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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)
LC Applied Maths Problem Set Solutions 3
(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.
LC Applied Maths Problem Set Solutions 4
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◦.
LC Applied Maths Problem Set Solutions 5
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◦.
LC Applied Maths Problem Set Solutions 6
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.
LC Applied Maths Problem Set Solutions 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)
LC Applied Maths Problem Set Solutions 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)
LC Applied Maths Problem Set Solutions 9
⇒ 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)
LC Applied Maths Problem Set Solutions 10
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.
LC Applied Maths Problem Set Solutions 11
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
LC Applied Maths Problem Set Solutions 12
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 =?
LC Applied Maths Problem Set Solutions 13
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
Using v2 = u2 + 2as;
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.
LC Applied Maths Problem Set Solutions 14
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 =?
Using v2 = u2 + 2as;
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
Using v2 = u2 + 2as;
s =v2 − u2
2a
⇒ s =0− 225
−5=
225
5= 45 m.
LC Applied Maths Problem Set Solutions 15
2.5 Problem: LC OL 1983
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
LC Applied Maths Problem Set Solutions 16
(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.
2.6 Problem: LC OL 1982
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.
LC Applied Maths Problem Set Solutions 17
2.7 Problem: LC OL 1981
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
LC Applied Maths Problem Set Solutions 18
2.8 Problem: LC OL 1980
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
LC Applied Maths Problem Set Solutions 19
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)
LC Applied Maths Problem Set Solutions 20
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
2.10 Problem: LC HL 2008
• 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
LC Applied Maths Problem Set Solutions 21
• 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
LC Applied Maths Problem Set Solutions 22
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.
LC Applied Maths Problem Set Solutions 23
2.11 Problem: LC HL 2007
• 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
LC Applied Maths Problem Set Solutions 24
• – 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.
LC Applied Maths Problem Set Solutions 25
2.12 Problem: LC HL 2006
• 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
LC Applied Maths Problem Set Solutions 26
• – 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.
LC Applied Maths Problem Set Solutions 27
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.
LC Applied Maths Problem Set Solutions 28
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
2.14 Problem: LC HL 2004 [Part (a)]
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.
LC Applied Maths Problem Set Solutions 29
(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.
LC Applied Maths Problem Set Solutions 30
2.15 Problem: LC HL 2003
(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
�
LC Applied Maths Problem Set Solutions 31
(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.
LC Applied Maths Problem Set Solutions 32
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.
�
2.16 Problem: LC HL 2002
(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
LC Applied Maths Problem Set Solutions 33
(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
�
LC Applied Maths Problem Set Solutions 34
2.17 Problem: LC HL 2001
(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
LC Applied Maths Problem Set Solutions 35
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
)�
LC Applied Maths Problem Set Solutions 36
(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
�
2.18 Problem: LC HL 2000
(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.
LC Applied Maths Problem Set Solutions 37
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)
LC Applied Maths Problem Set Solutions 38
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
LC Applied Maths Problem Set Solutions 39
(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
LC Applied Maths Problem Set Solutions 40
2.20 Problem: LC HL 1998
(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
LC Applied Maths Problem Set Solutions 41
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
LC Applied Maths Problem Set Solutions 42
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
LC Applied Maths Problem Set Solutions 43
3 Projectiles
3.1 Problem: LC HL 2009 [Part (a)]
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
LC Applied Maths Problem Set Solutions 44
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)
LC Applied Maths Problem Set Solutions 45
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)
�
3.3 Problem: LC HL 2007 [Part (a)]
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◦
LC Applied Maths Problem Set Solutions 46
3.4 Problem: LC HL 2006 [Part (a)]
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)
LC Applied Maths Problem Set Solutions 47
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
LC Applied Maths Problem Set Solutions 48
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
3.6 Problem: LC HL 2003: [Part (a)]
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)
LC Applied Maths Problem Set Solutions 49
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◦
3.7 Problem: LC HL 2002: [Part (a)]
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
�
LC Applied Maths Problem Set Solutions 50
(ii) Hence α is such that:
64 sin2 α− 12 = 20
⇒ 64 sin2 α = 32
⇒ sin2 α =1
2
⇒ sinα =1√2
⇒ θ = 45◦
3.8 Problem: LC HL 2001: [Part (a)]
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
LC Applied Maths Problem Set Solutions 51
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
LC Applied Maths Problem Set Solutions 52
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
LC Applied Maths Problem Set Solutions 53
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
�
LC Applied Maths Problem Set Solutions 54
3.11 Problem: LC HL 2008: [Part(b)]
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
LC Applied Maths Problem Set Solutions 55
3.12 Problem: LC HL 2006: [Part (b)]
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
Range, R, means sx when sy = 0.
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
LC Applied Maths Problem Set Solutions 56
3.13 Problem: LC HL 2005: [Part (b)]
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
Range, R, means sx when sy = 0.
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)
LC Applied Maths Problem Set Solutions 57
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
�
3.14 Problem: LC HL 2003: [Part (b)]
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
Range, R, means sx when sy = 0.
LC Applied Maths Problem Set Solutions 58
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)
3.15 Problem: LC HL 1997: [Part (b)]
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
LC Applied Maths Problem Set Solutions 59
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
LC Applied Maths Problem Set Solutions 60
3.16 Problem: LC HL 2007: [Part(b)]
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)
LC Applied Maths Problem Set Solutions 61
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
�
LC Applied Maths Problem Set Solutions 62
3.17 Problem: LC HL 2004: [Part(b)]
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
�
3.18 Problem: LC HL 2002: [Part(b)]
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
LC Applied Maths Problem Set Solutions 63
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
�
LC Applied Maths Problem Set Solutions 64
3.19 Problem: LC HL 2000: [Part(b)]
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 )
LC Applied Maths Problem Set Solutions 65
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
LC Applied Maths Problem Set Solutions 66
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
3.21 Problem: LC HL 1999: [Part (b)]
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)
LC Applied Maths Problem Set Solutions 67
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
LC Applied Maths Problem Set Solutions 68
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;
LC Applied Maths Problem Set Solutions 69
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
LC Applied Maths Problem Set Solutions 70
4.2 Problem: F.A.M. Q.4.A.7
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.
4.3 Problem: F.A.M. Q.4.A.8
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. �
LC Applied Maths Problem Set Solutions 71
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
4.4 Problem: F.A.M. Q.4.A.9
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
LC Applied Maths Problem Set Solutions 72
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.
4.5 Problem: F.A.M. Q.4.A.10
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.
LC Applied Maths Problem Set Solutions 73
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
LC Applied Maths Problem Set Solutions 74
4.6 Problem: F.A.M. Q.4.A.11
(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
LC Applied Maths Problem Set Solutions 75
4.7 Problem: F.A.M. Q.4.A.12
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
LC Applied Maths Problem Set Solutions 76
4.8 Problem: F.A.M. Q.4.A.13
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
LC Applied Maths Problem Set Solutions 77
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
LC Applied Maths Problem Set Solutions 78
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.
LC Applied Maths Problem Set Solutions 79
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
time =relative distance
relative speed
⇒ time =1120
|VAB|=
|VAB |=25
224
5s
LC Applied Maths Problem Set Solutions 80
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.
4.10 Problem: LC HL 2008: [Part(a)]
(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
Figure 21: The model triangle for θ.
cos θ =3
5=
L
100
⇒ L =300
5= 60 m
LC Applied Maths Problem Set Solutions 81
Using
time =relative distance
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
4.11 Problem: LC HL 2007: [Part(a)]
(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
LC Applied Maths Problem Set Solutions 82
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
time =relative distance
relative speed
⇒ time =9.504
24= 0.396 h = 0.396(60 min) = 23.76 min ≈ 24 min
LC Applied Maths Problem Set Solutions 83
4.12 Problem: LC HL 2005: [Part(b)]
(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
LC Applied Maths Problem Set Solutions 84
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
time =relative distance
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.
4.13 Problem: LC HL 2004: [Part(b)]
(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
LC Applied Maths Problem Set Solutions 85
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
time =relative distance
relative speed
⇒ time =75
2
1√65
≈ 4.6 s
4.14 Problem: LC HL 2003: [Part(b)]
•
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
LC Applied Maths Problem Set Solutions 86
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
time =relative distance
relative speed
⇒ time =375√13
.1√
2.52 + (5√3)2
⇒ time =375/
√13√
254+ 75
LC Applied Maths Problem Set Solutions 87
Figure 27: The model triangle for θ.
4.15 Problem: LC HL 2006: [Part(b)]
(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
LC Applied Maths Problem Set Solutions 88
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
LC Applied Maths Problem Set Solutions 89
Figure 31: Model Triangle for θ.
d = (5√3−
√7).
5
4√2≈ 5.316 m
4.16 Problem: LC HL 2005: [Part (a)]
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
speed in the j direction
⇒ t =d
u cos θ
⇒ t = (10u)�u
�u√u2 − v2
=10u√u2 − v2
LC Applied Maths Problem Set Solutions 90
Figure 32: The woman must head upstream at a speed u at an angle θ to counteract thecurrent.
Figure 33: The model triangle for θ.
4.17 Problem: LC HL 2004: [Part (a)]
(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◦
LC Applied Maths Problem Set Solutions 91
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
speed in the j direction
⇒ t =250
22 cos θ + 9√2
⇒ t =250√
322 + 9√2≃ 8.1507 ≈ 8.15 s
LC Applied Maths Problem Set Solutions 92
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.
LC Applied Maths Problem Set Solutions 93
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)
�
LC Applied Maths Problem Set Solutions 94
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
LC Applied Maths Problem Set Solutions 95
5.4 LC HL 2005: [Part (a)]
(i) The force diagrams are:
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:
Figure 39: Note R = 4g and hence µR = g.
Now to add these put them in an i-j basis:
F = −T i− T j
⇒ F = −10
3g i− 10
3g j
LC Applied Maths Problem Set Solutions 96
5.5 LC HL 2004: [Part (a)]
(i) The force diagrams are:
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)
LC Applied Maths Problem Set Solutions 97
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)
LC Applied Maths Problem Set Solutions 98
5.7 LC HL 2000: [Part (a)]
The force diagrams are:
Figure 40: Note R = 5g and hence µR = g.
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