Problem Sets: Solutions · PDF file04.02.2010 · LC Applied Maths Problem Set Solutions 2 1.1.2 Problem Show that the following simpliﬁes to a constant when x ̸= 2 3x−5 x−2

Problem Sets: Solutions

J.P. McCarthy

February 4, 2010

1 Introduction to Applied Mathematics

1.1 Algebra

1.1.1 Problem

Solve the simultaneous equations

x− y = 0,

(x+ 2)2 + y2 = 10.

Solution: Let

x− y = 0 (A)

(x+ 2)2 + y2 = 10 (B)

(A) ⇒ x = y.

(B) : (x+ 2)2 + x2 = 10

⇒ x2 + 4x+ 4 + x2 − 6 = 0

⇒ 2x2 + 4x− 6 = 0

⇒ x2 + 2x− 3 = 0

⇒ (x+ 3)(x− 1) = 0

⇒ x = −3, or 1

⇒ Sol. Set = {(1, 1), (−3,−3)}.

1

LC Applied Maths Problem Set Solutions 2

1.1.2 Problem

Show that the following simplifies to a constant when x ̸= 2

3x− 5

x− 2+

1

2− x

Solution:

3x− 5

x− 2+

1

2− x,

=3x− 5

x− 2− 1

x− 2,

=3x− 5− 1

x− 2=

3x− 6

x− 2= 3

(x− 2)

x− 2= 3.

1.1.3 Problem

Show that−1 +

√3

1 +√3

= 2−√3

Solution:

−1 +√3

1 +√3

× 1−√3

1−√3

=−1 +

√3 +

√3− 3

1− 3

=2√3− 4

−2= 2−

√3.

1.1.4 Problem

x2 − px+ q is a factor of x3 + 3px2 + 3qx+ r.

(i) Show that q = −2p2.

(ii) Show that r = −8p3.

(iii) Find the three roots of x3 + 3px2 + 3qx+ r = 0 in terms of p.

Solution: A cubic function f(x) = x3+ bx2+ cx+d has factors (x−α)(x−β)(x−γ) whereα, β and γ are the roots of f(x). Similarly a quadratic function g(x) = x2 − px + q hasfactors g(x) = (x − α1)(x − α2) where α1 and α2 are the roots of g(x). If a quadratic g(x)is a factor of a cubic f(x) then the roots of g are roots of f and f(x) = g(x).(x− α3) whereα3 is the third root of f(x). Hence let f(x) = x3 + 3px2 + 3qx+ r and g(x) = x2 − px+ q.

f(x) = (x2 − px+ q)(x− α3)

⇒ f(x) = x3 − px2 + qx− α3x2 + α3px− α3q

⇒ f(x) = x3 + (−α3 − p)x2 + (q + α3p) + (−α3q)


(i) Equating the x2-coefficients:

3p = −α3 − p

⇒ α3 = −4p.

Equating the x-coefficients:

3q = q + (−4p)p

⇒ 2q = −4p2

⇒ q = −2p2.

(ii) Equating the constant coefficient:

r = −α3q

⇒ r = −(−4p)(−2p2)

⇒ r = −8p3.

(iii) α3 = −4p is one root. The other roots of f(x) = g(x)(x− α3) are the roots of g(x);

g(x) = x2 − px− 2p2!= 0

⇒ x2 − 2px+ px− 2p2 = 0

⇒ x(x− 2p) + p(x− 2p) = 0

⇒ (x− 2p)(x+ p)

Hence the set of roots is

{x : f(x) = 0} = {−4p, 2p,−p}.

1.2 Vectors

1.2.1 Problem

s = 4i+ 3j and t = 2i− 5j.Find |st|Solution:

st = t− s

⇒ st = (2i− 5j)− (4i+ 3j) = −2i− 8j.

Where v = xi+ yj;|v| =

√x2 + y2. (1)

⇒ |st| =√

(−2)2 + (−8)2 =√68 =

√4(17) = 2

√17.


1.2.2 Problem

a = 2i+ (2k + 3)j and b = k2i+ 6j, where k ∈ Z.a is perpendicular to b.

(i) Find the value of k.

(ii) Using your value for k, write a+ b in terms of i and j.

(iii) Hence, find the measure of the angle between a and a+b correct to the nearest degree.

Solution:

(i)a ⊥ b ⇔ a · b = 0. (2)

⇒a⊥b

a · b = 0

⇒ 2k2 + 6(2k + 3)!= 0

⇒ 2k2 + 12k + 18 = 0

⇒ k2 + 6k + 9 = 0

⇒ (k + 3)2 = 0

⇒ k = −3

(ii)

⇒k=−3

a = 2i− 3j

b = 9i+ 6j

⇒ a+ b = 11i+ 3j

(iii) By the properties of the Dot Product:

a · (a+ b) = a · a︸︷︷︸=|a|2

+ a · b︸︷︷︸=0

Also, where θ is the angle between a and b.

a · (a+ b) = |a||a+ b| cos θ⇒ |a|2 = |a||a+ b| cos θ

⇒ cos θ =|a|2

|a||a+ b|=

|a||a+ b|

⇒ cos θ =

√4 + 9√121 + 9

=

√13√130

=1√10

⇒ θ = cos−1(1/√10) = 71.5651◦ ≈ 72◦.


1.2.3 Problem

rst is a triangle where r = −i+ 2j, s = −4i− 2j and t = 3i− j.

(i) Express rs, st and tr in terms of i and j.

(ii) Show that the triangle rst is right-angled at r

(iii) Find the measure of ∠rst.

Solution:

(i)

rs = s− r

= (−4i− 2j) + i− 2j

−3i− 4j.

st = ts

= 3i− j+ 4i+ 2j

= 7i+ j

tr = r− t

= −i+ 2j− 3i+ j

= −4i+ 3j

(ii) For ∆rst to be right-angled at r: rs ⊥ tr:

rs ⊥ tr ⇔ rs · tr = 0,

rs · tr = (−3i− 4j) · (−4i+ 3j) = 12− 12 = 0,

⇒ ∆rst right-angled at r

(iii) Where θ := ∠rst;

sr · st = |sr||st| cos θ

⇒ cos θ =sr · st|sr||st|

⇒sr=−rs

cos θ =21 + 4

5√50

⇒ cos θ =25

5√50

=5√50

=

√25√50

=

√1

2=

1√2

⇒ θ = 45◦.


1.3 Coordinate Geometry

1.3.1 Problem

The line B contains the points (6,−2) and (−4, 10).The line A with equation ax+ 6y + 21 = 0 is perpendicular to B.Find the value of the real number a.Solution: If (x1, y1) and (x2, y2) are two points on a line L, then the slope is given by:

mL =y2 − y1x2 − x1

. (3)

⇒ mB =10 + 2

−4− 6= −6

5.

For two lines L and K;L ⊥ K ⇔ mL.mK = −1. (4)

When a line is written in the form:

L ≡ y = bx+ c, (5)

then b = mL. Hence

B ≡ ax+ 6y + 21 = 0

⇒ B ≡ y = −a

6x− 21

6.

A ⊥ B

⇒ −6

5

(−a

6

)= −1

⇒ a = −5.

1.3.2 Problem

The equation of the line L is 14x+ 6y + 1 = 0.Find the equation of the line perpendicular to L that contains the point (3,−2).Solution:

L ≡ 14x+ 6y + 1 = 0

⇒ L ≡ y = −14

6x− 1

6

Hence mL = −7/3. Let K ⊥ L and (3,−2) ∈ K. K ⊥ L ⇒ mK = 3/7. The equation of aline A containing a point (x1, y1) of slope m is given by:

y − y1 = m(x− x1). (6)

⇒ K ≡ y + 2 =3

7(x− 3)

⇒ K ≡ y =3

7x− 9

7− 2 =

3

7− 9

7− 14

7

⇒ K ≡ y =3

7x− 23

7.


1.3.3 Problem

Show that the line 6x − 8y − 71 = 0 contains the midpoint of [ab] where a has coordinates(8,−6) and b has coordinates (5,−2).Solution: The mid-point of [pq] where p = (x1, y1), q = (x2, y2) is given by:(

x1 + x2

2,y1 + y2

2

). (7)

Hence the mid-point of [ab]: (13

2,−4

).

⇒ 6

(13

2

)− 8(−4)− 71 = 39 + 32− 71 = 0,

⇒(13

2,−4

)∈ L ≡ 6x− 8y − 71 = 0.

1.3.4 Problem

Find the equation of the line pq where p has coordinates (7,−6) and q has coordinates (−3, 2).Find the point of intersection of pq and the line 2x− 3y + 1 = 0.Determine the ratio in which the line 2x− 3y + 1 = 0 divides [pq].Solution:

mpq =2 + 6

−3− 7= −4

5.

⇒ pq ≡ y − 2 = −4

5(x+ 3)

⇒ pq ≡ 5y − 10 = −4x− 12

⇒ pq ≡ 4x+ 5y = −2.

To find the intersection between this line and the line 2x− 3y + 1 = 0 is the solution ofthe simultaneous equations:

4x+ 5y − 2 (A)

2x− 3y = −1 (B)

⇒(B)

2x = 3y − 1

⇒(A)

6y − 2 + 5y = −2

⇒ 11y = 0 ⇒ y = 0

⇒ 2x = −1 ⇒ x = −1

2

point of intersection =

(−1

2, 0

).

Suppose 2x−3y+1 = 0 divides [pq] in the ratio m : n at (−1/2, 0). Suppose a = (x1, y1),b = (x2, y2). Then the point that divides [a, b] in the ratio s : t is given by:(

mx2 + nx1

m+ n,my2 + ny1m+ n

). (8)


Thence (−1

2, 0

)!=

(m(−3) + n(y)

m+ n,m(2)− 6n

m+ n

)⇒ 2m− 6n

!= 0

⇒ m = 3n

⇒ m

n= m : n =

3n

n= 3 : 1.

1.4 Trigonometry

1.4.1 Problem

Find the value of θ for which

cos θ = −√3

2, 0◦ ≤ θ ≤ 180◦.

Solution: In the first instance cos 30◦ =√3/2. cos is negative in the second quadrant:

cos(180◦ − θ) = cos(180◦)︸︷︷︸=−1

cos(θ) + sin(180◦)︸︷︷︸=0

cos θ

⇒ cos(180◦ − θ) = − cos θ

⇒ θ = 180◦ − 30◦ = 150◦.

1.4.2 Problem

If tanA = 1/2, find tan 2A without evaluating A, where A is an acute angle.Express tanB in the form a/b, where a, b ∈ N, given that

tan(2A+B) =63

16.

Solution: The double-angle formula for tan:

tan 2A =2 tanA

1− tan2A. (9)

tan 2A =2

1− 14

=134

=4

3.

The addition formula for tan:

tan(A+B) =tanA+ tanB

1− tanA tanB(10)


⇒ tan(2A+B) =tan 2A+ tanB

1− tan 2A tanB!=

63

16

⇒ 63

16=

43+ tanB

1− 43tanB

⇒ 16

(4

3+ tanB

)= 63

(1− 4

3tanB

)⇒ 3(16)

(4

3+ tanB

)= 3(63)

(1− 4

3tanB

)⇒ 64 + 48 tanB = 189− 252 tanB

⇒ 300 tanB = 125

⇒ tanB =125

300=

5

12.

1.4.3 Problem

Express sin(135◦ − A) in terms of sinA and cosA.Express sin(135◦ − A) cos(135◦ + A) in the form k(1 + sin pA), where k, p ∈ R.Find the values of A for which

sin(135◦ − A) cos(135◦ + A) = −3

4

where 0◦ ≤ A ≤ 180◦.Solution: The subtraction formula for sin:

sin(A−B) = sinA cosB − cosA sinB (11)

sin(135◦ − A) = sin 135◦ cosA− cos 135◦ sinA

⇒ sin(135◦ − A) =1√2cosA+

1√2sinA =

1√2(cosA+ sinA)

The addition formula for cos:

cos(A+B) = cosA cosB − sinA sinB (12)

cos(135◦ + A) = cos 135◦ cosA− sin 135◦ sinA

⇒ cos(135◦ + A) = − 1√2cosA− 1√

2sinA = − 1√

2(sinA+ cosA)

⇒ sin(135◦ − A) cos(135◦ + A) = −1

2(cosA+ sinA)2

⇒ sin(135◦ − A) cos(135◦ + A) = −1

2(cos2A+ sin2A︸︷︷︸

=1

+2 sinA cosA︸︷︷︸=sin 2A

)2

⇒ sin(135◦ − A) cos(135◦ + A) = −1

2(1 + sin 2A)


sin(135◦ − A) cos(135◦ + A) = −3

4= −1

2(1 + sin 2A)

⇒ 1

2(1 + sin 2A) =

3

4

⇒ 1 + sin 2A =3

2

⇒ sin 2A =1

2⇒ 2A = 30◦

⇒ A = 15◦

2 Accelerated Linear Motion

2.1 Problem

A lift decelerates from 3 m s−1 to rest during the last 6 m of its motion. Find the decelerationand the time taken.

2.1.1 Solution

For this motion

s = 6

t =?

u = 3

v = 0

a =?

Using v2 = u2 + 2as;

⇒ a =v2 − u2

2s

⇒ a =0− 32

12

⇒ a = −3

4m/s2

Using

t =v − u

a

⇒ t =0− 3

−3/4=

4

4× 3

3/4= 4 s

Ans: Deceleration = 3/4 m/s2 and time taken = 4 s.


2.2 Problem

A train slows down from 70 m s−1 to 50 m s−1 over an eight-second time interval. Findthe deceleration and the distance covered. If the train continues to decelerate at the sameuniform rate, how much further will it travel before it comes to rest?

2.2.1 Solution

In the eight-second interval:

s =?

t = 8

u = 70

v = 50

a =?

Using

a =v − u

t

⇒ a =50− 70

8=

−20

8= −5

2m/s2

Using

s =

(u+ v

2

)t

⇒ s =

(70 + 50

2

)8

⇒ s = 60× 8 = 480 m

Examining now the motion as the train decelerates from 70 m/s to rest:

s =?

t =?

u = 70

v = 0

a = −5

2

Hence using:

v2 = u2 + 2as

⇒ s =v2 − u2

2a

⇒ s =0− 4900

2(−5/2)=

4900

5= 980 m


However

∴ (distance travelled from 50 m/s to rest) =

(distance travelled from 70 m/s to rest)− (distance travelled from 70 m/s to 50 m/s)

⇒ (distance travelled from 50 m/s to rest) = 980− 480 = 500 m.

Ans: 500 m.

2.3 Problem

(a) Convert 72 km/hour to metres per second

(b) A train decelerates from 72 km/hour to 48 km/hour over a distance of 1/2 km. Find inmetres/second2 the deceleration, and the time taken. If the train continues to decelerateat this rate find out how much further it will travel before it comes to rest.

2.3.1 Solution

(a)

72km

hr

⇒ 72km

hr= 72

1000 m

60mins= 72

1000 m

60(60 s)

⇒ 72km

hr=

72(1000)

3600= 20 m/s

(b) In the first instance units must be converted to SI units:

72 km/hr = 20 m/s

48 km/hr =2

372 km/hr =

40

3m/s

1

2km = 500 m

Hence over the 500 m:

s = 500

t =?

u = 20

v =40

3a =?


Hence using v2 = u2 + 2as;

a =v2 − u2

2s

⇒ a =(402/9)− 400

1000

⇒ a =9

9× (402/9)− 400

1000=

402 − 3600

9000=

−2000

9000

⇒ a = −2

9m/s2

Using

t =v − u

a

⇒ t =(40/3)− 20

−(2/9)=

9

9× (40/3)− 20

−(2/9)=

120− 180

−2= 30 s

Examining now the motion as the train decelerates from 72 km/hr to rest:

s =?

t =?

u = 20

v = 0

a = −2

9


s =v2 − u2

2a

⇒ s =0− 400

−(4/9)=

9

9× 400

(4/9)=

3600

4= 900 m

However

∴ (distance travelled from 48 km/hr to rest)

= (distance travelled from 72 km/hr to rest)− (distance travelled from 72 km/hr to 48 km/hr)

⇒ (distance travelled from 48 km/hr to rest) = 900− 500 = 400 m.

Ans: 500 m.


2.4 Problem: LC OL 1984

Define velocity and speed.Show that a speed of 1 km/hour is equivalent to 5/18 m/s. The speed of a car is reducedfrom 72 km/hour to 54 km/hour over a distance of 35 m. Find the retardation, assumingit is uniform throughout. If the retardation continues, how much farther will the car travelbefore coming to rest?

2.4.1 Solution

Velocity is speed in a given directionSpeed is the rate of change of distance with respect to time

1 km/hr = 11000 m

60 mins= 1

1000 m

60(60) s

⇒ 1 km/hr =1000

3600m/s =

5

18m/s.

First convert to SI units.

72 km/hr = 72

(5

18m/s

)= 20 m/s

⇒ 54 km/hr =3

4(72 km/hr) = 15 m/s

Examining the motion over the 35 m:

s = 35

t =?

u = 20

v = 15

a =?


a =v2 − u2

2s=

225− 400

70=

−175

70= −5

2m/s2.

Examining the motion from 15 m/s to rest:

s =?

t =?

u = 15

v = 0

a = −5/2


s =v2 − u2

2a

⇒ s =0− 225

−5=

225

5= 45 m.



A car starts from rest with a uniform acceleration and reaches a velocity of 27 m/s in 9 s.The brakes are then applied and it comes to rest with uniform deceleration after travelling afurther 54 m. Calculate:

(i) the uniform acceleration

(ii) the uniform deceleration

(iii) the average speed of the car for the journey

(iv) the two times that the velocity of the car will be 15 m/s

2.5.1 Solution

(i) Using

a =v − u

t

⇒ a =27− 0

9= 3 m/s2

(ii) In the decelerating part of the motion, u = 27, v = 0. Using

v2 = u2 + 2as

⇒ a =v2 − u2

2s

⇒272=729

a =0− 729

108= −6.75 m/s2

(iii) The average speed is given by:

v̄ =distance travelled

time taken(13)

The motion is acceleration followed immediately by deceleration hence

d : a = t1 : t2 (14)

where t1 is the time accelerating, t2 the time decelerating.

⇒ d

a=

t1t2

⇒ t2 = t1a

d= 9

3

6.75= 4 s

Therefore the time taken is 9+4=13 s. The distance travelled is the area under thegraph. The graph is a triangle of height 27 and base 13:

total distance =1

2(13)(27) = 175.5 m

⇒ v̄ =175.5

13= 13.5 m/s


(iv) Clearly the velocity will be 15 m/s once when t < 9 and once when 9 < t < 13. Tofind the first time, using

t =v − u

a=

15

3= 5 s

To find the second time, consider the decelerating part of the motion. u = 27, v = 15,a = −6.75:

t =v − u

a=

15− 27

6.75= 1

7

9s

But this represents the time after t = 9 s. Hence:Ans: 5 s and 107

9s.


Consider three points on a line, p, q and r, along the line in that order. A car travellingtowards p at a steady speed of 15 m/s, accelerated at a constant rate between p and q. At qits speed was 25 m/s. This speed was maintained as far as r.If |pr| = 980 m and the time from p to q was 40 seconds, draw a time-velocity graph of themotion and hence, or otherwise, calculate the acceleration.

2.6.1 Solution

To draw the time-velocity graph note it has the rough shape of Fig 1.By the theorem, the area under the graph must equal to the distance travelled. Let

acceleration from 15 m/s to 25 m/s take T s. Using, for the time t = 0 to t = T

s =

(u+ v

2

)t =

(15 + 25

2

)T = 20T

∴ 980 = 20T + 25(40− T )

⇒ 980 = 20T + 1000− 25T

⇒ 5T = 20

⇒ T = 4 s

Therefore the time-velocity graph is Fig 2.Using

a =v − u

t

⇒ a =25− 15

4=

5

2m/s2.



Define uniform acceleration in a straight line. A particle starts from rest with uniformacceleration 2 m/s2. After how many seconds will its speed be 30 km/hr?How far from its starting point will the particle be when its speed is 60 km/hr? The particleis then brought to rest in 2 m. Calculate the deceleration.

2.7.1 Solution

Uniform acceleration in a straight line is motion in a single direction with constant acceler-ation; acceleration is the rate of change of acceleration with respect to time.

30km

hr= 30

1000

3600m/s =

25

3m/s

Using

t =v − u

a=

25/3

2=

25

6s

60 km/hr =50

3m/s

Using

v2 = u2 + 2as

⇒ s =v2 − u2

2a

⇒ s =(50/3)2

4=

2500/9

4=

625

9

⇒ s =625

9m

Again using

v2 = u2 + 2as

⇒ a =v2 − u2

2s

⇒s=2,v=0

a =−2500/9

4= −625

9m/s2

⇒ d =625

9m/s2



p and q are points 162 m apart. A body leaves p with initial speed 5 m/s and travels towardq with uniform acceleration 3 m/s2. At the same instant another body leaves q and travelstowards p with initial speed 7 m/s and uniform acceleration 2 m/s2. After how many secondsdo they meet and what, then, is the speed of each body?

2.8.1 Solution

The particles meet after T s when the sum of the distance travelled by the particle atp-particle and the distance travelled by the q-particle is 162 m:

particles meet ⇔ sp + sq = 162 (15)

Using

s = ut+1

2at2

⇒ sp = 5T +3

2T 2

⇒122=1

sq = 7T + T 2

Hence solve for1 T

sp + sq = 162

⇒ 12T +5

2T 2 = 162

⇒ 24T + 5T 2 = 324

⇒ 5T 2 + 24T − 324 = 0

⇒ 5T 2 + 54T − 30T − 324 = 0

⇒ T (5T + 54)− 6(5T + 54) = 0

⇒ (T − 6) (5T + 54)︸︷︷︸see footnote

= 0

⇒ T = 6 s

The velocities of the p and q-particles after t s, using:

v = u+ at

⇒ vp = 5 + 3t = 23 m/s at t = 6

⇒ vq = 7 + 2t = 19 m/s at t = 6

1ignore t < 0; refers to time when distance between them was 324 m


2.9 Problem: LC HL 2009

• A particle is projected vertically upwards from a point p. At the same instant a secondparticle is let fall vertically from a point q directly above point p. The particles meet ata point r between them after 2sThe particles have equal speeds when they meet at rProve that |pr| = 3|rq|

• A train accelerates uniformly from rest to a speed v m/s with uniform acceleration fm/s2.It then declerates uniformly to rest with uniform retardation 2f m/s2.The total distance travelled is d metres.

– Draw a speed-time graph for the motion of the train

– If the average speed for the whole journey is√

d/3, find the value of f .

2.9.1 Solution

• Let v1 be the speed of the particle projected from p and v2 the speed of the particledropped from q. After 2 s, using

v = u+ at

v1 = u− 2g

v2 = 2g

⇒ u− 2g = 2g

⇒ u = 4g

Let s1 and s2 be the distance travelled by the particles. After 2s, using

s = ut+1

2at2

s1 = 4g(2) +1

2(−g)(4)

⇒ s1 = 6g , and

s2 = (0)(2) +1

2g(4) = 2g

But after 2 s, s1 = |pr| and s2 = |rq|. Hence as s1 = 3s2; |pr| = 3|rq|. �

• – See Fig 1.

– The average speed is given by:

v̄ =total distance

total time=

d

T(16)

where T = t1 + t2 is the total time. Hence

d

T=

√d

3

⇒ T = d

√3

d=

√3d (17)


Also the total distance is equal to the area under the time-velocity curve, in thiscase the triangle of width T and height v:

d =1

2vT (18)

Since the motion is acceleration from rest immediately followed by declaration torest:

d : a = t1 : t2 (19)

Hence

2f : f = t1 : t2

⇒ t1 : t2 = 2 : 1

⇒ t1 : t2 =2

3:1

3

⇒ t1 =2

3T

Using

v = u+ at

v = 0 + ft1

⇒ v =2

3fT

Hence using this and (17) in (18):

d =1

2vT

⇒ d =1

2

2

3f√3d.

√3d

⇒ d =1

3f3d = fd

⇒ f = 1 m/s2


• A ball is thrown vertically upwards with an initial velocity of 39.2 m/s.Find

– the time taken to reach the maximum height

– the distance travelled in 5 s


• Two particles P and Q, each having constant acceleration, are moving in the samedirection along parallel lines. When P passes Q the speeds are 23 m/s and 5.5 m/s,respectively. Two minutes later Q passes P , and Q is then moving at 65.5 m/s.Find

– the acceleration of P and the acceleration of Q

– the speed of P when Q overtakes it

– the distance P is ahead of Q when they are moving with equal speeds

2.10.1 Solution

• – The ball reaches the maximum when v = 0. Using

v = u+ at

v = 39.2− gt

⇒ ts=max =39.2

g= 4 s

– After 5 s, using

s = ut+1

2at2

⇒ s = 39.2(5)− 1

2g(25)

⇒ s = 4g(5)− 25

2g = 20g − 12.5g =

15

2g

• – Consider the motion of Q.

t = 120 (120 s = 2 min)

vQ = 65.5

uQ = 5.5

aQ =?

sQ =?

Using

a =v − u

t

aQ =65.5− 5.5

120=

1

2m/s2


After 120 s, sP!= sQ. Using

s = ut+1

2at2

sP = 23(120) +1

2aP (120)

2

sQ = 5.5(120) +1

4(120)2

⇒ sP = sQ

⇒ 1

2aP (120)

2 = 5.5(120)− 23(120) +1

4(120)2

⇒ 240aP = 22− 92 + 120 = 50

⇒ aP =5

25m/s2

– Using

v = ua+ at

⇒ vP = 23 +5

24(120) = 48 m/s

– After t s, using

v = u+ at

vP = 23 +5

24t

vQ = 5.5 +1

2t

For what t is vP = vQ?

23 +5

24t = 5.5 +

1

2t

⇒ 7

24t = 17.5

⇒ t =(24)(17.5)

7= 60 s

Hence look at sP and sQ after 60 s, using

s = ut+1

2at2

sP = 23(60) +1

2

5

24(602) = 1755

sQ = 5.5(60) +1

4(60)2 = 1230

Hence after 60 s, sP > sQ by 525 m. That is P is 525 m ahead of Q when theyare moving with equal speed.



• A particle is projected vertically downwards from the top of a tower with speed u m/s.It takes the particle 4 s to reach the bottom of the tower.During the third second of its motion the particle travels 29.9 m.Find

– the value of u

– the height of the tower

• A train accelerates uniformly from rest with a speed v m/s.It continues at this speed for a period of time and then decelerates uniformly to rest.In travelling a total distance d metres the train accelerates through a distance pd metresand decelerates through a distance qd metres, where p < 1 and q < 1.

– Draw a speed-time graph for the motion of the train

– If the average speed of the train for the whole journey is

v

p+ q + b,

find the value of b.

2.11.1 Solution

• – The particle moves under acceleration a = g. After 2 s and 3 s, using

v = u+ at

v(2) = u+ 2g

v(3) = u+ 3g

From t = 2 to t = 3, the particle travels 29.9 m where u = v(2) and v = v(3);using

s =

(u+ v

2

)t

⇒ 29.9 =

(2u+ 5g

2

)⇒ 59.8 = 2u+ 5g

⇒ u =59.8− 5g

2= 5.4 m/s

– The height of the tower is s after 4 s. Using

s = ut+1

2at2

⇒ s = (5.4)4 +1

2g(16)

⇒ s = 21.6 + 8g = 100 m


• – See figure 2

– Let T be the total time taken for the journey. Where d is the total distancetravelled, the average speed v̄ is given by:

v̄ =d

T(20)

v

p+ q + b=

d

T

⇒ d = T

(v

p+ q + b

)⇒ pd+ qd+ bd = Tv

⇒ bd = Tv − pd− qd

⇒ b =Tv − pd− qd

d(21)

Let t1 be the time spent accelerating and t2 the time spent decelerating. Exam-ining the time-velocity graph, the distance travelled in t1, the area of the trianglewith perpendicular height v and base t1 is:

1

2t1v = pd

⇒ t1 =2pd

v

Similarly

t2 =2qd

v

Let tc be the time spent at constant speed. From the time-velocity graph, thedistance travelled at constant speed is given by,

d− pd− qd = d(1− p− q)

Hence the distance travelled in tc is the area under the curve:

vtc = d(1− p− q)

⇒ tc =d(1− p− q)

v

Now T = t1 + t2 + tc;

T =1

v(2pd+ 2qd+ d− pd− qd)

⇒ T =1

v(pd+ qd+ d) =

d

v(p+ q + 1)

Substituting into (21):

b =d(p+ q + 1)− pd− qd

d= p+ q + 1− p− q = 1.



• A lift starts from rest. For the first part of its descent is travels with uniform accel-eration f . It then travels with uniform retardation 3f and comes to rest. The totaldistance travelled is d and the total time taken is t.

– Draw a speed-time graph for the motion

– Find d in terms of f and t

• Two trains P and Q, each of length 79.5 m, moving in opposite directions along parallellines, meet at o, when their speeds are 15 m/s and 10 m/s respectively.The acceleration of P is 0.3 m/s2 and the acceleration of Q is 0.2 m/s2. It takes thetrains t seconds to pass each other.

– Find the distance travelled by each train in t seconds.

– Hence, or otherwise, calculate the value of t

– How long does it take for 2/5 of the length of train Q to pass the point o?

2.12.1 Solution

• – See Figure 3

– Since the motion is uniform acceleration from rest followed immediately by uni-form deceleration from rest:

t1 : t2 = 3f : f

⇒ t1 : t2 =3

4:1

4

⇒ t1 =3

4t

Using

v = u+ at

⇒ v = ft1

⇒ v = f

(3

4t

)=

3

4ft

The total distance d is the area under the graph:

d =1

2vt

⇒ d =1

2

(3

4ft

)t

⇒ d =3

8ft2


• – Using

s = ut+1

2at2

⇒ sP = 15t+1

2

3

10t2 = 15t+

3

20t2

⇒ sQ = 10t+1

2

1

5t2 = 10t+

1

10t2

– The trains pass each other when the distance they travel adds up to twice theirlength: 159 m;

159 = 15t+3

20t2 + 10t+

1

10t2

⇒ 300t+ 3t2 + 200t+ 2t2 = 3180

⇒ 5t2 + 500t− 3180 = 0

⇒ t2 + 100t− 636 = 0

⇒ t2 + 106t− 6t− 636 = 0

⇒ t(t+ 106)− 6(t+ 106) = 0

⇒ (t+ 106)(t− 6) = 0

⇒ t = 6 s

The case t = −106 does not concern us.

– In this case t needs to be found such that:

sQ =2

579.5 = 31.8 m

⇒ 31.8 = 10t+1

10t2

⇒ 318 = 100t+ t2

⇒ t2 + 100t− 318 = 0

Using

roots of ax2 + bx+ c are x =−b±

√b2 − 4ac

2a(22)

t =−100±

√1002 − 4(1)(−318)

2

⇒ t =−100±

√11272

2⇒ t = 3.085 or − 103.085

Ignore t < 0.Ans: t = 3.085 s.


2.13 Problem: LC HL 2005 [Part (a)]

Car A and car B travel in the same direction along a horizontal straight road.Each car is travelling at a uniform speed of 20 m/s.Car A is at a distance d metres in front of car B.At a certain instant car A starts to brake with a constant retardation of 6 m/s2.0.5 s later car B starts to brake with a constant retardation of 3 m/s2.

Find

(i) the distance travelled by car A before it comes to rest

(ii) the minimum value of d for car B not to collide with car A

2.13.1 Solution

(i) With respect to car A, when at rest:

u = 20

v = 0

a = −6

s =?

Using

v2 = u2 + 2as

⇒ s =v2 − u2

2a

⇒ s =−400

−12=

100

3m

(ii) Car A has a greater deceleration than car B and also begins it deceleration before carB; therefore car A comes to rest before car B does. Therefore when B stops it musthave travelled d+ 100/3 m to just avoid a collision. Now B travels at 20 m/s for halfa second before decelerating. With respect to B decelerating, when at rest:

u = 20

v = 0

a = −3

t =?

Using

t =v − u

a

⇒ t =0− 20

−3=

20

3s.

Now before decelerating B travels at constant speed for a half second. In this halfsecond it travels:

s =1

2(20) = 10 m.


Hence when stopped car B has travelled:

s = 10 +

(20

(20

3

)− 3

2

(20

3

)2)

⇒ s =230

3

and this must equal d+ 100/3:

230

3= d+

100

3

⇒ d =130

3m


A ball is thrown vertically upwards with an initial velocity of 20 m/s. One second later,another ball is thrown vertically upwards from the same point with an initial velocity of um/s.The balls collide after a further 2 seconds.

(i) Show that u = 17.75.

(ii) Find the distance travelled by each ball before the collision, giving your answers correctto the nearest metre.

2.14.1 Solution

(i) Let s1(t) be the height of the first particle and s2(t) be the height of the second particle.For the particles to collide after 3 s:

s1(3)!= s2(3). (23)

s1(3) = 3(20)− 1

2g(32)

⇒ s1(3) = 60− 9

2g

Particle 2 is motionless for one of these seconds and thus

s2(3) = 2u− 1

2g(4) = 2u− 2g

⇒ 2u− 2g!= 60− 9

2g

⇒ 2u = 60− 5

2g

⇒ u =60− 5g/2

2= 17.75 m/s.


(ii) Take the distance to mean total distance in the sense that if a particle travels up, stops,then falls down the total distance is the distance travelled going up plus the distancetravelled going down. For the first particle the motion is upwards until v = 0. That isuntil, using

v = u+ at

⇒ 0 = 20− gt

⇒ t =20

g≃ 2.041 s

Hence the distance travelled up is given by, using:

s =

(u+ v

2

)t

⇒ s =

(20 + 0

2

)20

g=

200

g≃ 20.408 m

The distance travelled on the way down is, using:

s = ut+1

2at2

⇒ s =1

2g

(3− 20

g

)2

⇒ s ≃ 4.9(0.920) = 4.508 m.

Hence to the nearest metre the first particle travels 25 m.

For the second particle the motion is up until v = 0; using

v = u+ at

⇒17.75=71/4

0 =71

4− gt

⇒ t =71

4g≃ 1.811 s

Hence the distance travelled up is given by, using:

s =

(u+ v

2

)t

⇒ s =

(71

2(4)

)71

4g=≃ 16.075 m

The distance travelled on the way down is, using:

s = ut+1

2at2

⇒ s =1

2g

(2− 71

4g

)2

⇒ s ≃ 4.9(0.0356) = 0.175 m.

Hence to the nearest metre the first particle travels 16 m.



(a) The points p, q and r all lie on a straight line.A train passes point p with speed u m/s. The train is travelling with uniform retardationf m/s2. The train takes 10 s to travel from p to q and 15 s to travel from q to r, where|pq| = |qr| = 125 m.

(i) Show that f = 1/3

(ii) The train comes to rest s metres after passing r.Find s, giving your answer correct to the nearest metre.

(b) A man runs at constant speed to catch a bus.At the instant the man is 40 m away from the bus, it begins to accelerate uniformlyfrom rest away from him.The man just catches the bus 20 s later.

(i) Find the constant speed of the man

(ii) If the constant speed of the man had instead been 3 m/s, show that the closest hegets to the bus is 17.5 m

2.15.1 Solution

(a) (i) Considering the motion from p to q, using:

s = ut+1

2at2

⇒ 125 = 10u− 1

2f(100)

⇒ 125 = 10u− 50f

⇒ 25 = 2u− 10f (24)

Now considering the motion from p to r:

250 = 25u− 1

2f(625)

⇒ 10 = u− 25

2f

⇒ u = 10 +25

2f (25)

Plugging into (24):

25 = 20 + 25f − 10f

⇒ 15 = 5f

⇒ f =1

3

�


(ii) From (25),

u = 10 +25

6=

85

6.

From p, the particle comes to rest when v = 0, using:

v2 = u2 + 2as

⇒ s =v2 − u2

2a

⇒ s =02 − (85/6)2

(−2/3)≃ 301.042 m.

The particle travels 250 m from p to r hence travels s = 51 m after passing rbefore coming to rest.

(b) (i) If the man just catches the bus then when his constant speed u is equal to that ofthe bus vb and he has travelled as far as the bus has plus the 40 m between them.That is if sm is the distance travelled by the man and sb the distance travelled bythe bus the condition to just catch the bus is

u = vb when sm = sb + 40. (26)

Let a be the acceleration of the bus. After 20 s, using:

v = u+ at

⇒ vb = 20a

⇒vb

!=u

a =u

20

Using, after 20 s,

s = ut+1

2at2

⇒ sm = 20u

⇒ sb =1

2a(400) = 200a = 10u

⇒sm

!=sb+40

20u = 10u+ 40

⇒ u = 4 m/s

(ii) If u = 3 m/s, the distance travelled by the man after t s is given by:

sm = 3t

As u = 4 above, a = 1/5 m/s2. With respect to the man, after t s, the bus hastravelled a distance sb + 40 away;

sb + 40 =1

2

1

5t2 + 40 =

t2

10+ 40.


Hence in terms of t, the distance between the man and bus is given by the distancetravelled by the bus less the distance travelled by the man:

t2

10+ 40− 3t. (27)

To minimise this function differentiate with respect to t and solve equal to 0:

t

5− 3 = 0

⇒ tmin = 15 s

To show this is a min note the second derivative is1

5> 0 ⇒ t = 15 a local minimum.

Hence the minimum separation is:

⇒t=15

225

10+ 40− 45 = 17.5 m.

�


(a) A stone is thrown vertically upwards under gravity with a speed of u m/s from a point30 m above the horizontal ground.The stone hits the ground 5 s later.

(i) Find the value of u

(ii) Find the speed with which the stone hits the ground.

(b) A particle, with initial speed u, moves in a straight line with constant acceleration.During the time interval from 0 to t, the particle travels a distance p.During the time interval from t to 2t, the particle travels a distance q.During the time interval from 2t to 3t, the particle travels a distance r.

(i) Show that 2q = p+ r

(ii) Show that the particle travels a further distance 2r − q in the time interval from3t to 4t.

2.16.1 Solution

(a) (i) With respect to the point the stone was thrown with s = −30 after t = 5 s. Thestone is under acceleration −g. Using

s = ut+1

2at2

⇒ −30 = 5u− 1

2g(25)

⇒ 5u =25

2g − 30

⇒ u =5

2g − 6 = 18.5 m/s


(ii) Using

v = u+ at

⇒ v = 18.5− g(5)

⇒ v = −30.5 m/s

Hence the stone hits the ground with speed |v| = 30.5 m/s.

(b) Considering the motion in the first t seconds, using

s = ut+1

2at2

⇒ p = ut+1

2at2 (28)

In the first 2t seconds:

p+ q = 2ut+4

2at2 (29)

In the first 3t seconds:

p+ q + r = 3ut+9

2at2 (30)

Now r = (30)− (29):

r = 3ut+9

2at2 − 2ut− 4

2at2

⇒ r = ut+5

2at2

⇒p=(28)

p+ r = 2ut+ 3at2

Now q = (29)− (28):

q = 2ut+4

2at2 − ut− 1

2at2

⇒ q = ut+3

2at2

⇒ 2q = 2ut+ 3at2 = p+ r.

�

(ii) Suppose the particle travels a distance s further in the next second. Hence in the first4t seconds:

p+ q + r + s = 4ut+16

2at2

⇒p+q+r=(30)

s = ut+7

2at2

Now

2r − q = 2ut+10

2at2 − ut− 3

2at2

⇒ 2r − q = ut+7

2at2 = s

�



(a) Points p and q lie in a straight line, where |pq| = 1200 m.Starting from rest at p, a train accelerates at 1 m/s2 until it reaches the speed limit of20 m/s. It continues at this speed of 20 m/s and then decelerates at 2 m/s2, coming torest at q.

Find the time it takes the train to go from p to q.

Find the shortest time it takes the train to from rest at p to rest at q if there is no speedlimit, assuming that the acceleration and deceleration remain unchanged at 1 m/s2 and2 m/s2, respectively.

(b) A particle is projected vertically upwards with an initial velocity of u m/s and anotherparticle is projected vertically upwards from the same point and with the same initialvelocity T seconds later.

Show that the particles

(i) will meet (T

2+

u

g

)seconds from the instant of projection of the first particle

(ii) will meet at a height of

4u2 − g2T 2

8gmetres.

2.17.1 Solution

(a) Let sa be the distance travelled whilst accelerating and sd be the distance travelledwhile decelerating. Using

v2 = u2 + 2as

s =v2 − u2

2a

sa =400

2= 200 m

sb =−400

−4= 100 m

Let sc be the distance travelled at constant speed v = 20 and tc be the time spenttravelling at constant speed. The total distance travelled is 1200 m:

1200 = 200 + 100 + 20tc

⇒ tc =900

20= 45 s


To travel a distance from rest to rest in the shortest possible times implies accelerationfollowed by immediate deceleration such that if t1 is the time spent accelerating andt2 the time spent decelerating:

t1 : t2 = d : a (31)

⇒ t1 : t2 = 2 : 1

⇒ t1 : t2 =2

3:1

3⇒ t1 = 2T/3, and t2 = T/3

Also the maximum speed reached is given by, using

v = u+ at

⇒ v =2

3T

Now distance travelled is the area under the graph:

1200!=

1

2vT

⇒ 1200 =1

2

(2

3T

)T

⇒ T 2 =3(2)(1200)

2= 3600

⇒ T = 60 s

(b) (i) If the particles meet at a time t after the first particle is emitted, then they willhave equal heights at that time:

s1(t)!= s2(t) (32)

Using

s = ut+1

2at2

⇒ s1(t) = ut− g

2t2

The second particle is only in motion after a time T so in terms of t it is in motionfor a time t− T :

s2(t) ≡ s2(t− T )

⇒ s2(t) = u(t− T )− g

2(t− T )2

⇒s1

!=s2

ut− g

2t2 = ut− uT − g

2(t2 − 2tT + T 2)

⇒ uT = gtT − T 2g

2

⇒ t =uT

gT+

T 2g

2gT

⇒ t =

(T

2+

u

g

)�


(ii) To find the height they meet at is to find s1 or s2 at the time t (s1(t) = s2(t)):

s1

(T

2+

u

g

)= u

(T

2+

u

g

)− 1

2g

(T

2+

u

g

)2

⇒ s1

(T

2+

u

g

)=

Tu

2+

u2

g− 1

2g

(T 2

4+

Tu

g+

u2

g2

)⇒ s1

(T

2+

u

g

)=

Tu

2+

u2

g− gT 2

8− Tu

2− u2

2g

⇒ s1

(T

2+

u

g

)=

4Tug + 8u2 − g2T 2 − 4uTg − 4u2

8g

⇒ s1

(T

2+

u

g

)=

4u2 − g2T 2

8g

�


(a) A stone projected vertically upwards with an initial speed of u m/s rises 70 m in thefirst t seconds and another 50 m in the next t seconds.

Find the value of u.

(b) A car, starting from rest and travelling from p to q on a straight level road, where|pq| = 10 000 m, reaches its maximum speed 25 m/s by constant acceleration in thefirst 500 m and continues at this maximum speed for the rest of the journey.

A second car, starting from rest and travelling from q to p, reaches the same maximumspeed by constant acceleration in the first 250 m and continues at this maximum speedfor the rest of the journey.

(i) If the two cars start at the same time, after how many seconds do the two carsmeet?Find, also, the distance travelled by each car in that time.

(ii) If the start of one car is delayed so that they meet each other at exactly halfwaybetween p and q, find which car is delayed and by how many seconds.


2.18.1 Solution

(a) Examining separately the motion in the first t seconds and in the first 2t seconds;using:

s = ut+1

2at2

⇒ 70 = ut− 1

2gt2 (33)

⇒ 120 = 2ut− 2gt2 (34)

⇒(33)

2ut = 140 + gt2 (35)

⇒(34)

120 = 140 + gt2 − 2gt2

⇒ gt2 = 20

⇒ t =

√20

g

⇒(35)

2u

√20

g= 140 + g

20

g

⇒ u =

√g

20(80) = 8

√100g

20= 8√

5g (36)

(b) (i) The cars meet whens1(t) + s2(t) = 10000 m (37)

How long does it take car 1 to accelerate to 25 m/s? Using

s =

(u+ v

2

)t

⇒ 500 =

(25

2

)ta,1

⇒ ta,1 = 40 s

Similarly,

250 =

(25

2

)ta,2

ta,2 = 20 s

Therefore the motion of the cars in terms of t after they take off (because the carscertainly don’t meet in less than 40 s - s1(40)+s2(40) = 500+250+20(25) = 1250m) is given by the distance whilst accelerating plus the distance travelled atconstant speed 25 m/s for a time (t−the time spent accelerating):

s1(t) = 500 + 25(t− 40) (38)

s2(t) = 250 + 25(t− 20) (39)


Now (38) + (39)!= 10000:

⇒ 750 + 25(2t− 60) = 10000

⇒ 25(2t− 60) = 9250

⇒ 2t− 60 = 370

⇒ t = 215 s

Also

s1(215) = 500 + 25(215− 40) = 4875 m

s2(215) = 250 + 25(215− 20) = 5125 m

(ii) Car 1 travels 500 m in 40 s. At constant speed, a = 0, using:

s = vt

⇒ t =s

v

⇒ t =4500

25= 180 s

So it takes car 1 40 + 180 = 220 s to travel 5000 m.Similarly car 2 travels 250 m in 20 s. At constant speed, a = 0, using:

s = vt

⇒ t =s

v

⇒ t =4750

25= 190 s

So it takes car 2 20 + 190 = 210 s to travel 5000 m. Hence if car 2 is delayed by10 s, they will meet at half way.

2.19 Problem: LC HL 1999 [Part (b)]

A particle travels in a straight line with constant acceleration f for 2t seconds and covers 15metres. The particle then travels a further 55 metres at constant speed in 5t seconds. Finallythe particle is brought to rest by a constant retardation 3f .

(i) Draw a speed-time graph for the motion of the particle.

(ii) Find the initial velocity of the particle in terms of t.

(iii) Find the total distance travelled in metres, correct to two decimal places.

(ii) First examining the constant speed motion, using:

s = vt

⇒ v =s

t=

55

5t=

11

t


(i)

Now using:

s =

(u+ v

2

)t

⇒ u =2s

t− v

⇒t=2t

u =15

t− 11

t

⇒ u =4

t

(iii) Using

a =v − u

t

⇒ f =11t− 4

t

2t

⇒ f =7t

2t=

7

2t2

Now for the decelerating motion, using:

v2 = u2 + 2as

⇒ s =v2 − u2

2a

⇒ s =−121

t2

−6f=

121

t26(

72t2

)⇒ s =

121

21≃ 5.762 m

Hence the total distance travelled, d, is:

d = 15 + 55 + 5.76 = 75.76 m



(a) A train accelerates uniformly from rest to a speed v m/s. It continues at this constantspeed for a period of time and then decelerates uniformly to rest. If the average speedfor the whole journey is 5v/6, find what fraction of the whole distance is described atconstant speed.

(b) Car A, moving with uniform acceleration 3b/20 m/s2 passes a point p with speed 9um/s. Three seconds later car B, moving with uniform acceleration 2b/9 m/s2 passesthe same point with speed 5u m/s. B overtakes A when their speeds are 6.5 m/s and5.4 m/s respectively. Find

(i) the value u and the value b

(ii) the distance travelled from p until overtaking occurs.

2.20.1 Remark

I think Q.1 in 1998 was particularly difficult. This is certainly the most difficult AM Q. 1I’ve ever seen.

2.20.2 Solution

(a) Let s1, s2 and s3 be the distances travelled at acceleration, constant speed and deceler-ation respectively. Similarly let t1, t2 and t3 be the time spent at acceleration, constantspeed and deceleration respectively. Hence as the area under the time-velocity graphis the distance travelled:

s1 =1

2vt1

s2 = vt2

s3 =1

2vt3

Using:

average speed =total distance

total time

⇒ 5v

6=

12vt1 + vt2 +

12vt3

t1 + t2 + t3

⇒ 5(t1 + t2 + t3) = 6

(1

2t1 + t2 +

1

2t3

)⇒ 5t1 + 5t2 + 5t3 = 3t1 + 6t2 + 3t3

⇒ t2 = 2t1 + 2t3


Now the fraction travelled at constant speed:

s2s1 + s2 + s3

=vt2

12vt1 + vt2 +

12vt3

=2t2

2t2 + (t1 + t3)

=(t2=2t1+2t3)

2t22t2 +

12t2

=4t2

4t2 + t2=

4t25t2

=4

5

(b) (i) In terms of a t after car A starts moving; using;

v = u+ at

⇒ vA(t) = 9u+

(3b

20

)t

Car B is stationary for three of these seconds:

vB(t) ≡ vB(t− 3) , where

vB(t− 3) = 5u+

(2b

9

)(t− 3)

For overtaking to occur after T seconds, vA(T ) = 5.4 m/s, vB(T ) = 6.5 m/s and

sA(T ) =: s := sB(T ) (40)

Now using;

v2 = u2 + 2as

⇒ v2A(T )!= 5.42 = 81u2 + 2

(3b

20

)s

⇒ 29.16 = 81u2 +3bs

10(41)

⇒ v2B(T )!= 6.52 = 25u2 + 2

(2b

9

)s

⇒ 42.25 = 25u2 +4bs

9(42)

40× (41) = 1166.4 = 3240u2 + 12bs

−27× (42) = −1140.75 = −675u2 − 12bs

⇒40×(41)+−27×(42)

25.65 = 2565u2

⇒ u2 = 0.01 =1

100

⇒ u =

√1

100=

1√100

=1

10m/s


Now note vA(T ) = 5.4;

vA(T ) = 5.4 = 9u+3bT

20

⇒ 5.4 = 0.9 +3bT

20

⇒ 4.5 =3bT

20⇒ bT = 30 (43)

Similarly

vB(T ) = 6.5 = 5u+2b

9(T − 3)

⇒ 6.5 = 0.5 +2bT

9+

6b

9

⇒ 6 =2bT

9− 6b

9⇒ 54 = 2bT − 6b

⇒ 6b = 2(bT )− 54

⇒(43)

6b = 60− 54 = 6

⇒ b = 1

(ii) Note the distance from overtaking is

s1(T ) =: s := s2(T )

Taking (41):

29.16 = 81u2 +3bs

10

⇒ s = (29.16− 81u2)10

3b

⇒ s = (29.16− 81(0.01))10

3⇒ s = 94.5 m


3 Projectiles


A straight vertical cliff is 200 m high.A particle is projected from the top of the cliff.The speed of projection is 14

√10 m/s at an angle α to the horizontal.

The particle strikes the level ground at a distance 200 m from the foot of the cliff.

(i) Find, in terms of α, the time taken for the particle to hit the ground.

(ii) Show that the two possible directions of projection are at right angles to each other.

3.1.1 Solution

(i) Let T be the time when sx = 200 and sy = −200. Using

sx(t) = uxt

⇒ 200 = 14√10 cosαT

⇒ T =200

14√10

secα

(×√10√10

)⇒ T =

10

7

√10 secα

(ii) Now sy(T ) = −200, using:

sy(t) = uyt−1

2gt2

⇒ −200 = 14√10 sinα

(10

7

√10 secα

)− 1

2g

(1000

49sec2 α

)Now sinα secα = tanα and

sec2 α ≡ 1 + tan2 α (44)

⇒ −200 = 200 tanα− g

98(1000(1 + tan2 α))

⇒ −200 = 200 tanα− 1

10(1000(1 + tan2 α))

⇒ −200 = 200 tanα− 100(1 + tan2 α)

⇒ −2 = 2 tanα− (1 + tan2 α)

Let u := tanα:

−2 = 2u− 1− u2

⇒ u2 − 2u− 1 = 0

Using the formula for the roots of a quadratic:

u =2±

√4 + 4

2=

2±√8

2

⇒√8=

√4.2=

√4√2tanα =

2± 2√2

2= 1±

√2


Now if mL and mK are the slopes of lines L and K then

mL ×mL = −1 ⇒ L ⊥ K (45)

Equivalently, two directions are at right angles if:

tanα1 × tanα2 = −1 (46)

⇒ tanα1 × tanα2 = (1 +√2)(1−

√2)

⇒ tanα1 × tanα2 = 1− 2 = −1

∴ the two directions are at right angles. �

3.2 Problem: LC HL 2008 [Part (b)]

A ball is projected from a point on the ground at a distance of a from the foot of a verticalwall of height b, the velocity of projection being u at angle 45◦ to the horizontal. If the balljust clears the wall prove that the greatest height reached is

a2

4(a− b).

3.2.1 Solution

In the first instance maximum height is sy when vy = 0. Using

vy = uy − gt

⇒ 0 = uy − gtmax

⇒ tmax =uy

g

⇒ symax = uy

(uy

g

)− 1

2g

(u2y

g2

)⇒ symax =

u2y

g− 1

2

u2y

g=

u2y

2g(47)

Now uy = u sin 45◦ = u/√2:

symax =u2

4g(48)


Now at a time, say T , sx(T ) = a and sy(T ) = b. Using:

sx = uxt

⇒ a = u cos 45◦T

⇒ T =

√2a

u

⇒sy(T )=b

b =u√2

(√2a

u

)− 1

2g

(2a2

u2

)⇒ b = a− ga2

u2

⇒ a− b =ga2

u2

⇒ u2 =ga2

a− b

⇒ symax =u2

4g=

ga2

4g(a− b)

⇒ symax =a2

4(a− b)

�


A particle is projected with a speed of 7√5 m/s at an angle α to the horizontal.

Find the two values of α that will give a range of 12.5 m.

3.3.1 Solution

The range is sx when sy = 0. Using

sy = uyt−1

2gt2

⇒ t(uy −

g

2t)= 0

⇒ t =2uy

g

⇒ R = sx(t) = ux

(2uy

g

)⇒ R =

u2

g2 sinα cosα

⇒2 sinx cosx=sin 2x

R =u2

gsin 2α

!=

25

2

⇒ sin 2α =25g

2u2=

25g

2(49)(5)=

1

2

⇒ 2α = 30◦ or 150◦

⇒ α = 15◦ or 75◦



A particle is projected from a point o with velocity 9.8 i + 29.4 j m/s where i and j are unitperpendicular vectors in the horizontal and vertical directions, respectively.

(i) Express the velocity and displacement of the particle after t seconds in terms of i andj.

(ii) Find, in terms of t, the direction in which the particle is moving after t seconds.

(iii) Find the two times when the direction of the particle is at right angles to the line joiningthe particle to o.

3.4.1 Solution

(i) Noting first that 9.8 = g and 29.4 = 3g. The velocity vector is given by:

v(t) = vx(t)i+ vy(t)j (49)

and using

vx(t) = ux , and

vy = uy − gt

⇒ v(t) = gi+ (3g − gt)j

The displacement vector is given by:

r(t) = sx(t)i+ sy(t)j (50)

Using:

sx(t) = uxt , and

sy(t) = uyt−1

2gt2

⇒ r(t) = gti+

(3gt− 1

2gt2)j

(ii) The direction the particle is travelling in is the slope of the tangent to the displacementat t; the tangent being given by the derivative:

r′(t) = vxi+ vyj

⇒ direction after t =j-component r′(t)

i-component r′(t)=

vyvx

=3g − gt

g= 3− t

(iii) The line joining the particle to o has slope:

sysx

For two lines L and K to be perpendicular the slopes mL and mK must satisfy

mL ×mL = −1. (51)


Hence solve:sysx

× (3− t)!= −1

⇒3gt− 1

2gt2

gt(3− t) = −1

⇒÷gt

(3− 1

2t

)(3− t) = −1

⇒ 9− 3

2t+

1

2t2 − 3t = −1

⇒ 18− 3t+ t2 − 6t = −2

⇒ t2 − 9t+ 20 = 0

⇒ t2 − 4t− 5t+ 20 = 0

⇒ t(t− 4)− 5(t− 4) = 0

⇒ (t− 4)(t− 5) = 0

Ans: At t = 4s, and 5 s.

3.5 Problem: LC HL 2004: [Part (a)]

A particle is projected from a point on the horizontal floor of a tunnel with maximum heightof 8 m. The particle is projected with an initial speed of 20 m/s inclined at an angle α tothe horizontal floor.Find, to the nearest metre, the greatest range which can be attained in the tunnel.

3.5.1 Solution

The range R is sx when sy = 0. Solving this gives:

R =u2

gsin 2α

Next the angle of projection which gives the max height as the height of the tunnel, 8 m, isfound. Max height is sy when vy = 0. Solving this gives:

symax =u2y

2g=

u2

2gsin2 α

⇒ u2

2gsin2 α = 8

⇒ sin2 α =16g

400

⇒ sinα =4√g

20=

√g

5⇒ α = arcsin(

√g/5) ≈ 38.763◦

Now if α > arcsin(√g/5) then max height is bigger than 8 m, so this motion is not in the

tunnel as required. Hence α < arcsin(√g/5). Now looking at the range:

R =u2

gsin 2α


For θ ∈ [0, 45◦], sin 2θ is increasing, as

d

dθsin 2θ = 2 cos 2θ > 0 , θ ∈ [0, 45◦]

Hence as 0 ≤ α ≤ arcsin(√g/5), sin 2α attains it maximum at α = arcsin(

√g/5):

R =202

g2 sinα cosα

Now the model triangle gives:

R =800

g2

√g

5

√25− g

5=

800

25

√g

g

√15.2

⇒ R =32√g(3.8987) = 39.85 ≃ 40 m


A particle is projected from a point on level horizontal ground at an angle θ to the horizontalground.Find θ, if the horizontal range of the particle is five times the maximum height reached bythe particle.

3.6.1 Solution

The range, R, is sx when sy = 0:

R =u2

gsin 2θ (52)

The maximum height, symax, of the particle is sy when vy = 0:

symax =u2 sin2 θ

2g(53)


Hence θ is the angle such that R = symax:

u2

gsin 2θ

!=

5u2 sin2 θ

2g

⇒u̸=0

2 sin θ cos θ =5

2sin2 θ

⇒θ ̸=0

2 cos θ =5

2sin θ

⇒ sin θ

cos θ=

4

5

⇒ tan θ =4

5

⇒ θ = arctan

(4

5

)≃ 38.66◦


A particle is projected from a point on the horizontal ground with a speed of 39.2 m/s inclinedat an angle α to the horizontal ground. The particle is at a height of 14.7 m above thehorizontal ground at times t1 and t2 seconds, respectively.

(i) Show that

t2 − t1 =√64 sin2 α− 12

(ii) Find the value of α for which t2 − t1 =√20.

3.7.1 Solution

(i) First note 14.7 = 3g/2 and 39.2 = 4g. t1 and t2 are times when sy = 14.7; hence arethe solutions of the quadratic equation:

3

2g = 4gt sinα− 1

2gt2

⇒×2,÷g

3 = 8t sinα− t2

⇒ t2 − 8t sinα− 3 = 0

Using the formula for the roots of a quadratic, letting t1 be the ‘+’ solution and t2 bethe ‘-’ solution:

t =8±

√64 sin2 α− 12

2

t1 − t2 =

(8 +

√64 sin2 α− 12

2

)−

(8−

√64 sin2 α− 12

2

)

⇒ t1 − t2 = 4 +

√64 sin2 α− 12

2− 4 +

√64 sin2 α− 12

2

⇒ t1 − t2 =√64 sin2 α− 12

�


(ii) Hence α is such that:

64 sin2 α− 12 = 20

⇒ 64 sin2 α = 32

⇒ sin2 α =1

2

⇒ sinα =1√2

⇒ θ = 45◦


A player hits a ball with an initial speed of u m/s from a height of 1 m at an angle of 45◦

to the horizontal ground. A member of the opposing team, 21 m away, catches the ball at aheight of 2 m above the ground.Find the value of u.

3.8.1 Solution

sy = 1 when sx = 21. Suppose the catcher catches the ball at a time T , noting sin 45◦ =1/√2 = cos 45◦:

sx =u√2T

⇒ T =21√2

uT

Now sy(T ) = 1, using:

sy = uyt−1

2gt2

⇒ 1 =u√2

21√2

u− 1

2g(21)2(2)

u2

⇒×u2, cancelling

u2 = 21u2 − 212g

⇒ 20u2 = 212g

⇒ u2 =212

20g

⇒ u = 21

√g

20= 14.7 m/s


3.9 Problem: LC HL 2000: [Part (b)]

A particle is projected with a velocity u m/s at an angle β to the horizontal ground. Showthat the particle hits the ground at a distance

u2

gsin 2β

from the point of projection. Find the angle of projection which gives maximum range.

3.9.1 Solution

Range, R, is sx when sy = 0;

sy = uyt−1

2gt2

!= 0

⇒ t =2uy

g

Now

sx = uxt

⇒ R = ux

(2uy

g

)⇒ R =

2u2 sin β cos β

g

⇒ R = u22 sin β cos β

g

⇒2 sinx cosx=sin 2x

R =u2

gsin 2β

�

Taking u to be fixed, to maximise R vary β. The maximum value of sine is 1. Thisoccurs when

2β = 90◦ (sin 90◦ = 1)

⇒ β = 45◦ for maximum range


3.10 Problem: LC HL 2009: [Part(b)]

A plane is inclined at an angle 60◦ to the horizontal. A particle is projected up the planewith initial speed u at an angle θ to the inclined plane. The plane of projection is verticaland contains the line of greatest slope.The particle strikes the plane at right angles.Show that the range on the inclined plane is

4√3u2

13g.

3.10.1 Solution

Let T be the time of flight. If the particle lands at right angles, sy(T ) = 0 and vx(T ) = 0.Using

vx = u cos θ − g sin 60◦ t

⇒ T =u cos θ

g sin 60◦

Figure 1: If the particle lands at right angles, the final velocity is entirely in the y-direction

Using

sy = u sin θ t− 1

2g cos 60◦ t2

⇒ T =2u sin θ

cos 60◦

⇒ u cos θ

g sin 60◦=

2u sin θ

cos 60◦

⇒ tan θ =1

2.cos 60◦

sin 60◦=

1

2. cot 60◦

⇒ tan θ =1

2√3


Figure 2: The model triangles for 60◦ and θ

Now range means sx(T ). Using

sx = uxt−1

2gxt

2

⇒ R = u cos θ

(u cos θ

g sin 60◦

)− 1

2�g��sin 60◦

(u2 cos2 θ

g�2 sin�2 60◦

)

⇒ R =u2 cos2 θ

2g sin 60◦(2− 1)

⇒ R =u2

�2g.4(3)

13.�2√3

⇒ R =4√3u2

13g

�



A particle is projected down an inclined plane with initial velocity u m/s. The line of pro-jection makes an angle 2θ◦ with the inclined plane and the plane is inclined at θ◦ to thehorizontal. The plane of projection is vertical and contains the line of greatest slope.The range of the particle on the inclined plane is ku2 sin θ/g.Find the value of k.

3.11.1 Solution

Range, R, means sx when sy = 0.

Using

sy = u sin 2θ t− 1

2g cos θ t2

⇒ T =2u sin 2θ

g cos θ

⇒sin 2x=2 sinx cosx

4u sin θ

g

Using

sx = uxt+1

2gxt

2

⇒ R = u cos 2θ

(4u sin θ

g

)+

1

�2.�g sin θ

8

��16u2 sin2 θ

g�2

⇒ R =

4u2 sin θ

g(cos 2θ + 2 sin2 θ)

⇒cos 2x=cos2 x−sin2 x

R =4u2 sin θ

g(cos2 θ − sin2 θ + 2 sin2 θ)︸︷︷︸

=cos2 θ+sin2 θ

⇒cos2 x+sin2 x=1

R =4u2

gsin θ

⇒ k = 4



A particle is projected up an inclined plane with initial speed u m/s. The line of projectionmakes an angle 30◦ with the plane and the plane is inclined at 30◦ to the horizontal. Theplane of projection is vertical and contains the line of greatest slope.Find in terms of u, the range of the particle on the inclined plane.

3.12.1 Solution


Using

sy = u sin 30◦ t− 1

2g cos 30◦ t2

⇒ T =2u

gtan 30◦

⇒ R = u cos 30◦(2u

gtan 30◦

)− 1

�2�g sin 30◦.

2

�4u2 tan2 30◦

g�2

⇒ R =2u2 tan 30◦

g(cos 30◦ − sin 30◦ tan 30◦)

⇒using Figure 2

R =�2u2

g.1√3

(√3

�2− 1

�2.1√3

)

⇒ R =u2

√3 g

(3− 1√

3

)⇒ R =

2u2

3g



A plane is inclined at an angle β to the horizontal. A particle is projected up the plane withinitial speed u at an angle α to the inclined plane. The plane of projection is vertical andcontains the line of greatest slope.

(i) Find the range of the particle on the inclined plane in terms of u, α and β.

(ii) Show that for a constant value of u the range is a maximum when

α = 45◦ − β

2

3.13.1 Solution


Using

sy = u sinα t− 1

2g cos β t2

⇒ T =2u sinα

g cos β

⇒ R = u cosα

(2u sinα

g cos β

)− 1

�2�g sin β

2

�4u2 sin2 α

g�2 cos2 β

⇒ R =

2u2 sinα

g cos2 β(cosα cos β − sinα sin β)

⇒cos(x+y)=cosx cos y−sinx sin y

R =u2

g cos2 β2 sinα cos(α+ β)

⇒2 sinx cos y=sin(x+y)+sin(x−y)

R =u2

g cos2 β(sin(2α+ β) + sin(−β))

⇒sin(−x)=− sinx

R =u2

g cos2 β(sin(2α+ β)− sin β)


To maximise R the only term which may be varied is sin(2α+ β). Sine has a maximumvalue of 1; namely sin 90◦ = 1. Hence for Rmax;

2α+ β = 90◦

⇒ 2α = 90◦ − β

⇒ α = 45◦ − β

2

�


A particle is projected up the inclined plane with initial speed u m/s. The line of projectionmakes an angle α with the horizontal and the inclined plane makes an angle β with thehorizontal. ( The plane of projection is vertical and contains the line of greatest slope).Find in terms of u, g, α and β, the range of the particle up the inclined plane.

3.14.1 Solution



Using

sy = u sin(α− β) t− 1

2g cos β t2

⇒ T =2u sin(α− β)

g cos β

⇒ R = u cos(α− β)

(2u sin(α− β)

g cos β

)− 1

�2�g sin β

2

�4u2 sin2(α− β)

g�2 cos2 β

⇒ R =

2u2 sin(α− β)

g cos2 β(cos(α− β) cos β − sin(α− β) sin β)

⇒cos(x+y)=cosx cos y−sinx sin y

R =u2

g cos2 β2 sin(α− β) cos (α− β + β)︸︷︷︸

=α

⇒2 sinx cos y=sin(x+y)+sin(x−y)

R =u2

g cos2 β(sin(α− β + α) + sin(α− β − α))

⇒sin(−x)=− sinx

R =u2

g cos2 β(sin(2α− β)− sin β)


A particle is projected from a point p with initial speed 15 m/s, down a plane inclined at anangle of 30◦ to the horizontal. The direction of projection is at right angles to the inclinedplane. (The plane of projection is vertical and contains the line of greatest slope). Find

(i) the perpendicular height of the particle above the plane after t seconds and hence, orotherwise, show that the vertical height h of the particle above the plane after t secondsis

10√3 t− 4.9t2

(ii) the greatest vertical height it attains above the plane (i.e. the maximum value of h)correct to two places of decimals.

3.15.1 Solution

Using

sy = uyt−1

2gyt

2

⇒ sy(t) = 15t− 4.9 cos 30◦t2

⇒cos 30◦ from Fig 2

sy(t) = 15t− 4.9√3

2t2


Due to the geometry,

sin 60◦ =syh

⇒ h =sy

sin 60◦

⇒sin 60◦ from Fig 2

h =2√3

(15t− 4.9

√3

2t2

)⇒ h =

30√3t− 4.9t2

⇒ h = 10√3t− 4.9t2

To maximise h(t) note that it is a concave down2 quadratic and hence has a single localmaximum when

dh

dt= 0

⇒ 10√3− gtmax = 0

⇒ tmax =10√3

g

⇒ hmax = h(tmax) = 10√3

(10√3

g

)− g

2

(300

g2

)⇒ hmax =

300

g− 1

2.300

g=

150

g⇒ hmax ≃ 15.31 m

2or sad



A plane is inclined at an angle 45◦ to the horizontal. A particle is projected up the planewith initial speed u at an angle θ to the horizontal. The plane of projection is vertical andcontains the line of greatest slope. The particle is moving horizontally when it strikes theinclined plane.Show that tan θ = 2.

3.16.1 Solution

If the particle lands horizontally then the landing angle is 45◦.

Figure 3: If the particle lands horizontally then as alternate angles the landing angle is equalto the angle the plane makes with the horizontal.

The landing angle is given by:

tan l =−vyvx

(54)

⇒ −vyvx

= tan 45◦ = 1 (55)

⇒ −vy = vx (56)


where vy, vx are the final speeds in the y- and x-directions. Using

sy(T ) = 0 = u sin(θ − 45◦)T − 1

2g cos 45◦.T 2

⇒ T =2u sin(θ − 45◦)

g cos 45◦

⇒ vy = u sin(θ − 45◦)−��g cos 45◦.2u sin(θ − 45◦)

��g cos 45◦

⇒ vy = −u sin(θ − 45◦)

⇒sin 45◦=1/

√2=cos 45◦

vx = u cos(θ − 45◦)−��g sin 45◦(2u sin(θ − 45◦)

��g cos 45◦

)⇒ vx = u(cos(θ − 45◦)− 2 sin(θ))

⇒(56)

sin(θ − 45◦) = cos(θ − 45◦)− 2 sin(θ − 45◦)

⇒ 3 sin(θ − 45◦) = cos(θ − 45◦)

⇒ tan(θ − 45◦) =1

3

⇒tan(A−B)= tanA−tanB

1+tanA tanB

tan θ − 1

1 + tan θ=

1

3

⇒ 1 + tan θ = 3 tan θ − 3

⇒ 2 tan θ = 4

⇒ tan θ = 2

�



A particle is projected up an inclined plane with initial speed u m/s. The line of projectionmakes an angle α with the horizontal and the inclined plane makes an angle θ with thehorizontal. (The plane of projection is vertical and contains the line of greatest slope.)If the particle strikes the inclined plane at right angles, show that

tanα =1 + 2 tan2 θ

tan θ

3.17.1 Solution

If the particle strikes the plane at right angles vx(T ) = 0 where T is the time of flight.

T =2u sin(α− θ)

g cos θ

⇒ vx(T ) = u cos(α− θ)−�g sin θ

(2u sin(α− θ)

�g cos θ

)= 0

⇒×1/u

cos(α− θ) = 2 tan θ sin(α− θ)

⇒ 1 = 2 tan θ tan(α− θ)

⇒ 1 = 2 tan θ

(tanα− tan θ

1 + tanα tan θ

)⇒ 1 + tanα tan θ = 2 tan θ tanα− 2 tan2 θ

⇒ tanα tan θ = 1 + 2 tan2 θ

⇒ tanα =1 + 2 tan2 θ

tan θ

�


A particle is projected with speed u m/s at an angle θ to the horizontal, up a plane inclinedat an angle β to the horizontal. (The plane of projection is vertical and contains the line ofgreatest slope). The particle strikes the plane at right angles,

(i) Show that 2 tan β tan(θ − β) = 1

(ii) Hence, or otherwise, show that if θ = 2β, the range of the particle up the inclined planeis u2/(g

√3)

3.18.1 Solution

(i) This is shown in four lines in the above solution. �

(ii) If θ = 2β;

2 tanβ tan(2β − β) = 1

⇒ 2 tan2 β = 1

⇒ tan β =1√2


Figure 4: The Model Triangle for β.

Range is sx(T ):

R = u cos β.2u sin β

g cos β− 1

�2�g sin β.

2

�4u2 sin2 β

g�2 cos2 β

⇒ R =2u2 sin β

g cos2 β(cos2 β − sin2 β)

⇒ R =2u2

g.1√3.3

2

(2

3− 1

3

)⇒ R =

�2u2

g√3.�3

�2

(1

�3

)⇒ R =

u2

g√3

�



A particle is projected at an angle α = tan−1 3 to the horizontal up a plane inclined atan angle θ to the horizontal. (The plane of projection is vertical and contains the line ofgreatest slope). The particle strikes the plane at right angles.Find the two possible values for θ.

3.19.1 Solution

From Problem 3.17:

tanα =1 + 2 tan2 θ

tan θ

⇒ 3 =1 + 2 tan2 θ

tan θ⇒ 3 tan θ = 1 + 2 tan2 θ

⇒ 2 tan2 θ − 3 tan θ + 1 = 0

⇒ 2 tan2 θ − 2 tan θ − tan θ + 1 = 0

⇒ 2 tan θ(tan θ − 1)− 1(tan θ − 1) = 0

⇒ (2 tan θ − 1)(tan θ − 1) = 0

⇒ θ = arctan(1/2) ≃ 26.565◦ , or 45◦.

3.20 Problem: LC HL 1999 [Parts (a) & (b)(i)]

A particle is projected from a point p up an inclined plane with a speed of 4g√2 m/s at an

angle tan−1(1/3) to the inclined plane. The plane is inclined at an angle θ to the horizontal.(The plane of projection is vertical and contains the line of greatest slope). The particle ismoving horizontally when it strikes the plane at the point q.

(a) Find the two possible values for θ.

(b)(i) If tan θ = 0.5 then find the magnitude of the velocity with which the particle strikes theinclined plane at q.

3.20.1 Solution

(a) Let α := tan−1(1/3). As in Figure 18, the landing angle is θ as the particle landshorizontally. Hence

tan θ =−vy(T )

vx(T )


where T is the time of flight;

T =2u sinα

g cos θ

⇒ sy(T ) = −u sinα , and

sx(T ) = u cosα−�g sin θ.2u sinα

�g cos θ

⇒ sx(T ) = u cosα− 2u tan θ sinα

⇒ tan θ =�u sinα

�u cosα− 2�u tan θ sinα

⇒ tan θ cosα− 2 tan2 θ sinα = sinα

⇒×1/ sinα

tan θ

tanα− 2 tan2 θ = 1

⇒tanα=1/3

2 tan2 θ − 3 tan θ + 1 = 0

⇒ 2 tan2 θ − 2 tan θ − tan θ + 1 = 0

⇒ 2 tan θ(tan θ − 1)− 1(tan θ − 1) = 0

⇒ (2 tan θ − 1)(tan θ − 1) = 0

⇒ θ = arctan(1/2) ≃ 26.565◦ , or 45◦.

(b)(i) The final speed of the particle is given by:

|v(T )| =√

v2x(T ) + v2y(T ) (57)

Figure 5: The model triangles for θ & α


Now from (a);

vy(T ) = −u sinα

⇒ vy(T ) = −4g

√2√10

⇒ v2y(T ) =16g2

5, and

vx(T ) = u cosα− 2u tan θ sinα

⇒ vx(T ) = 4g√2

3√10

− 4g√2

1√10

⇒ vx(T ) =4g√2√

10(3− 1) =

8g√5

⇒ v2x(T ) =64g2

5

⇒ |v(T )| =√

80g2

5=√16g2

⇒ |v(T )| = 4g


A particle is projected down a slope which is inclined at 45◦ to the horizontal. The particleis projected from a point on the slope and has an initial velocity of 7

√2 m/s at an angle α

to the inclined plane. Find the value of α if

(i) the particle first hits the slope after 2 seconds

(ii) the landing angle with the slope is tan−1(1/3)

3.21.1 Solution

(i) The particle hits the slope when sy = 0:

sy = u sinαt− 1

2g sin 45◦t2

⇒ 0!= 7

√2 sinα.(2)− 1

�2g sin 45◦(2)�2

⇒ 0 = 14√2 sinα− 2√

2g

⇒ sinα =2g

28=

19.6

28

⇒ α = sin−1

(19.6

28

)≃ 44.4◦

(ii) The landing angle l is given by:

tan l =−vy(T )

vx(T )(58)


where T is the time of flight and vy(T ) and vx(T ) are the final velocities in the y- andx-directions respectively. The time of flight is when sy = 0:

sy = u sinαt− 1

2g cos 45◦t2

!==

⇒ T =2u sinα

g cos 45◦

⇒ vy(T ) = u sinα−�g��cos 45◦(2u sinα

�g��cos 45◦

)⇒ vy(T ) = u sinα− 2u sinα = −u sinα

and

vx(T ) = u cosα+��g sin 45◦(2u sinα

��g cos 45◦

)⇒ vx(T ) = u cosα+ 2u sinα

Now if l = tan−1(1/3):

⇒ 1

3=

−vy(T )

vx(T )

⇒ vx(T ) = −3vy(T )

⇒�u cosα+ 2�u sinα = 3�u sinα

⇒ cosα = sinα

⇒ tanα = 1

⇒ α = 45◦

3.22 Problem: F.A.M. Exercises 3.D 3(iii) [LC 1979]

A plane is inclined at an angle α to the horizontal. A particle is projected up the plane witha speed u at an angle θ to the plane. The plane of projection is vertical and contains the lineof greatest slope.Prove that the particle will strike the plane horizontally if

tan θ =sinα cosα

(2− cos2 α)

3.22.1 Solution

Let l be the landing angle.

tan l =−vy(T )

vx(T )(59)

Suppose tan l = tanα. As both l and α are greater than 0◦ and less than 90◦, and tan−1 isone-to-one on [0◦, 90◦];

tan−1(tan l) = tan−1(tanα)

⇒ l = α


which implies that the particle strikes horizontally. Now from before

vy(T ) = −u sin θ

vx(T ) = u cos θ − 2u tanα sin θ

⇒ tan l =sin θ

cos θ − 2 tanα sin θ

Now if

tan θ =sinα cosα

(2− cos2 α)

Figure 6: The model triangle for θ

Hence, where λ is the hypotenuse of the model triangle, by Pythagoras Theorem;

λ2 = sinα cos2 α+ (2− cos2 α)2

⇒ λ =√

sinα cos2 α+ (2− cos2 α)2

⇒ tan l =

sinα cosα

�λ(2−cos2 α)

�λ− 2 tanα sinα cosα

�λ

⇒tan sin cos=sin2

tan l =sinα cosα

2− cos2−2 sin2 α

⇒cos2 +sin2=1

tan l =sinα cosα

2 cos2 α+��2 sin2 α− cos2 α−��

2 sin2 α

⇒ tan l =sinα��cosα

cos�2 α=

sinα

cosα⇒ tan l = tanα

Hence the particle strikes the plane horizontally. �

3.22.2 Remark

Questions of this subtlety rarely come up. I wouldn’t expect anything of this difficulty inLC 2010 Q.3(b). This question is taken from LC 1979. Consider these statements P & Q;


P. the particle strikes the plane horizontally

Q.

tan θ =sinα cosα

(2− cos2 α)

Ordinarily one is asked to show that P ⇒ Q (read P implies Q). That is assuming P , proveQ; and this is straightforward. However in this example one is asked to show Q ⇒ P . Notethe difference. If one proved P ⇒ Q one would not receive full marks as the question hasbeen fundamentally misunderstood. Note our approach here. It is shown for l, α ∈ [0◦, 90◦]:

If tan l = tanα, then l = α

and tan l is computed; and shown to be equal to tanα. Thence l = α and implies that theparticle lands horizontally.

4 Relative Velocity

4.1 Problem: F.A.M. Q.4.A.6

With O as the origin the position vectors of P , Q, S are rP = i − 2j, rQ = −4i + j,rS = −3i+ 5j. Find rQP , the displacement of Q relative to P . Find in terms of i and j theposition vector of T if rTS = rQP .

4.1.1 Solution

Using

rAB = rA − rB

⇒ rQP = rQ − rP

⇒ rQP = −5i+ 3j

Let

rT = xi+ yj

⇒ rTS = rT − rS

⇒ rTS = (x+ 3)i+ (y − 5)j

Now for rTS = rQP , comparing components:

−5i+ 3j!= (x+ 3)i+ (y − 5)j

⇒ rT = −8i+ 8j



A train is travelling on a straight track with velocity 30j and a car, visible from the train,is travelling on a straight road with velocity 10i + 6j where speeds are measured in m/s.Calculate the magnitude and direction of the car’s velocity as it appears to a person sittingon the train.

4.2.1 Solution

Using

VCT = VC −VT

⇒ VCT = 10i− 24j

⇒ |VCT | =√102 + 242 =

√100 + 576

⇒ |VCT | = 26m/s

Figure 7: VCT in the plane. Note θ. The direction of VCT is E θ S.

Now

tan θ =24

10=

12

5⇒ θ = tan−1(12/5) ≈ 67.38◦

Hence the direction of VCT is E 67.38◦ S.


A particle P is 100 m due West of another particle Q. The velocity of P is 6i+2j m/s, andthe velocity of Q is −4i + 2j m/s. Show that P and Q are on collision course. How muchtime will pass before the collision occurs?

4.3.1 Solution

Clearly RPQ = −100i. Using

VPQ = VP −VQ

⇒ VPQ = 10i

∴ the particles are on collision course. �


Figure 8: Particle P is due West of particle Q; in fact its position relative to particle Q isalong the negative i-axis at −100i m. The velocity of particle P relative to Q is +10i, backalong the i axis towards particle Q. Therefore, they must be on collision course.

Using

time =relative distance

relative speed

⇒ t =100

10= 10 s


K is a particle which is 20 m due West of another particle T . Their velocities are i + 2jm/s and −2i − 2j m/s respectively. Find the velocity of K relative to T . Find the shortestdistance between them in subsequent motion.

4.4.1 Solution

Now RKT = −20i. Using

VKT = VK −VT

⇒ VKT = 3i+ 4j

Now consider the position of K relative to T , RKT :Now tan θ = 4/3:Now from Figure 9:

sin θ =d

20!=

4

5

⇒ d =80

5= 16m


Figure 9: The shortest distance between K and T in the subsequent motion is the perpen-dicular distance, d, between T and the path of K relative to T .

Figure 10: The model triangle for θ.


A particle P is moving with velocity −8i+12j while another particle Q is moving with velocity7i+4j. Find the velocity of P relative to Q. P is at the point 119i when Q is at the origin O.Show the positions of P and Q on a diagram and show the path of P relative to Q. Calculatethis distance of O from this path. What does this distance represent?

4.5.1 Solution

Using

VPQ = VP −VQ

⇒ VPQ = −15i+ 8j

Now from Figure 11:

sin θ =d

119!=

8

17

⇒ d = 1198

17= 56

This distance represents the shortest distance between P and Q is subsequent motion.


Figure 11: The position RPQ and path VPQ of P relative to Q. α = tan−1(8/15). d is thedistance of O from the path.

Figure 12: The model triangle for α. Sides 8 & 15 are given by α = tan−1(8/15). Thehypotenuse is

√82 + 152 =

√289 = 17



(a) If |ti+ 3j| = 5, find the value of t > 0, t ∈ R.

(b) (i) A ship K is 60 km due West of another ship M which is travelling with velocity−2i + 3j km/hr, find in terms of i and j its velocity if it is to intercept (collidewith) M . (Hint: K will have to travel with the same j-speed as ship M in orderto keep on collision course.)

(ii) When will collision occur?

4.6.1 Solution

(a) For any vector a := xi+ yj;

|a| =√

x2 + y2

⇒ |ti+ 3j| =√t2 + 9

!

= 5

⇒ t2 + 9 = 25

⇒ t2 = 16

⇒ t = ±4

⇒t>0

t = 4.

(b) (i) RKM = −60i hence for collision VKM = ki with k > 0. Let

VK = xi+ yj , with√x2 + y2 = 5

Using

VKM = VK −VM

⇒ VKM = (x+ 2)i+ (y − 3)j

For collision course, y − 3 = 0, y = 3; and x+ 2 > 0, x > −2:

√x2 + 9 = 5

⇒ x = ±4

⇒x>−2

x = 4

⇒ VK = 4i+ 3j

(ii) Now VKM = 6i. Using

t =relative distance

relative speed

⇒ t =60

6= 10 h



A ship P is 3.4 km due West of another ship Q. P is moving with speed 5√2 m/s in a NE

direction. Q can travel at 13 m/s. If they are on collision course find the velocity of Q interms of i and j. When will collision occur?

4.7.1 Solution

Let VQ = xi+ yj with

|VQ| = 13√x2 + y2

!= 13

RPQ = −3400 i, hence for P to collide with Q, it is required that:

VPQ = k i , k > 0 (60)

Now VP :

Figure 13: NE is the direction E 45◦ N as shown. As sin 45◦ = 1/√2 = cos 45◦, VP = 5 i+5 j.

Using

VPQ = VP −VQ

⇒ VPQ = (5− x)i+ (5− y)j

Hence for collision, y = 5. Now |VP | = 13:√x2 + 52 = 13

⇒ x2 = 132 − 52 = 144

⇒ x = ±12

⇒(5−x)

!>0

x = −12

⇒ VQ = −12 i+ 5 j

Now VPQ = 17 i. Using

t =relative distance

relative speed

⇒ t =

200��3400

��17= 200 s



Ship T is 100 km due West of ship Q. T is travelling at 10 km/hr in a direction E 30◦

S. Q is travelling at 20 km/hr in direction W 45◦ N. Find in terms of i and j the velocityof T , the velocity of Q, and the velocity of T relative to Q. Find, also, the magnitude anddirection of the velocity of T relative to Q and hence find the shortest distance between themin subsequent motion correct to one decimal place.

4.8.1 Solution

Consider VT :

Figure 14: VT and the model triangle for 30◦. VT = 10 cos 30◦ i− 10 sin 30◦ j.

VT = 5√3 i− 5 j (61)

Consider now VQ:

Figure 15: VQ = −20 cos 45◦ i+ 20 sin 45◦ j.

VQ = −201√2i+ 20

1√2j

⇒ VQ = −201√2·√2√2i+ 20

1√2·√2√2j

⇒ VQ = −20

2

√2 i+

20

2

√2 j

⇒ VQ = −10√2 i+ 10

√2 j


Using

VTQ = VT −VQ

⇒ VTQ = (5√3 + 10

√2)i+ (−5− 10

√2)j

⇒ VTQ = 5(√3 + 2

√2)i− 5(1 + 2

√2)j

For a general vector a = x i+ y j:

|a| =√

x2 + y2

⇒(−x)2=x2

|VTQ| =√52(

√3 + 2

√2)2 + 52(1 + 2

√2)2

⇒ |VTQ| = 5

√3 + 4

√6 + 8 + 1 + 4

√2 + 8

⇒ |VTQ| = 5

√20 + 4

√2 + 4

√6

⇒a√b2x=ab

√x|VTQ| = 10

√5 +

√2 +

√6 ≈ 29.772 km/hr

Figure 16: The direction of VTQ is given by θ = tan−1((�5(1 + 2√2))/(�5(

√3 + 2

√2)))

tan θ =1 + 2

√2√

3 + 2√2≈ 40.01◦

Hence the direction of VTQ is given by E 40.01◦ S.

Now RTQ = −100 i.Now from Figure 17;

sin 40.01 =d

100⇒ d = 100 sin 40.01

⇒ d = 100(0.64291) ≈ 64.3 m


Figure 17: Ship T is initially 100 km West of ship Q. Relative to Q, T travels along VTQ;hence the shortest distance between T and Q in the subsequent motion is given by d.

4.9 Problem: LC HL 2009: [Part(a)]

(i) VA = 15 i m/s and VB = 20 j m/s. Using

VAB = VA −VB

⇒ VAB = 15 i− 20 j

When both cars are 800 m from the intersection RA = −800 i and RB = −800 j. Using

RAB = RA −RB

⇒ RAB = −800 i+ 800 j

Now

Figure 18: The position of A relative to B and the subsequent motion of A relative to B.

As a line, the slope of VAB is m = − tan θ = −4/3. Also (−800, 800) ∈ L := VAB.


Using

L ≡ y − y1 = m(x− x1)

⇒ L ≡ y − 800 = −4

3(x+ 800)

⇒ L ≡ 3y − 2400 = −4x− 3200

⇒ L ≡ 4x+ 3y + 800 = 0

Using, where d is the perpendicular distance between a point p(x1, y1) and the lineax+ by + c = 0;

d =|ax1 + by1 + c|√

a2 + b2=

x1=0=y1

|c|√a2 + b2

⇒ d =800

5= 160 m

(ii) Using, where θ is the angle between two lines of slope m1 and m2, with respect to theangle α between the lines RAB (m1 = −1) and VAB (m2 = −4/3);

tan θ = ± m1 −m2

1 +m1m2

⇒ tan θ = ±−1 + 4/3

1 + 4/3= ±−1/�3

7/�3=

α∈[0,90◦]

1

7

Hence

Figure 19: tanα = d/D

tanα =1

7=

d

D⇒ D = 7d = 1120 m

Using


relative speed

⇒ time =1120

|VAB|=

|VAB |=25

224

5s


Hence in this time, using s = ut:

sA = 15.224

5= 672 m , and

sB = 20.224

5= 896 m

Hence A is 128 m from the intersection and B is 96 m from the intersection.


(i) VC = 1.5 i m/s and VD = 2 j m/s. Using

VCD = VC −VD

⇒ VCD = 1.5 i− 2 j

(ii) When D passes the intersection RCD = −100 i m. Consider

Figure 20: cos θ = L/100.

Now tan θ = −4/3


cos θ =3

5=

L

100

⇒ L =300

5= 60 m


Using


relative speed

⇒ time =60

|VCD|=

60√9/4 + 4

=60√25/4

= 24 s

Using

s = ut

⇒ sC = 1.5(24) = 36 m

⇒ 64 m from the intersection


(i) VB = −24 i km/hr and VA = 32 j km/hr. Using

VAB = VA −VB

⇒ VAB = 24 i+ 32 j

(ii) If ship B is 8 km NE of ship A, ship A is 8 km SW of ship B:

RAB = −8 cos 45◦ i− 8 sin 45◦ j

⇒ RAB = − 8√2i− 8√

2j

⇒×√2/

√2RAB = −4

√2 i− 4

√2 j

Considered as a line, V := VAB has slope 32/24 = 4/3; and noting (−4√2,−4

√2) ∈ L,

using:

L ≡ y − y1 = m(x− x1)

⇒ V ≡ y + 4√2 =

4

3(x+ 4

√2)

⇒ V ≡ 3y + 12√2 = 4x+!6

√2

⇒ V ≡ y =4x+ 4

√2

3

⇒ V ≡

{(x,

4x+ 4√2

3

): x ≥ −4

√2

}=: {px : x ≥ −4

√2}

Now consider the following:

V is the set of points {px}. The distance between ship A and ship B is 8 km when the


Figure 22: R denotes RAB. The circle represents a circle of radius 8 km about B. Twiceship A will be exactly 8 km from ship B; initially and again at a later time.

distance from px to (0, 0), d, is 8:

d =

√√√√(x− 0)2 +

(4x+ 4

√2

3− 0

)2

= |px|!= 8

⇒

√x2 +

16x2 + 32√2x+ 32

9!= 8

⇒ x2 +16x2 + 32

√2x+ 32

9= 64

⇒ 9x2 + 16x2 + 32√2x+ 32 = 576

⇒ 25x2 + 32√2x− 544 = 0

Using

x± =−b±

√b2 − 4ac

2a

⇒ x± =−32

√2±

√2048 + 54400

50

⇒ x± =−32

√2± 237.588

50⇒ x = 3.847 or − 5.657 km

The x = −5.657 km is the initial (−5.657 ≃ −4√2). Hence shipA travels 4

√2+3.847 ≃

9.504 km in the i-direction. The speed of A relative to B in this direction is 24 km/hr.Using


relative speed

⇒ time =9.504

24= 0.396 h = 0.396(60 min) = 23.76 min ≈ 24 min



(i)

VA = −p cos 45◦ i− p sin 45◦ j

⇒×√2/

√2VA = −p

√2

2i− p

√2

2j

VB = −8 i

Using

VAB = VA −VB

⇒ VAB =

(8− p

√2

2

)i− p

√2

2j

!= −2 i− 10 j

⇒ p√2

2= 10

⇒ p =20√2

=×√2/

√2

20√2

2= 10

√2

(ii) Initially

RA = 220√2 cos 45◦ i+ 220 sin 45◦ j = 220 i+ 220 j , and

RB = 136 i

Using

RAB = RA −RB

⇒ RAB = 84 i+ 220 j

Figure 23: R denotes RAB. D is the distance A has to travel relative to B up to the instantthe cars are d-close together. θ is the angle between V and R.

As a line VAB =: V has slope −10/− 2 = 5, (84, 220) ∈ V and using

L ≡ y − y1 = m(x− x1)

⇒ V ≡ y − 220 = 5(x− 84)

⇒ V ≡ 5x− y − 200 = 0


Using

d =|ax+ by + c|√

a2 + b2=

|c|√a2 + b2

⇒ d =200√26

m

The slope of RAB =: R as a line is of slope 220/84 = 55/21. Using

tan θ = ± m1 −m1

1 +m1m2

⇒ tan θ = ± 5− 55/21

1 + 275/21= ± 50/21

296/21=

θ∈[0,90◦]

25

148

Now

tan θ =d

D

⇒ D =d

tan θ=

8

��200√26

.148

��25

⇒ D = 592

√2

13

Using


relative speed

⇒ time = 592

√2

13.

1√4 + 100

=296

13s

Using

s = ut

⇒ sA = 10√2

(296

13

)≃ 322.006 m

But A is initially 220√2 ≃ 311.127 m from the intersection; hence A is now 322 −

311.1 = 10.9 m≈ 11 m from the intersection.


(i) Using

VAB = VA −VB

⇒ VQP = i− 8 j

(ii) As a line VQP =: V has slope −8. Initially RQP = 20 i + 40 j; hence (20, 40) ∈ V .Using

L ≡ y − y1 = m(x− x1)

⇒ V ≡ y − 40 = −8(x− 20)

⇒ V ≡ y − 40 = −8x+ 160

⇒ V ≡ 8x+ y − 200 = 0


Using

d =|ax+ by + c|√

a2 + b2=

(x,y)=(0,0)

|c|√a2 + b2

⇒ d =200√65

≈ 25 m

(iii)

Figure 24: R denotes RQP . D is the distance Q has to travel relative to P up to the instantthe particles are d-close together. θ is the angle between V and R.

Using, noting as a line RAB =: R has slope 40/20 = 2;

tan θ = ± m1 −m2

1 +m1m2

⇒ tan θ = ±−8− 2

1− 16=

θ∈[0,90◦]

2

3

From Figure 24;

tan θ =d

D

⇒ D =d

tan θ= 25

3

2

Using


relative speed

⇒ time =75

2

1√65

≈ 4.6 s


•

VA = 7.5 i

VB = 10 cos 60◦ i+ 10 sin 60◦ j

⇒ VB = 5 i+ 5√3 j

⇒ VAB = VA −VB

⇒ VAB = 2.5 i− 5√3 j


Figure 25: The model triangle for 60◦

• Initially

RAB = −375 i

Figure 26: D is the distance A has to travel relative to B up to the instant the particles ared-close together.

From VAB = 2.5 i− 5√3 j, tan θ = (�5

√3)/(�5/2) = 2

√3.

From Figures 27 and 26,

cos θ =1√13

=D

375

⇒ D =375√13

m

Using


relative speed

⇒ time =375√13

.1√

2.52 + (5√3)2

⇒ time =375/

√13√

254+ 75




(i) Using

s = ut

sB = 10(0.8) = 8m

sA = 10(0.4) = 4 m

(ii) Consider the following diagram:

Figure 28: Straight away it should be clear A = 30◦, h = 10 sin 60◦ and a = 10 cos 30◦.

Now a = 10 sin 60◦ = 5 m (see Figure 25). Hence b = 3 m and thus α = arctan(√7/3).

(iii) From Figure 28, c =√7, h = 10 sin 60◦ = 10

√3/2 = 5

√3 and

r = h− c = 5√3−

√7


Figure 29: The model triangle for α.

Now

VG = −2

5cosα i− 2

5sinα j

⇒ VG = − �2

5.32

�2

i− �2

5.

√7

2

�4

j

⇒ VG = − 3

10i−

√7

10j

VB = −4

5i

Using

VGB = VG −VB =

(4

5− 3

10

)i−

√7

10j

⇒ VGB = VG −VB =1

2i−

√7

10j

Now

Figure 30: Clearly B = θ and so d = (5√3−

√7) cos θ.

tan θ =−j component of VGB

i component of VGB

=

√7

10.2 =

√7

5


Figure 31: Model Triangle for θ.

d = (5√3−

√7).

5

4√2≈ 5.316 m


To cross the river in the shortest time she must head straight across:

VWC = u j

VC = v i

⇒ VW = VWC +VC

⇒ VW = v i+ u j

The time taken to cross is:distance in the j direction

speed in the j direction

⇒ 10 =d

u⇒ d = 10u

In order to cross by the shortest path she must head upstream at an angle θ as shown tocounteract the current:

VWC = −u sin θ i+ u cos θ j

To counteract the current:

u sin θ = v

⇒ sin θ =v

uNow using

t =distance in the j direction


⇒ t =d

u cos θ

⇒ t = (10u)�u

�u√u2 − v2

=10u√u2 − v2


Figure 32: The woman must head upstream at a speed u at an angle θ to counteract thecurrent.



(i) If the wind blows from SE then it blows in a NW direction:

VW = −18 cos 45◦ i+ 18 sin 45◦ j

⇒ VW = −9√2 i+ 9

√2 j

In order for the bird to fly due North, it must head at an angle θ against the wind asshown:

VBW = 22 sin θ i+ 22 cos θ j

⇒ VB = (22 sin θ − 9√2) i+ (22 cos θ + 9

√2)

To counteract the current:

22 sin θ = 9√2

⇒ sin θ =9√2

22

⇒ θ = arcsin(9√2/22) ≈ 35◦


Figure 34: The bird must head into the wind at a speed u at an angle θ to counteract thewind.

(ii)

Figure 35: The model triangle for θ. Using Pythagoras Theorem, the adjacent side is√222 − 81(2) =

√322

Now using

t =distance in the j direction


⇒ t =250

22 cos θ + 9√2

⇒ t =250√

322 + 9√2≃ 8.1507 ≈ 8.15 s


5 Newton’s Laws and Connected Particles

5.1 LC HL 2009: [Part (a)]

(i) The force diagrams are:

Figure 36: The force diagrams for particles A & B

Hence

10g − T = 10a (62)

T − 5g = 5a (63)

⇒ 5g = 15a

⇒ a =g

3m/s2 (64)

Using

v2 = u2 + 2as (65)

⇒ v =√u2 + 2as

⇒ v =√

2 (g/3) =

√2g

3

(ii) When particle A hits the ground, particle B is 1 m above ground and travelling at thesame speed as particle A. When particle A hits the ground the string is no longer tautand particle B travels freely under gravity. Hence the speed of B is a maximum whenv = 0:

s =v2 − u2

2a

⇒ s =−2g/3

−2g=

1

3m

Hence particle B reaches a height 4/3 m above the ground.


5.2 LC HL 2008: [Part (a)]

The force diagrams are as follows:

Figure 37: Note that there are two tensions acting on pulley A. Also the acceleration of theparticle is twice that of pulley A because if pulley A is raised a distance x, the particle willbe lowered a distance 2x.

Hence

2T −mg = ma (66)

m1g − T = 2m1a (67)

⇒ 2m1g − 2T = 4m1a

⇒ 2m1g −mg = 4m1a+ma

⇒ (2m1 −m)g = a(m+ 4m1)

⇒ a =(2m1 −m)g

4m1 +m(68)

�


5.3 LC HL 2006: [Part (a)]

(i) The force diagrams:

Hence

T − 2

5g =

2

5a (69)

g

2− T =

a

2(70)

⇒ 9

10g =

g

10

⇒ a =g

9m/s2

(ii) First find the speed when the 0.5 kg mass strikes the horizontal surface. Using

v2 = u2 + 2as (71)

⇒ v =√u2 + 2as =

√2g

9m/s

Once the 0.5 kg particle strikes the horizontal surface, the 0.4 kg particle acts asa projectile from this point and the string returns to taut when the 0.4 kg particlereturns back down to this height. Now the constant acceleration is −g, the initialspeed is

√2g/9 and by symmetry the final speed is −

√2g/9. Now using

v = u+ at

⇒ t =v − u

a

⇒ t =−√

2g9−√

2g9

−g

⇒ t =2√

2g9√

g2=

√8

9g≈ 0.30 s


5.4 LC HL 2005: [Part (a)]


Figure 38: Note R = 4g and hence µR = g.

Hence

T − g = 4a (72)

8g − T = 8a (73)

⇒ 7g = 12a

⇒ a =7

12g m/s2

⇒ T = g + 4a = g +7

3g =

10

3g N

(ii) The force diagram is:


Now to add these put them in an i-j basis:

F = −T i− T j

⇒ F = −10

3g i− 10

3g j


5.5 LC HL 2004: [Part (a)]


Hence

2mg − T = 2ma (74)

T −mg = ma (75)

⇒ mg = 3ma

⇒ a =g

3m/s2

(ii) When the speed of the first particle is v, using

v2 = u2 + 2as

⇒ s =v2 − u2

2a

⇒ s =3v2

2g

However the second particle will travel the same distance up as the first particle travelsthis distance down:

∴ vertical separation =3v2

g(76)


5.6 LC HL 2003: [Part (a)]

The force diagrams are:

Hence

T − g = a (77)

S − T = 3a (78)

6g − S = 6a (79)

⇒ 5g = 10a

⇒ a =g

2m/s2 (80)


5.7 LC HL 2000: [Part (a)]

The force diagrams are:


Hence

T − g = 5a (81)

3g − T = 3a (82)

⇒ 2g = 8a

⇒ a =g

4

Using

s = ut+1

2at2

⇒ s =1

2

(g4

)4 =

g

2m

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Problem Sets: Solutions · PDF file04.02.2010 · LC Applied Maths Problem Set Solutions 2 1.1.2 Problem Show that the following simpliﬁes to a constant when x ̸= 2 3x−5 x−2