion, 60 cos 25°, accelerates the sledge; the vertical component only changes the normal reaction.
Figure 1 shows two forces P and Q acting at a point O, drawn on a square grid in which one square represents 1 N. The unit vectors i and j are parallel to the gridlines, as shown.
(a)
Write P and Q in the form (ai + bj) N.
(2)
(b)
Find the resultant R of P and Q, and find |R| in exact form.
(2)
(c)
Find the angle between R and the unit vector j.
(2)
(d)
A force S has magnitude 10 N and acts in the same direction as Q. Find S, giving each component to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
P = (5i − 2j) N
B1
Q = (−i + 4j) N
B1
Part (b)
R = (4i + 2j) N
B1
|R| = √20 = 2√5 N (4.47 N)
B1
Part (c)
tan−1(4 ÷ 2)
M1
= 63.4°
A1
Part (d)
|Q| = √17, so S = (10 ÷ √17)(−i + 4j)
M1
S = (−2.43i + 9.70j) N
A1
Exam tip: The angle with j uses the i-component as the opposite side: tan θ = 4 ÷ 2. Mark the angle on a quick sketch of the vector before choosing which component goes on top.
[In this question i and j are unit vectors due east and due north respectively.] Figure 1 shows the route of a kayak. It leaves a harbour H and paddles in a straight line to a rock A, then in a straight line to a beach B. The distances and bearings of the two stages are shown in Figure 1.
(a)
Write the displacement from H to A in the form (pi + qj) km, giving p and q to 3 significant figures.
(2)
(b)
Write the displacement from A to B in the same form.
(1)
(c)
Find the distance HB.
(2)
(d)
The kayak returns directly from B to H. Find the bearing on which it must travel.
(2)
(e)
The kayak returns at a constant speed of 5.0 km h−1. Find the time taken, to the nearest minute.
(1)
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Answer
Mark
Notes
Part (a)
4.0 sin 60° i + 4.0 cos 60° j
M1
= (3.46i + 2.00j) km
A1
Part (b)
2.5 sin 165° i + 2.5 cos 165° j = (0.647i − 2.41j) km
B1
Part (c)
HB = (4.111i − 0.4148j) km, then Pythagoras
M1
HB = 4.13 km
A1
Part (d)
angle of HB below east = tan−1(0.4148 ÷ 4.111) = 5.76°, so B is on a bearing of 095.8° from H
M1
bearing of H from B = 095.8° + 180° = 276° (275.8°)
A1
Part (e)
4.132 ÷ 5.0 = 0.826 h = 50 minutes (49.6)
B1
Exam tip: A return bearing differs from the outward bearing by exactly 180°. Find the bearing of B from H first, then add or subtract 180° so the answer lies between 000° and 360°.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O.] A ball rolls across a smooth horizontal floor with constant velocity and is modelled as a particle. Figure 1 shows its position A at time t = 0 and its position B at time t = 4 s, on a grid of 1 m squares.
(a)
Write down the position vectors of A and B.
(1)
(b)
Find the velocity of the ball.
(2)
(c)
Find the speed of the ball.
(1)
(d)
Find the position vector of the ball at time t seconds.
(1)
(e)
Find the time at which the ball is due north of O, and its distance from O at that time.
(2)
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Answer
Mark
Notes
Part (a)
a = (−3i + j) m and b = (5i + 7j) m
B1
Part (b)
v = (b − a) ÷ 4 = (8i + 6j) ÷ 4
M1
= (2i + 1.5j) m s−1
A1
Part (c)
√(22 + 1.52) = 2.5 m s−1
B1
Part (d)
r = ((−3 + 2t)i + (1 + 1.5t)j) m
B1
Part (e)
i-component zero: −3 + 2t = 0 ⇒ t = 1.5 s
M1
j-component = 1 + 2.25 = 3.25 m (positive, so north ✓)
A1
Common error: Dividing the position vector of B by 4 to get the velocity. The velocity is the change in position divided by the time taken: subtract b − a first, then divide by 4 s.
A small ring rests on a smooth horizontal table. Three horizontal forces act on it: F1 = (7i − 2j) N, F2 = (−4i + 6j) N and F3. The ring is in equilibrium. The ring has mass 0.4 kg and is modelled as a particle.
(a)
Find F3.
(2)
(b)
Find the magnitude of F3 and the angle between F3 and i.
(3)
(c)
The force F2 is now removed. Find the magnitude of the acceleration of the ring.
(3)
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Answer
Mark
Notes
Part (a)
F1 + F2 + F3 = 0
M1
F3 = (−3i − 4j) N
A1
Part (b)
|F3| = 5 N
B1
tan−1(4 ÷ 3) = 53.1° below −i
M1
angle with i = 180° − 53.1° = 127° (126.9°)
A1
Part (c)
resultant = −F2 = 4i − 6j N
B1
a = (4i − 6j) ÷ 0.4 = 10i − 15j
M1
|a| = √325 = 18.0 m s−2
A1
Exam tip: When one force is removed from a body in equilibrium, the resultant of the others is equal and opposite to the removed force. That shortcut saves adding the remaining forces.
Two forces F and G act on a particle. F = (4i − 3j) N. G has magnitude 13 N and acts in the direction of the vector (5i + 12j).
(a)
Find G in the form (ai + bj) N.
(2)
(b)
Find the magnitude of the resultant R of F and G, giving your answer in exact form.
(2)
(c)
Write down the angle between R and i.
(1)
(d)
A third force H = (ki − 2kj) N, where k is a constant, is added. The resultant of F, G and H is parallel to i. Find k and the magnitude of H.
(3)
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Answer
Mark
Notes
Part (a)
|5i + 12j| = 13, so G = 13 × (5i + 12j) ÷ 13
M1
G = (5i + 12j) N
A1
Part (b)
R = 9i + 9j
M1
|R| = 9√2 N (12.7 N)
A1
Part (c)
45°
B1
Part (d)
j-component of the new resultant is zero: 9 − 2k = 0
M1
k = 4.5
A1
|H| = √(4.52 + 92) = 10.1 N
A1
Exam tip: A vector of given magnitude in a given direction is magnitude × unit vector. Divide the direction vector by its own magnitude first, then multiply by the size you need.
A model hovercraft of mass 0.5 kg moves on a smooth horizontal surface. It is modelled as a particle acted on by two constant horizontal forces, (2i + j) N and (−0.5i + j) N. At time t = 0 its velocity is (2i − 6j) m s−1.
(a)
Find the resultant force on the hovercraft and its magnitude.
(2)
(b)
Find the acceleration of the hovercraft.
(1)
(c)
Find the angle between the acceleration and i.
(1)
(d)
Find the speed of the hovercraft when t = 2 s.
(3)
(e)
Find the time at which the hovercraft is moving parallel to i.
(2)
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Answer
Mark
Notes
Part (a)
R = (1.5i + 2j) N
B1
|R| = √(2.25 + 4) = 2.5 N
B1
Part (b)
a = R ÷ 0.5 = (3i + 4j) m s−2
B1
Part (c)
tan−1(4 ÷ 3) = 53.1°
B1
Part (d)
v = u + at = (2i − 6j) + 2(3i + 4j)
M1
= 8i + 2j
A1
speed = √68 = 8.25 m s−1
A1
Part (e)
j-component of v is zero: −6 + 4t = 0
M1
t = 1.5 s
A1
Common error: Setting the i-component to zero for “parallel to i”. Moving parallel to i means there is no j-component of velocity.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are relative to a fixed point O.] Two small robots, A and B, move on a level floor, each with constant velocity. At time t = 0, A is at O and B is at (10i − 4j) m. The velocity of A is (2i + 3j) m s−1 and the velocity of B is (−0.5i + 4j) m s−1. The robots are modelled as particles.
(a)
Find the speed of each robot.
(2)
(b)
Write down the position vectors of A and B at time t seconds.
(2)
(c)
Show that the robots collide, and find the time and the position vector of the point where they collide.
(3)
(d)
Find the distance between the robots when t = 2 s.
(2)
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Answer
Mark
Notes
Part (a)
A: √13 = 3.61 m s−1
B1
B: √16.25 = 4.03 m s−1
B1
Part (b)
rA = 2ti + 3tj
B1
rB = (10 − 0.5t)i + (−4 + 4t)j
B1
Part (c)
i-components equal: 2t = 10 − 0.5t ⇒ t = 4
M1
j-components at t = 4: 3 × 4 = 12 and −4 + 16 = 12, equal, so they collide at t = 4 s
A1
at (8i + 12j) m
A1
Part (d)
rB − rA = (10 − 2.5t)i + (−4 + t)j = 5i − 2j at t = 2
M1
distance = √29 = 5.39 m
A1
Exam tip: To show a collision, find t from one component and then check that the other component agrees at that same t. Equal i-components alone only show that one is due north or south of the other.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are relative to a fixed point O.] A radio-controlled toy car of mass 1.2 kg moves across a horizontal car park with constant acceleration (i − 2j) m s−2. At time t = 0 it is at the point with position vector (−4i + 6j) m and has velocity (2i + j) m s−1. The toy car is modelled as a particle.
(a)
Find the magnitude of the resultant force on the toy car.
(2)
(b)
Find the position vector of the toy car at time t seconds.
(2)
(c)
Find the time at which the toy car is due east of O, and its distance from O at that time.
(3)
(d)
Find the speed of the toy car at this time.
(2)
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Answer
Mark
Notes
Part (a)
F = 1.2(i − 2j) = 1.2i − 2.4j
M1
|F| = 2.68 N
A1
Part (b)
r = r0 + ut + ½at2
M1
r = (−4 + 2t + 0.5t2)i + (6 + t − t2)j
A1
Part (c)
j-component zero: t2 − t − 6 = 0
M1
(t − 3)(t + 2) = 0, t ≥ 0 so t = 3 s
A1
i-component = −4 + 6 + 4.5 = 6.5 m (positive, so east ✓)
A1
Part (d)
v = (2i + j) + 3(i − 2j) = 5i − 5j
M1
speed = 5√2 = 7.07 m s−1
A1
Exam tip: A quadratic in t usually gives two roots. Reject a negative time, and say why, so that the choice is part of your working.
[In this question i and j are unit vectors due east and due north respectively.] A light aircraft flies with a velocity relative to the air of 180 km h−1 on a bearing of 070°. A wind blows from due north at 40 km h−1, so the velocity of the air is −40j km h−1. The velocity of the aircraft over the ground is the resultant of its velocity relative to the air and the velocity of the air. The aircraft is modelled as a particle and the wind velocity is assumed constant.
Figure: the aircraft's velocity relative to the air and the velocity of the air; not to scale.
(a)
Write the velocity of the aircraft relative to the air in the form (pi + qj) km h−1.
(2)
(b)
Find the velocity of the aircraft over the ground.
(1)
(c)
Find the speed of the aircraft over the ground and the bearing on which it travels.
(3)
(d)
Find the distance the aircraft travels over the ground in 1 hour 30 minutes.
(1)
(e)
Explain how the answer to part (d) would be affected if the wind speed increased during the flight.
the northward component of the ground velocity would be smaller, so the ground speed, and hence the distance in (d), would be less than the model predicts
B1
Exam tip: Write each velocity in components before adding. Adding the speeds 180 and 40 directly ignores the directions and gives nothing useful.
Figure 1 is a plan view of three horizontal forces acting on a puck at rest on smooth horizontal ice. A force of 14 N acts in the direction of −j. Forces of magnitude P N and Q N act at the angles shown, where i and j are perpendicular unit vectors in the horizontal plane. The puck has mass 2.5 kg, is modelled as a particle and is in equilibrium.
(a)
By resolving in the directions of i and j, write down two equations connecting P and Q.
(2)
(b)
Find P and Q.
(3)
(c)
The force of magnitude Q N is removed. Find the magnitude of the acceleration of the puck and the angle that the acceleration makes with i.
(3)
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Answer
Mark
Notes
Part (a)
i: P cos 35° = Q cos 25°
B1
j: P sin 35° + Q sin 25° = 14
B1
Part (b)
substituting Q = P cos 35° ÷ cos 25° into the second equation
M1
P = 14.7 (14.65)
A1
Q = 13.2 (13.24)
A1
Part (c)
the resultant of the remaining forces is equal and opposite to the removed force: magnitude 13.24 N
M1
a = 13.24 ÷ 2.5 = 5.30 m s−2
A1
direction opposite to that force: 25° below i (i.e. at −25° to i)
B1
Exam tip: Write each resolved equation with its direction label before solving. When one force is removed from a body in equilibrium, the resultant of the rest is the reverse of the removed force, so its direction comes straight from the figure.
[In this question i and j are unit vectors parallel and perpendicular to the river banks, as shown.] Figure 1 shows a straight river with parallel banks 40 m apart. The water flows with a constant velocity of 0.8i m s−1. A swimmer starts at S on the south bank and swims at 1.5 m s−1 relative to the water, heading upstream at the angle to the bank shown. The velocity of the swimmer over the river bed is the resultant of the swimmer's velocity relative to the water and the velocity of the water. The swimmer is modelled as a particle.
(a)
Write the swimmer's velocity relative to the water in the form (pi + qj) m s−1.
(2)
(b)
Find the resultant velocity of the swimmer and the speed of the swimmer over the river bed.
(2)
(c)
Find the time taken to cross the river and how far downstream from the point directly opposite S the swimmer lands.
(3)
(d)
Find the angle to the bank, upstream, at which the swimmer should head in order to land directly opposite S.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
−1.5 cos 70° i + 1.5 sin 70° j
M1
= (−0.513i + 1.41j) m s−1
A1
Part (b)
(0.287i + 1.41j) m s−1
B1
speed = √(0.2872 + 1.4102) = 1.44 m s−1
B1
Part (c)
only the j-component crosses the river: t = 40 ÷ 1.4095
M1
t = 28.4 s
A1
downstream distance = 0.2870 × 28.38 = 8.14 m
A1
Part (d)
i-component of the resultant zero: 1.5 cos θ = 0.8
M1
θ = 57.8°
A1
Exam tip: Split every velocity into components along and across the river. The time to cross depends only on the component across the river; the drift depends only on the component along it.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O.] A ship moves with constant velocity. Figure 1 shows the position of a fixed buoy and the positions of the ship at times t = 0, 1, 2, 3 and 4 hours, on a grid of 1 km squares. The ship and buoy are modelled as particles.
(a)
Write down the position vector of the buoy, and find the velocity of the ship.
(2)
(b)
Find the speed of the ship and the bearing on which it sails.
(2)
(c)
Show that the position vector of the ship relative to the buoy at time t hours is (3t − 4)i + (5t − 10)j.
(2)
(d)
Find the time at which the ship is due east of the buoy and the distance between them at that time.
(2)
(e)
Find the time at which the ship is directly north-east of the buoy, and the distance between them at that time.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
buoy: (4i + j) km
B1
ship moves (3i + 5j) km each hour, so v = (3i + 5j) km h−1
B1
Part (b)
speed = √34 = 5.83 km h−1
B1
bearing = tan−1(3 ÷ 5) = 031° (031.0°)
B1
Part (c)
rship = 3ti + (−9 + 5t)j
B1
subtracting (4i + j) gives (3t − 4)i + (5t − 10)j
B1
Part (d)
j-component zero: 5t − 10 = 0 ⇒ t = 2 h
M1
i-component = 2, positive, so due east at a distance of 2 km
A1
Part (e)
north-east ⇒ the two components are equal (and positive): 3t − 4 = 5t − 10
M1
t = 3 h, giving 5i + 5j (both positive ✓)
A1
distance = 5√2 = 7.07 km
A1
Common error: Setting one component to zero for “north-east”. North-east means the relative position is a positive multiple of (i + j), so the two components must be equal and both positive.
A puck of mass 0.25 kg slides on a horizontal air table under the action of a constant horizontal force, and is modelled as a particle. Figure 1 is a velocity diagram, with i and j perpendicular unit vectors in the plane of the table. It shows the velocity of the puck at time t = 0 and at time t = 5 s, drawn from O.
(a)
Write down the velocity of the puck at t = 0 and at t = 5 s.
(2)
(b)
Find the acceleration of the puck.
(2)
(c)
Find the magnitude of the force acting on the puck and the angle between the force and i.
(3)
(d)
Find the time at which the puck is moving in the direction of the vector (i + j).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
u = (−2i + 4j) m s−1
B1
v = (8i − j) m s−1
B1
Part (b)
a = (v − u) ÷ 5 = (10i − 5j) ÷ 5
M1
= (2i − j) m s−2
A1
Part (c)
F = 0.25(2i − j) = (0.5i − 0.25j) N
M1
|F| = 0.559 N
A1
tan−1(0.25 ÷ 0.5) = 26.6° below i
A1
Part (d)
u + at = (−2 + 2t)i + (4 − t)j; equal components: −2 + 2t = 4 − t
M1
t = 2 s (velocity 2i + 2j, both components positive ✓)
A1
Exam tip: “In the direction of i + j” needs equal components that are both positive. At the same time check the sign: equal negative components would mean the direction of −(i + j).
A bead of mass 0.2 kg is threaded on a fixed straight horizontal wire. The bead is pulled by a force of magnitude 1.5 N acting in the vertical plane containing the wire, at the angle below the horizontal shown in Figure 1. The wire is modelled as smooth. Take g = 9.8 m s−2.
(a)
Explain what is meant by modelling the bead as a particle threaded on the wire, and state what the assumption that the wire is smooth tells you about the force exerted on the bead by the wire.
(2)
(b)
Find the magnitude of the force exerted on the bead by the wire.
(2)
(c)
Find the acceleration of the bead.
(2)
(d)
The bead starts from rest. Find the distance it moves along the wire in the first 1.2 s.
(1)
(e)
State, with a reason, how the answer to part (c) would change if the wire were modelled as rough.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the bead is a particle that can move only along the wire
B1
no friction, so the force from the wire is perpendicular to the wire (here vertical)
B1
Part (b)
resolving vertically: R = 0.2 × 9.8 + 1.5 sin 30°
M1
R = 2.71 N
A1
Part (c)
horizontally: 1.5 cos 30° = 0.2a
M1
a = 6.50 m s−2
A1
Part (d)
s = ½ × 6.495 × 1.22 = 4.68 m
B1
Part (e)
it would be smaller (possibly zero), because friction would act along the wire against the motion
B1
Common error: Taking the reaction from the wire as just the weight, 1.96 N. The applied force pulls the bead downwards as well, so the wire must push up with the weight plus the downward component 1.5 sin 30°.
Figure 1 is a plan view of a barge in a straight canal. The barge is pulled by two horizontal ropes attached to the same point at its front. One rope goes to a horse on the north towpath and has tension 900 N; the other goes to a horse on the south towpath and has tension T N. The angles between the ropes and the direction of the canal are shown in Figure 1. The barge moves along the canal, parallel to the banks. The barge has mass 2400 kg and is modelled as a particle; the ropes are light and inextensible. The water exerts a constant resistance of 600 N opposing the motion.
(a)
Explain why the components of the two tensions perpendicular to the canal must cancel.
(1)
(b)
Find T.
(2)
(c)
Find the resultant pull of the two ropes along the canal.
(2)
(d)
Find the acceleration of the barge.
(2)
(e)
State one feature of the real barge that is ignored by modelling it as a particle.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the barge moves parallel to the banks, so it has no acceleration across the canal and the resultant across the canal is zero
B1
Part (b)
T sin 32° = 900 sin 20°
M1
T = 581 (580.9)
A1
Part (c)
900 cos 20° + 580.9 cos 32°
M1
= 1340 N (1338)
A1
Part (d)
1338 − 600 = 2400a
M1
a = 0.308 m s−2
A1
Part (e)
any one: its size / its turning or rotation (the ropes could turn it) / the ropes being attached at a point of a large body
B1
Exam tip: Resolve across the canal first: it gives the unknown tension directly. Then resolve along the canal for the driving force, and only then use F = ma with the resistance.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are relative to a fixed point O.] At 09:00 yacht A is at (−6i + 2j) km and yacht B is at (18i + 14j) km. A sails with constant velocity (8i + 6j) km h−1 and B sails with constant velocity (−4i + 10j) km h−1. At time t hours after 09:00 the distance between the yachts is d km.
(a)
Find the speed of A and the bearing on which it sails.
(3)
(b)
Find the position vectors of A and B at time t, and hence show that the position vector of B relative to A is (24 − 12t)i + (12 + 4t)j.
(3)
(c)
Find the time at which B is due north of A, and the distance between the yachts at that time.
(2)
(d)
Show that d2 = 160t2 − 480t + 720.
(2)
(e)
Hence find the least distance between the yachts and the time at which it occurs.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
speed = √(64 + 36) = 10 km h−1
B1
bearing = tan−1(8 ÷ 6)
M1
= 053° (053.1°)
A1
Part (b)
rA = (−6 + 8t)i + (2 + 6t)j
B1
rB = (18 − 4t)i + (14 + 10t)j
B1
rB − rA = (18 − 4t + 6 − 8t)i + (14 + 10t − 2 − 6t)j, which simplifies to the given result
B1
Part (c)
i-component zero: 24 − 12t = 0 ⇒ t = 2, i.e. at 11:00
M1
j-component = 12 + 8 = 20 km (positive, so north ✓)
Exam tip: Completing the square answers both questions at once: the bracket is zero at the time of closest approach, and the constant left over is the least value of d2. Remember to square-root it.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are relative to a fixed point O.] A particle P of mass 2 kg moves on a smooth horizontal plane under the action of two constant forces, F1 = (3i − 2j) N and F2. At time t = 0, P is at the point with position vector (i − 12j) m and has velocity (5i − 2j) m s−1. When t = 4 s, the velocity of P is (−3i + 14j) m s−1.
(a)
Find the acceleration of P.
(2)
(b)
Find F2 and its magnitude.
(3)
(c)
Find the time at which P is moving parallel to j.
(2)
(d)
Find the position vector of P at time t seconds.
(2)
(e)
Find the time at which P is due east of O, and its distance from O at that time.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
a = (v − u) ÷ t = (−8i + 16j) ÷ 4
M1
= (−2i + 4j) m s−2
A1
Part (b)
resultant = 2(−2i + 4j) = −4i + 8j
M1
F2 = (−4i + 8j) − (3i − 2j) = (−7i + 10j) N
A1
|F2| = √149 = 12.2 N
A1
Part (c)
i-component of v zero: 5 − 2t = 0
M1
t = 2.5 s
A1
Part (d)
r = (i − 12j) + (5i − 2j)t + ½(−2i + 4j)t2
M1
= (1 + 5t − t2)i + (−12 − 2t + 2t2)j
A1
Part (e)
j-component zero: 2t2 − 2t − 12 = 0, so t2 − t − 6 = 0
M1
t = 3 s (rejecting t = −2)
A1
i-component = 1 + 15 − 9 = 7 m (positive ✓)
A1
Exam tip: When an unknown force is asked for, find the resultant from F = ma first, then subtract the known forces. Writing “resultant = …” as a separate line keeps the method visible.
A crate of mass 6 kg rests on smooth horizontal ice. Three light horizontal ropes are attached to it. The tension in the first rope is 40 N, acting in the direction of i. The tension in the second rope is 30 N, acting at 120° to i, measured anticlockwise. The tension in the third rope has magnitude T N and acts at an angle θ° to i, measured anticlockwise, where 0 ≤ θ < 360. The crate is modelled as a particle and is in equilibrium.
Figure: the first two rope tensions on the crate; the third rope (T N at θ° to i) is not shown; not to scale.
(a)
Find T and θ.
(4)
(b)
The second rope breaks. The other two tensions do not change. Find the magnitude of the resultant force on the crate and the magnitude of its acceleration.
(3)
(c)
The crate starts from rest at the point O when the rope breaks. Find its distance from O after 2 s.
(2)
(d)
State what is meant by modelling the ropes as light, and explain how the answer to part (c) would change if the ice were modelled as rough.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
40i + 30 cos 120° i + 30 sin 120° j = 25i + 25.98j
M1
third tension = −25i − 25.98j
M1
T = √1300 = 36.1
A1
θ = 180 + tan−1(25.98 ÷ 25) = 226 (226.1)
A1
Part (b)
resultant = −(second tension) = 15i − 25.98j
M1
magnitude = 30 N
A1
a = 30 ÷ 6 = 5 m s−2
A1
Part (c)
r = ½at2, so distance = ½ × 5 × 22
M1
= 10 m
A1
Part (d)
light: the mass of each rope is ignored (so its tension is the same along its length)
B1
friction would oppose the motion, so the acceleration and the distance in 2 s would be smaller
B1
Common error: Quoting tan−1(25.98 ÷ 25) = 46.1° as the direction of the third tension. Both components are negative, so the force points into the third quadrant: add 180°.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are relative to a fixed point O.] Two drones, P and Q, fly at the same constant height and are modelled as particles moving in a horizontal plane. At time t = 0, P is at (−10i + 4j) m and Q is at (8i − 6j) m. P flies with constant velocity (6i + 2j) m s−1 and Q flies with constant velocity (−3i + 7j) m s−1.
(a)
Find the speed of P and the angle between its direction of motion and i.
(3)
(b)
Write down the position vectors of P and Q at time t seconds.
(2)
(c)
Show that, if both drones keep these velocities, they will collide. Find the time and the position vector of the point of collision.
(4)
(d)
To avoid the collision, Q is instead launched from the same point with the same velocity but 0.5 s later, so that it is at (8i − 6j) m when t = 0.5. P is unchanged. Find the distance between the drones when t = 2.
(3)
(e)
In this new situation, show that Q is never due east of P while both drones are flying.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
speed = √40 = 6.32 m s−1
B1
tan−1(2 ÷ 6)
M1
= 18.4°
A1
Part (b)
rP = (−10 + 6t)i + (4 + 2t)j
B1
rQ = (8 − 3t)i + (−6 + 7t)j
B1
Part (c)
rQ − rP = (18 − 9t)i + (−10 + 5t)j
M1
both components zero requires 18 − 9t = 0 and −10 + 5t = 0
M1
both give t = 2 s, so they collide
A1
at (2i + 8j) m
A1
Part (d)
rQ = (8 − 3(t − 0.5))i + (−6 + 7(t − 0.5))j for t ≥ 0.5
M1
at t = 2: rQ − rP = (3.5i + 4.5j) − (2i + 8j) = 1.5i − 3.5j
M1
distance = √14.5 = 3.81 m
A1
Part (e)
j-component of rQ − rP = −13.5 + 5t = 0 only when t = 2.7
M1
then the i-component = 19.5 − 9 × 2.7 = −4.8 < 0, so Q is due west of P, never due east
A1
Exam tip: When one object starts later, replace t by (t − delay) in its position vector only. Keep a single clock for both objects so that “at the same time” means the same t.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O.] Figure 1 shows a radar station R and the starting position S of a ship at 12:00, on a grid in kilometres. The ship sails with constant velocity (3i + 4j) km h−1 in the direction shown. At time t hours after 12:00 the distance between the ship and R is d km. The ship and the station are modelled as particles.
(a)
Find the speed of the ship and the bearing on which it sails.
(2)
(b)
Using Figure 1, find the position vector of the ship at time t, and show that the position vector of the ship relative to R is (3t − 18)i + (4t − 4)j.
(3)
(c)
Show that d2 = 25t2 − 140t + 340.
(2)
(d)
By completing the square, find the least distance between the ship and R, and the time at which it occurs.
(3)
(e)
The radar detects the ship whenever it is within 13 km of R, the dashed circle in Figure 1. Find the times at which the ship enters and leaves this region.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
speed = 5 km h−1
B1
bearing = tan−1(3 ÷ 4) = 037° (036.9°)
B1
Part (b)
S = (−12i + 6j) and R = (6i + 10j) from the figure
enters at 13:48 and leaves at 15:48 (detected for 1 hour)
A1
Exam tip: Once d2 is written as a(t − h)2 + k, a range condition such as d ≤ 13 becomes a simple inequality for (t − h)2. Convert decimal hours to clock times at the end: 0.8 h is 48 minutes.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O.] A particle P of mass 0.5 kg moves on a smooth horizontal plane under the action of a single constant force. At time t = 0, P is at the point A with position vector r0, and it has velocity u m s−1 and constant acceleration a m s−2. Figure 1 shows the path of P, with its positions at t = 0, 1, 2, 3 and 4 s marked on a grid of 1 m squares.
(a)
Using the positions at t = 0, 1 and 2 shown in Figure 1 and r = r0 + ut + ½at2, show that u = i + 5j and a = 2i − 2j.
(4)
(b)
Verify that the position at t = 4 shown in Figure 1 is consistent with these values.
(1)
(c)
Find the magnitude of the force acting on P.
(2)
(d)
Find the time at which P is moving parallel to i, and the position vector of P at that time.
v = (1 + 2t)i + (5 − 2t)j; j-component zero ⇒ t = 2.5 s
M1
r = (−6 + 2.5 + 6.25)i + (−3 + 12.5 − 6.25)j
M1
= (2.75i + 3.25j) m
A1
Part (e)
v = 9i − 3j
M1
speed = √90 = 9.49 m s−1
A1
Exam tip: Each marked point gives a vector equation, which is two scalar equations. Subtracting the equation for one time from the equation for another removes the unknowns step by step, exactly like simultaneous equations.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O.] A puck of mass 0.5 kg moves on smooth horizontal ice and is modelled as a particle. At time t = 0 it passes through O with velocity (4i − j) m s−1. For 0 ≤ t ≤ 4 a constant horizontal force (−0.5i + 0.5j) N acts on the puck; at t = 4 the force is removed. The puck then moves with constant velocity until it reaches a straight barrier. Figure 1 shows the path of the puck and the barrier on a grid of 1 m squares.
(a)
Find the acceleration of the puck for 0 ≤ t ≤ 4.
(1)
(b)
Find the position vector and the velocity of the puck at t = 4.
(4)
(c)
Find the time, with 0 < t ≤ 4, at which the puck is due east of O, and its distance from O at that time.
(3)
(d)
Using Figure 1, find the time at which the puck reaches the barrier and its position vector at that time.
(2)
(e)
State the speed of the puck when it reaches the barrier, and give one reason why the real puck would be moving more slowly.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
a = (−0.5i + 0.5j) ÷ 0.5 = (−i + j) m s−2
B1
Part (b)
r = (4i − j)t + ½(−i + j)t2
M1
at t = 4: r = (8i + 4j) m
A1
v = (4i − j) + 4(−i + j)
M1
= 3j m s−1
A1
Part (c)
j-component zero: −t + ½t2 = 0
M1
t = 2 s (rejecting t = 0)
A1
i-component = 8 − 2 = 6 m (positive ✓)
A1
Part (d)
for t ≥ 4: r = 8i + (4 + 3(t − 4))j; barrier at y = 10 ⇒ 4 + 3(t − 4) = 10
M1
t = 6 s, at (8i + 10j) m
A1
Part (e)
3 m s−1 (constant velocity after the force is removed)
B1
the ice is not perfectly smooth (friction) / air resistance acts
B1
Exam tip: When a force stops acting, the end of the first stage becomes the start of the next. Use the position and velocity at t = 4 as the new r0 and u, and measure time from 4 s.
[In this question i and j are unit vectors due east and due north respectively, and position vectors are given relative to a fixed point O.] Figure 1 shows the position S of a cargo ship at 12:00 and the position of a patrol base B, on a grid in kilometres. The ship sails with constant velocity (6i − 4j) km h−1 in the direction shown. A patrol boat leaves B at 12:30 and moves with constant velocity so that it meets the ship at 14:00. Both vessels are modelled as particles. Let t be the time in hours after 12:00.
(a)
Find the speed of the ship and the bearing on which it sails.
(3)
(b)
Using Figure 1, find the position vector of the point where the two vessels meet.
(2)
(c)
Find the velocity of the patrol boat, its speed and the bearing on which it travels.
(4)
(d)
Find the distance between the two vessels at 13:00.
(3)
(e)
Give one reason why modelling the vessels as particles is reasonable here.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
speed = √52 = 7.21 km h−1
B1
bearing = 90° + tan−1(4 ÷ 6)
M1
= 124° (123.7°)
A1
Part (b)
rS = (−8 + 6t)i + (20 − 4t)j, with t = 2
M1
= (4i + 12j) km
A1
Part (c)
displacement from B = (4i + 12j) − (16i + 3j) = −12i + 9j in 1.5 h
the sizes of the vessels are very small compared with the distances involved (km)
B1
Common error: Using the full 2 hours for the patrol boat. It leaves at 12:30, so it covers its displacement in 1.5 h, and its position at time t is B + (t − 0.5)v for t ≥ 0.5.
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