APPLIED MATHEMATICS
MECHANICS
TOPIC 1: MOTION ON SMOOTH INCLINED PLANE
Introduction:
𝐴
𝑃
𝛼
𝐵
𝑔
𝐶
•
•
P - particle of mass, 𝑚.
ABC – vertical cross-section of smooth plane inclined at an angle, 𝛼.
•
•
AB – represents a line of greatest slope on the plane.
The acceleration 𝑔 of the particle P falling freely (sliding) can be resolved into two
components:
e.g.
(i)
𝒈 𝐬𝐢𝐧 𝜶 - down along the plane
(ii)
𝒈 𝐜𝐨𝐬 𝜶 - perpendicular to the plane.
The plane prevents motion perpendicular to itself, so the particle, P can only move downwards
(or upwards) along the plane with acceleration 𝒈 𝐬𝐢𝐧 𝜶.
If ̅̅̅̅
𝐴𝐵 = 𝑙 metres, the speed acquired in sliding from rest to 𝑣 (from A to B) is obtained as
follows:
From 3rd equation of linear motion:
𝑣 2 = 𝑢2 + 2𝑎𝑠 but 𝑢 = 0 𝑚𝑠 −1 ; 𝑎 = 𝑔 sin 𝛼 and 𝑠 = 𝑙
∴ 𝑣 2 = 2𝑔𝑙 sin 𝛼
∴ 𝑣 = √2𝑔ℎ where ℎ = 𝑙 sin 𝛼 = ̅̅̅̅
𝐴𝐶
which is the same as the speed acquired falling freely through a vertical height equal to that
of the plane.
Prepared by Elsam Senjobe @ GHS 2021.
1
Time:
The time taken to slide down AB is not the same as that taken to fall freely through AC.
e.g.
the time taken to slide down AB is obtained from the equation:
1
𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2 .
1
Implying: 𝑙 = 2 𝑔 sin 𝛼 𝑡 2 i.e. 𝑢 = 0 𝑚𝑠 −1 and 𝑠 = 𝑙.
𝟐𝒍
∴ 𝒕 = √𝒈 𝐬𝐢𝐧 𝜶.
NOTE:
the time taken to fall freely from A to C through height, ℎ is given by:
1
2ℎ
2 𝑙 sin 𝛼
ℎ = 2 𝑔𝑡1 2 ∴ 𝑡1 = √ 𝑔 = √
𝑔
.
∴ 𝑡 ≠ 𝑡1 .
EXAMPLES:
1. A particle is projected (a) upwards (b) downwards on a plane inclined to the horizontal at
30𝑜 .if the initial velocity is 5 𝑚𝑠 −1 in each case, find the distances and the velocities
acquired after 4 𝑠.
Solution:
For motion up the plane:
30𝑜
Using 1st equation of linear motion:
𝑣 = 𝑢 + 𝑎𝑡
∴ 𝑣 = 𝑢 − (𝑔 sin 30𝑜 )𝑡
1
∴ 𝑣 = 5 − 9.8 × 2 × 4
∴ 𝑣 = −𝟏𝟒. 𝟔 𝒎𝒔−𝟏 .
Using 2nd equation of motion:
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
1
∴ 𝑠 = 5 × 4 + (−9.8 × ) × 42 = 20 − 39.2 = −𝟏𝟗. 𝟐 𝒎
2
2
Conclusion: After 4 s, the body is moving down the plane with velocity of 14.6 𝑚𝑠 −1
and is 19.2 𝑚 below the point of projection.
Prepared by Elsam Senjobe @ GHS 2021.
2
For downward motion:
1
𝑠 = 𝑢𝑡 + 2 (𝑔 sin 𝑠300 )𝑡 2
1
30𝑜
1
∴ 𝑠 = 5 × 4 + 2 × 9.8 × 2 × 42 = 20 + 39.2
∴ 𝑠 = 𝟓𝟗. 𝟐 𝒎
Also 𝑣 = 𝑢 + 𝑔𝑡 sin 30𝑜
1
∴ 𝑣 = 5 + 9.8 × 4 × 2 = 5 + 19.6 = 𝟐𝟒. 𝟔 𝒎𝒔−𝟏 .
Conclusion: After 4 s, the body is moving down the plane with velocity of 24.6 𝑚𝑠 −1
and is 59.2 𝑚 below the point of projection.
2. A particle is projected with a velocity of 20 𝑚𝑠 −1 up a smooth inclined plane of
inclination 30𝑜 . Find the distance described up the plane, and the time that elapses
before it comes to rest.
Solution:
𝒖 = 𝟐𝟎 𝒎𝒔−𝟏 , 𝒗 = 𝟎 𝒎𝒔−𝟏
𝒗𝟐 = 𝒖𝟐 − 𝟐𝒈 𝒔𝒊𝒏 𝟑𝟎𝒐 × 𝒔
𝟏
∴ 𝟎 = (𝟐𝟎)𝟐 − 𝟐 × 𝟗. 𝟖 × 𝟐 × 𝒔
∴𝒔=
30𝑜
𝟒𝟎𝟎
𝟗.𝟖
= 𝟒𝟎. 𝟖𝟏𝟔 𝒎.
Also:
𝒗 = 𝒖 − 𝒈 𝒔𝒊𝒏 𝟑𝟎𝒐 × 𝒕
𝟏
∴ 𝟎 = 𝟐𝟎 − 𝟗. 𝟖 × 𝟐 × 𝒕
∴𝒕=
𝟐𝟎×𝟐
𝟗.𝟖
= 𝟒. 𝟎𝟖𝟏𝟔 𝒔
3. A particle is sliding down a smooth plane, 3.6 𝑚 long, acquiring a velocity of 6 𝑚𝑠 −1 from
rest. Find the inclination of the plane.
Solution:
𝑣 2 = 𝑢2 + 2𝑔𝑙 sin 𝛼
∴ 62 = 0 + 2 × 9.8 × 3.6 sin 𝛼
36
∴ sin 𝛼 = 70.56
36
𝛼
Prepared by Elsam Senjobe @ GHS 2021.
∴ 𝛼 = sin−1 (70.56) = 30.68𝑜 .
3
4. A particle slides from rest down a smooth inclined plane which is 12 m long and 2.7 m
high. What is its velocity on reaching the bottom of the plane, and how long does it take
to get there?
Solution:
𝛼
𝑙 = 12 𝑚
𝑣 2 = 𝑢2 + 2𝑙𝑔 sin 𝛼
∴ 𝑣 2 = 𝑢2 + 2𝑔ℎ
𝑠𝑖𝑛𝑐𝑒 ℎ = 𝑙 sin 𝛼
∴ 𝑣 2 = 0 + 2 × 9.8 × 2.7
∴ 𝑣 2 = 52.92
∴ 𝒗 = 𝟕. 𝟐𝟕𝟒𝟔𝒎𝒔−𝟏 .
ℎ = 2.7 𝑚
𝑣
𝑣
𝑙
From 𝑣 = 𝑢 + 𝑔𝑡 sin 𝛼
⇒ 𝑡 = 𝑔 sin 𝛼 = 𝑔 × ℎ
7.2746 × 12
∴𝑡=
= 𝟑. 𝟐𝟗𝟗𝟏 𝒔
9.8 × 2.7
TOPIC 2: FORCE, NEWTON’S LAWS OF MOTION ON LEVEL AND INCLINED PLANES
A force is a pull or push that produces or tends to produce a change in a body’s state of rest or
uniform motion in a straight line.
SI unit of a force:
The SI unit of force is the newton (N).
The newton is that force which produces an acceleration of 1 𝑚𝑠 −2 on a mass of 1 𝑘𝑔.
NEWTON’S LAWS OF MOTION:
LAW 1:
Everybody continues in its state of rest or uniform motion in a straight line
unless it is compelled by an external force to do otherwise.
LAW 2:
The rate of change of momentum is proportional to the force acting on the
body and takes place in a straight line along which the force acts.
𝐹𝛼
(𝑚𝑣 − 𝑚𝑢)
𝑡
⇒𝐹=𝑘𝑚
(𝑣 − 𝑢)
= 𝑘 𝑚𝑎
𝑡
Prepared by Elsam Senjobe @ GHS 2021.
4
If 𝐹 = 1𝑁, 𝑚 = 1 𝑘𝑔 ; 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎 = 1𝑚𝑠 −2 then 𝑘 = 1 and
𝑭 = 𝒎𝒂.
NOTE: Acceleration, 𝑎 always occurs in the same direction of the force, 𝐹 producing it.
LAW 3:
Action and reaction are equal but opposite.
Resultant force on a moving body:
If a number of forces act on a body, their resultant is a single force that produces the
same effect as they would produce on the body.
Examples:
1. UPWARD MOTION ON A HORIZONTAL FLOOR OR SUPPORT:
𝑅
𝒂
𝑹 = reaction between the body and the level
plane.
𝒂 = upward acceleration.
𝑚𝑔 = weight of the body.
In this case, 𝑅 > 𝑚𝑔 and the resultant force, 𝐹 is
given by;
𝑅 − 𝑚𝑔 = 𝐹
∴ 𝑹 − 𝒎𝒈 = 𝒎𝒂.
𝑚𝑔
NOTE: 𝑅 = 𝑚(𝑔 + 𝑎), and the body’s resultant weight seems to have increased.
2. DOWNWARD MOTION ON A HORIZONTAL FLOOR OR SUPPORT:
𝑹 = reaction between the body and the level
𝑅
plane.
𝒂 = downward acceleration.
𝒂
𝑚𝑔
𝑚𝑔 = weight of the body.
In this case, 𝑚𝑔 > 𝑅 and the resultant force, 𝐹 is
given by;
𝑚𝑔 − 𝑅 = 𝐹
∴ 𝒎𝒈 − 𝑹 = 𝒎𝒂.
NOTE: 𝑅 = 𝑚(𝑔 − 𝑎), and the body’s resultant weight seems to have decreased.
If 𝑔 = 𝑎, then 𝑅 = 0 i.e. there is no reaction on the body by the plane. The body is said
to be weighless.
Prepared by Elsam Senjobe @ GHS 2021.
5
3. MOTION OF AN ENGINE ON A LEVEL PLANE/GROUND
𝑅
𝒂
RESISTANCE, 𝐹𝑟
DRIVING FORCE (TRACTIVE FORCE), P
𝑚𝑔
When the engine of mass 𝑚 is accelerating with an acceleration of 𝑎 𝑚𝑠 −2 then;
𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 − 𝑅𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
∴ 𝑷 − 𝑭𝒓 = 𝒎𝒂
When acceleration 𝑎 = 0 𝑚𝑠 −2 , then 𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑅𝑖𝑠𝑡𝑎𝑛𝑐𝑒 i.e. 𝑃 = 𝐹𝑟 .
Power of the engine:
𝑷𝒐𝒘𝒆𝒓 =
𝒅𝒓𝒊𝒗𝒊𝒏𝒈 𝒇𝒐𝒓𝒄𝒆 × 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒎𝒐𝒗𝒆𝒅
= 𝒅𝒓𝒊𝒗𝒊𝒏𝒈 𝒇𝒐𝒓𝒄𝒆 × 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚
𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏
4
MOTION OF AN ENGINE ON AN INCLINED PLANE
(a)
UP THE PLANE
𝛼
𝑚𝑔
Resolving forces along the plane:
𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 − (𝑚𝑔 sin 𝛼 + 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒) = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
Therefore:
𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑔 sin 𝛼 + 𝑅𝑒𝑠𝑖𝑠𝑖𝑡𝑎𝑛𝑐𝑒 + 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
Prepared by Elsam Senjobe @ GHS 2021.
6
Power of the engine:
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒 = 𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ( in watts).
(b)
DOWN THE PLANE
𝛼
𝑚𝑔
Resolving forces along the plane:
𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 + 𝑚𝑔 sin 𝛼 − 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
Therefore:
𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 + 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 − 𝑚𝑔 sin 𝛼
Power of the engine:
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒 = 𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ( in watts).
EXAMPLES:
1. A rifle bullet passes through two planks in succession and the average resistance of the
second plank is 50% more than that of the first. The initial velocity is 800 𝑚𝑠 −1 and the
bullet loses 160 𝑚𝑠 −1in passing through each plank. Show that the thickness of the
planks are as 27: 14.
SOLUTION:
𝑠1
𝑢 = 800 𝑚𝑠 −1
Prepared by Elsam Senjobe @ GHS 2021.
𝑣1 = 640 𝑚𝑠 −1
𝑠2
𝑣2 = 480 𝑚𝑠 −1
7
Motion of bullet in 1st plank:
𝑣1 2 = 𝑢2 − 2𝑎1 𝑠1
∴ 6402 = 8002 − 2𝑎1 𝑠1
∴ 𝑠1 =
8002 −6402
2𝑎1
=
(800+640)(800−640)
2𝑎1
=
Motion of bullet in 2nd plank:
𝑣2 2 = 𝑣1 2 − 2𝑎2 𝑠2
∴ 4802 = 6402 − 2𝑎2 𝑠2
∴ 𝑠2 =
6402 −4802
2𝑎2
=
(640+480)(640−480)
2𝑎2
=
1440×160
2𝑎1
1120×160
2𝑎2
…………………………(i)
……………………………(ii)
Resistance force in the first plank = 𝐹1 = 𝑚𝑎1
Resistance force in the second plank = 𝐹2 = 𝑚𝑎2
150
3
3
But 𝐹2 = 100 𝐹1
∴ 𝑚𝑎2 = 2 𝑚𝑎1
∴ 𝑎2 = 2 𝑎1
, where 𝑚 = mass of bullet.
Now:
𝑠1 1440 × 160
2𝑎2
9𝑎2
=
×
=
𝑠2
2𝑎1
1120 × 160 7𝑎1
3
𝑠1 9 2 𝑎1 27
∴
= ×
=
𝑠2 7
𝑎1
14
∴ 𝑠1 : 𝑠2 = 27: 14
2. A ship of 10,000 Mg slows, with engines stopped, from 12 kmh-1 to 10 kmh-1 in a distance
of 90 m. Assuming the resistance to be uniform, calculate its value in newtons.
SOLUTION:
Mass of ship, 𝑚 = 10000 × 103 𝑘𝑔
12×1000
10
Initial speed, 𝑢 = 3600 = 3 𝑚𝑠 −1
Final speed, 𝑣 =
10×1000
3600
=
25
Distance moved, 𝑠 = 90 𝑚
9
𝑚𝑠 −1
From 2nd equation of linear motion:
𝑢2 − 𝑣 2 (𝑢 + 𝑣)(𝑢 − 𝑣)
𝑣 2 = 𝑢2 − 2𝑎𝑠 ∴ 𝑎 =
=
2𝑠
2𝑠
The resistance force, 𝐹 = 𝑚𝑎
Prepared by Elsam Senjobe @ GHS 2021.
8
∴ 𝐹 = 10000 × 103 ×
10 25 10 25
( 3 + 9 )( 3 − 9 )
2 × 90
=
(55)(5)
× 107
2 × 90 × 9 × 9
∴ 𝐹 = 1.8861 × 105 𝑁
3. A truck is found to travel with uniform speed down a slope which falls 1 m vertically for
every 112 m length of the slope. If the truck starts from the bottom of the slope with a
speed of 18 kmh-1, how far up will it travel before coming to rest.
SOLUTION:
Downward motion:
1
112
𝛼
𝑚𝑔 sin 𝛼 − 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑎 ……………………(i)
1
But sin 𝛼 =
; 𝑎 = 0 𝑚𝑠 −2 , therefore (i) becomes:
112
1
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑐𝑒 = 𝑚𝑔 × 112 ………………………..(ii)
Upward motion:
1
112
𝛼
𝑅𝑒𝑡𝑎𝑟𝑑𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑔 sin 𝛼 + 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
1
1
∴ 𝑚𝑎 = 𝑚𝑔 ×
+ 𝑚𝑔 ×
112
112
𝑔
∴ 𝑎 = 56 =
9.8
56
= 0.175 𝑚𝑠 −2 is the retardation.
From 3rd equation of motion: 𝑣 2 = 𝑢2 − 2𝑎𝑠
but 𝑢 = 18 𝑘𝑚ℎ−1 and 𝑣 = 0 𝑚𝑠 −1
Prepared by Elsam Senjobe @ GHS 2021.
9
18 × 1000 2
) − 2 × 0.175 × 𝑠
∴0=(
3600
∴𝑠=
25
= 71.4286 𝑚
0.35
4. Find in newtons per kilogram the force exerted by the brakes of a train travelling at
60 kmh-1 which will bring it to rest in half a kilometer, and find the time during which the
brakes act.
SOLUTION:
60×1000
50
Initial velocity of the train, 𝑢 = 60 𝑘𝑚ℎ−1 = 3600 = 3 𝑚𝑠 −1
Final velocity of train, 𝑣 = 0 𝑚𝑠 −1
1
1000
Distance covered, 𝑠 = 2 𝑘𝑚 = 2 = 500 𝑚
Applying 3rd equation of linear motion: 𝑣 2 = 𝑢2 − 2𝑎𝑠
50 2
0 = ( ) − 2𝑎 × 500
3
2500
5
∴ 𝑎 = 9×2×500 = 18 𝑚𝑠 −2 is the retardation.
The retarding force i.e. force exerted by the brakes in 𝑁𝑘𝑔−1 is:
𝐹
𝑚
5
= 𝑎 = 18 𝑚𝑠 −2 .
From the 1st equation of motion: 𝑣 = 𝑢 − 𝑎𝑡
50 5
∴0=
−
×𝑡
3 18
50
18
∴ 𝑡 = 3 × 5 = 60 𝑠 ( 1 𝑚𝑖𝑛)
5. A force equal to the weight of 10𝑔 acts on a mass of 218 g for 5 s. Find the velocity
generated and distance moved in this time.
SOLUTION:
𝑣−𝑢
)
𝑡
10
49
218
But 𝐹 = 𝑚𝑔 = 1000 × 9.8 = 500 and 𝑀 = 218𝑔 = 1000 𝑘𝑔
𝐹 = 𝑀𝑎 = 𝑀 (
Prepared by Elsam Senjobe @ GHS 2021.
10
∴
49
218 𝑣 − 0
(
)
=
500 1000
5
49×5×1000
∴𝑣=
218×500
= 2.2477 𝑚𝑠 −1 .
Distance moved is determine using 2nd equation of linear motion:
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
𝐹
49
1000
49
But 𝐹 = 𝑀𝑎 ∴ 𝑎 = 𝑀 = 500 × 218 = 109
∴𝑠 = 0×5+
∴𝑠=
1 49
×
× 52
2 109
1225
= 5.6193 𝑚
218
6. A body of mass 5 kg, initially at rest on a smooth horizontal surface is pulled along the
surface by a constant force P inclined at 45𝑜 above the horizontal. In the first 5 s of
motion, the body moves a distance of 10 m along the surface. Find
(a)
The acceleration of the body.
(b)
The value of P.
SOLUTION:
𝑃
45𝑜
𝑃 cos 45𝑜
Initial velocity, 𝑢 = 0 𝑚𝑠 −1 ; time 𝑡 = 5 𝑠; distance, 𝑠 = 10 𝑚, mass, 𝑚 = 5 𝑘𝑔
𝑃
Horizontal component of the pulling force, = 𝑃 cos 45𝑜 = .
From 2nd equation of motion:
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
∴ 10 = 0 × 5 + × 𝑎 × 25
2
10 × 2 4
𝑎=
= = 0.8 𝑚𝑠 −2 .
25
5
(b)
Now: 𝐹 = 𝑚𝑎
(a)
Prepared by Elsam Senjobe @ GHS 2021.
√2
11
∴
𝑃
= 5 × 0.8 = 4
√2
∴ 𝑃 = 4√2 𝑁
7. Find the magnitude of the force which, acting on a mass of 1 kg for 5 s causes the mass
to move through 10 m from rest in that time.
SOLUTION:
1
1
𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2 ∴ 10 = 0 × 5 + 2 𝑎 × 25
10 × 2
∴𝑎=
= 0.8 𝑚𝑠 −2
25
But 𝐹 = 𝑚𝑎 ∴ 𝐹 = 1 × 0.8 = 0.8 𝑁
8. The resistance to the motion a train due to friction etc. is equal to
1
160
of the weight of
the train. If the train is travelling on a level road at 72 kmh-1 and comes to the foot of
an incline of 1 in 150 the engine is then cut off, how far will the train go up the incline
before it comes to rest?
SOLUTION:
1
At the foot of the incline, the initial velocity 𝑢 = 72 𝑘𝑚ℎ−1 =
1
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 160 𝑀𝑔 , where 𝑀 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛
1
sin 𝛼 =
150
72×1000
3600
= 20 𝑚𝑠 −1.
𝑅𝑒𝑡𝑎𝑟𝑑𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 + 𝑀𝑔 sin 𝛼
∴ 𝑀𝑎 =
1
1
𝑀𝑔 +
𝑀𝑔
160
150
160 + 150
31
)𝑔 =
∴𝑎=(
𝑔
150 × 160
2400
Final velocity of the train , 𝑣 = 0 𝑚𝑠 −1
Prepared by Elsam Senjobe @ GHS 2021.
12
From 3rd equation of motion: 𝑣 2 = 𝑢2 − 2𝑎𝑠
31
∴ 0 = 202 − 2 ×
𝑔×𝑠
2400
400 × 2400
= 1579.9868 𝑚 ≈ 1.57998 𝑘𝑚
2 × 31 × 9.8
∴𝑠=
∴ 𝑠 ≈ 1.6 𝑘𝑚.
9. A train travelling uniformly on the level at the rate of 80 kmh-1 begins an ascent of
1
in 75. The tractive force that the engine exerts during the ascent is constant and equal
to
1
100
of the weight of the train; the resistance (due to friction etc.) is constant and
equal to
1
150
of the weight of the whole train. Show that the train will come to rest
(standstill) after climbing for 2.5 km.
SOLUTION:
𝑅
𝛼
𝑀𝑔
1
75
1
Driving force, 𝐷 = 100 𝑀𝑔
sin 𝛼 =
1
Resistance, 𝐹𝑟 = 150 𝑀𝑔
Component of weight of the train along the incline, 𝑀𝑔 sin 𝛼 =
𝑀𝑔
75
Resolving forces along the plane:
𝐷 − (𝐹𝑟 + 𝑀𝑔 sin 𝛼) = 𝑀𝑎
∴
1
1
𝑀𝑔
𝑀𝑔 −
𝑀𝑔 −
= 𝑀𝑎
100
150
75
∴𝑎=
(15−10−20)𝑔
1500
𝑔
9.8
= − 100 = 100 𝑚𝑠 −2 i.e. a retardation.
Prepared by Elsam Senjobe @ GHS 2021.
13
Applying the 3rd equation of motion:
𝑣 2 = 𝑢2 + 2𝑎𝑠 where 𝑣 = 0 𝑚𝑠 −1 ; 𝑢 = 80 𝑘𝑚ℎ−1 =
Substituting:
200 2
9.8
) −2×
0=(
×𝑠
9
100
∴𝑠=
80×1000
3600
=
200
9
𝑚𝑠 −1
40000 × 100
= 2,519.52633 𝑚
81 × 2 × 9.8
∴ 𝑠 ≈ 2.5 𝑘𝑚
10. A body of mass 25g is observed to travel in a straight line through 369 cm, 615 cm and
861 cm in successive seconds. Prove that this is consistent with a constant force acting
on the body. What is the value of this force?
SOLUTION:
369 𝑐𝑚
615 𝑐𝑚
25
861 𝑐𝑚
1
Mass of body, 𝑀 = 25𝑔 = 1000 = 40 𝑘𝑔
1
Using 2nd equation of linear motion: 𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2
where 𝑡 = 1 𝑠 in each case, then
1
369 = 𝑢 + 2 𝑎 ………………(i)
Also
1
(369 + 615) = 2𝑢 + 𝑎(22 )
2
∴ 984 = 2𝑢 + 2𝑎
∴ 492 = 𝑢 + 𝑎 ……………(ii)
And
1
(369 + 615 + 861) = 3𝑢 + 𝑎(32 )
2
9
∴ 1845 = 3𝑢 + 𝑎
2
3
∴ 615 = 𝑢 + 2 𝑎 …………….(iii)
Prepared by Elsam Senjobe @ GHS 2021.
14
(𝑖𝑖𝑖) − (𝑖): 615 − 369 = 𝑎
∴ 𝑎 = 246 𝑐𝑚𝑠 −2 = 2.46 𝑚𝑠 −2
From (ii): 𝑢 = 492 − 𝑎 = 492 − 246 = 246 𝑐𝑚𝑠 −1
∴ 𝑢 = 2.46 𝑚𝑠 −1
1
The constant force: 𝐹 = 𝑀𝑎 = 40 × 2.46
∴ 𝐹 = 0.0615 𝑁
11. A certain truck starting from rest on an incline of 1 in 140 acquired a speed of 25 kmh-1
in 10 minutes. Calculate the resistance to the motion of the truck in N per kg mass of
the truck.
SOLUTION:
𝑅
𝛼
𝑀𝑔
sin 𝛼 =
1
25 × 1000 125
; 𝑢 = 0 𝑚𝑠 −1 ; 𝑣 = 25 𝑘𝑚ℎ−1 =
=
𝑚𝑠 −1 ; 𝑡 = 10 𝑚𝑖𝑛𝑠 = 600𝑠
140
3600
18
Resolving forces along the plane:
𝑀𝑔 sin 𝛼 − 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑀𝑎 …………………..(i)
𝑣−𝑢
125
5
But 𝑎 =
=
=
𝑡
8×600
18×24
Therefore, equation (i) becomes:
𝑔
− 𝑎)
140
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑀𝑔 sin 𝛼 − 𝑀𝑎 = 𝑀 (
Prepared by Elsam Senjobe @ GHS 2021.
15
∴
𝐹𝑟
9.8
5
=
−
= 0.07 − 0.011574 = 0.05843 𝑁𝑘𝑔−1
𝑀 140 18 × 24
12. A force equal to the weight of 1 Mg acts for 3 seconds on a mass of 5 Mg. Find the
velocity produced and space passed over.
SOLUTION:
𝐹 = 1𝑀𝑔 = 1 × 102 × 9.8 𝑁 , time 𝑡 = 3 𝑠 and mass, 𝑀 = 5𝑀𝑔 = 5 × 103 𝑘𝑔
𝐹
But 𝐹 = 𝑀𝑎 ∴ 𝑎 = 𝑀 =
∴ acceleration, 𝑎 =
9.8
5
9.8×103
5×103
𝑚𝑠 −1
Using equation: 𝑣 = 𝑢 + 𝑎𝑡
9.8
29.4
∴𝑣 = 0+
×3=
𝑚𝑠 −1 = 5.88 𝑚𝑠 −1
5
5
Distance covered:
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1 9.8
∴𝑠= ×
× 32
2
5
∴ 𝑠 = 8.82 𝑚
LIFT PROBLEMS:
13. A mass of 10 kg rests on a horizontal plane which is made to ascend
(i)
With a constant a velocity of 5 𝑚𝑠 −1 ;
(ii)
With a constant acceleration of 5 𝑚𝑠 −2 ;
Find in each case the reaction of the plane.
SOLUTION:
𝑅
(i)
𝑅 − 𝑀𝑔 = 𝑀𝑎
But 𝑎 = 0 𝑚𝑠 −2
𝑅 = 𝑀𝑔 = 10 × 9.8 = 98 𝑁
10𝑔
Prepared by Elsam Senjobe @ GHS 2021.
16
(ii)
𝑅 − 𝑀𝑔 = 𝑀𝑎
∴ 𝑅 = 𝑀(𝑔 + 𝑎) = 10(9.8 + 5) = 10 × 14.8
∴ 𝑅 = 148 𝑁
14. A man of mass 70 kg, stands on the floor of a lift. Find the reaction of the floor when
the lift is (i) ascending (ii) descending, with a uniform acceleration of 4 ms-2.
SOLUTION
(a)
𝑅
𝑅 − 𝑚𝑔 = 𝑚𝑎
∴ 𝑅 − 70𝑔 = 70 × 4
∴ 𝑅 − 70 × 9.8 = 280
∴ 𝑅 = 280 + 686
∴ 𝑅 = 966 𝑁
70𝑔
(b)
𝑅
𝒂
𝑚𝑔 − 𝑅 = 𝑚𝑎
∴ 70 × 9.8 − 𝑅 = 70 × 4
∴ 𝑅 = 686 − 280
∴ 𝑅 = 406 𝑁
70𝑔
15. A scale pan on which rests a mass of 50g is drawn upwards with a constant acceleration
and the reaction between the mass and the pan is found to be 0.5 N. Find the
acceleration of the scale pan.
SOLUTION
𝑅
𝑅 − 𝑚𝑔 = 𝑚𝑎
50
50
∴ 0.5 − 1000 × 9.8 = 1000 × 𝑎
∴𝑎=
50
𝑔
1000
Prepared by Elsam Senjobe @ GHS 2021.
0.5−0.49
0.005
∴ 𝑎 = 0.2 𝑚𝑠 −2
17
16. A body whose weight was 13 N appeared to weigh 12 N when weighed by means of a
spring balance in a moving lift. What was the acceleration of the lift at that instant of
weighing?
SOLUTION
Reaction, 𝑅 = 12 𝑁. Real weight. 𝑊 = 𝑚𝑔 = 13 𝑁 since 𝑊 > 𝑅 the lift is moving
downwards.
Now 𝑊 − 𝑅 = 𝑚𝑎
𝑊
∴ 13 − 12 = × 𝑎
𝑔
13
∴1=
×𝑎
𝑔
𝑔
∴ 𝑎 = 13 𝑚𝑠 −2 downwards.
17. A train of mass 160 Mg starts from a station, the engine exerting a tractive force of
of the weight in excess of the resistance until a speed of 60 kmh-1 is attained. This
1
64
speed is then maintained until the brakes, causing a retardation of 0.75 ms-2, bring the
train to rest 8 km away. Find the time
(i)
(ii)
(iii)
the train took during accelerating,
during retardation
when its speed was constant.
Determine the distance the trained covered in each case.
SOLUTION:
𝑣
𝑠1
𝑡1
𝑠2
𝑡2
𝑠3
𝑡3
𝑡
𝑇𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 − 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑀𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
1
∴ 𝑀𝑎 =
𝑀𝑔
64
Prepared by Elsam Senjobe @ GHS 2021.
18
𝑔
𝑚𝑠 −2
64
1000
50
The maximum speed attained, 𝑣 = 60 𝑘𝑚ℎ−1 = (60 × 3600) = 3 𝑚𝑠 −1
∴𝑎=
From 1st equation of motion: 𝑣 = 𝑢 + 𝑎𝑡1
50
𝑔
∴
=0+
× 𝑡1
3
64
50
64
∴ 𝑡1 = 3 × 9.8 = 108.84 𝑠
Distance covered during the acceleration:
1
1
50
64
50
2500×32
𝑠1 = 2 × 𝑡1 × 𝑣 = 2 × 3 × 9.8 × 3 = 9×9.8 = 907.03 𝑚
During the retardation:
𝑣 = 𝑢 − 𝑑𝑡3
50
∴0=
− 0.75 × 𝑡3
3
50
1
∴ 𝑡3 =
×
= 22.22 𝑠
3 0.75
Distance covered: 𝑣 2 = 𝑢2 − 2𝑑𝑠3
50 2
∴ 0 = ( ) − 2 × 0.75 × 𝑠3
3
2500
∴ 𝑠3 =
= 185.19 𝑚
9 × 1.5
Distance covered when velocity was constant:
𝑠2 = 8000 − (𝑠1 + 𝑠3 ) = 8000 − (907.03 + 185.19) = 6907.78 𝑚
Now: 𝑠2 = 𝑣𝑡2
∴ 6907.78 =
∴ 𝑡2 =
50
× 𝑡2
3
6907.78 × 3
= 414.47 𝑠
50
18. The pull exerted by an engine is
1
80
of the weight of the whole train, and the maximum
brake force which can be exerted is
1
30
of the weight of the train. Find the time in which
the train travels from rest to maximum velocity 𝑉𝑖 up a slope of 1 in 240 and 4.8 km long,
before the brakes being applied when steam is shut off and it comes to rest again.
Determine the distance it would have covered during its acceleration and retardation
respectively.
Prepared by Elsam Senjobe @ GHS 2021.
19
SOLUTION
𝑉𝑓 = 0 𝑚𝑠 −1
𝑉𝑖
𝑆
𝛼
1
The component of the trains weight along the slope; 𝑀𝑔 sin 𝛼 = 𝑀𝑔 × 240.
1
The pull of the train engine, 𝑃 = 80 𝑀𝑔
1
Maximum braking force, 𝐹𝑏 = 30 𝑀𝑔
Before the brakes are applied, the accelerating force is determined from the
equation:
𝑃 − 𝑀𝑔 sin 𝛼 = 𝑀𝑎
1
1
∴
𝑀𝑔 −
𝑀𝑔 = 𝑀𝑎
80
240
2𝑔
𝑔
∴
=
= 𝑎 𝑖𝑛 𝑚𝑠 −2
240 120
The maximum velocity attained during acceleration is calculated from:
𝑉𝑖 2 = 𝑈 2 + 2𝑎𝑆
𝑔
𝑔𝑆
∴ 𝑉𝑖 2 = 2 ×
×𝑆 =
120
60
When the brakes are applied, then
𝑅𝑒𝑡𝑎𝑟𝑑𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝐹𝑏 + 𝑀𝑔 sin 𝛼
1
1
𝑀𝑑 =
𝑀𝑔 +
𝑀𝑔
30
240
3𝑔
∴ 𝑑 = 80 𝑚𝑠 −2 i.e. the deceleration.
From the equation
𝑉𝑓 2 = 𝑉𝑖 2 − 2𝑑(4800 − 𝑆)
𝑔𝑆
3𝑔
∴0=
−2×
× (4800 − 𝑆)
60
80
𝑆 × 40 2𝑆
∴ 4800 − 𝑆 =
=
3 × 60
9
11𝑆
∴
= 4800
9
Prepared by Elsam Senjobe @ GHS 2021.
20
∴𝑆=
4800 × 9
= 3927.3 𝑚
11
Maximum velocity of the motion:
𝑔𝑆 9.8 4800 × 9
𝑉𝑖 2 =
=
×
= 641.4546
60 60
11
∴ 𝑉𝑖 = √641.4546 = 25.33 𝑚𝑠 −1
Time taken for train to accelerate. 𝑡1 :
𝑉𝑖 = 𝑈 + 𝑎𝑡1
9.8
× 𝑡1
120
25.33 × 120
∴ 𝑡1 =
= 310.16 𝑠
9.8
Time taken for train to decelerate. 𝑡2 :
∴ 25.33 = 0 +
𝑉𝑓 = 𝑉𝑖 − 𝑑𝑡2
3 × 9.8
× 𝑡2
80
25.33 × 80
∴ 𝑡2 =
= 68.93 𝑠
3 × 9.8
∴ 0 = 25.33 −
Total time taken to complete the motion:
𝑇 = 𝑡1 + 𝑡2 = (310.16 + 68.93 ) 𝑠
∴ 𝑇 = 379.09 𝑠
19. In a lift, accelerated upwards at a certain rate, a spring balance indicates a weight to
have a mass of 10 kg. When the lift is accelerated downwards at twice the rate, the
mass appears to be 7 kg. Find the actual mass and the upward acceleration of the lift.
SOLUTION:
Let 𝑎 = 𝑢𝑝𝑤𝑎𝑟𝑑 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛,
For upward motion:
Reaction, 𝑅 = 10𝑔
Applying: 𝑅 − 𝑚𝑔 = 𝑚𝑎
∴ 10𝑔 − 𝑚𝑔 = 𝑚𝑎 …………..(i)
Prepared by Elsam Senjobe @ GHS 2021.
2𝑎 = 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛.
21
For downward motion:
Reaction, 𝑅 = 7𝑔
And: 𝑚𝑔 − 𝑅 = 2𝑚𝑎
∴ 𝑚𝑔 − 7𝑔 = 2𝑚𝑎 ……………………(ii)
𝑔
Adding: 3𝑔 = 3𝑚𝑎 ∴ 𝑎 = 𝑚
𝑔
Substituting in (i): 10𝑔 − 𝑚𝑔 = 𝑚 (𝑚) = 𝑔
∴ 10𝑔 − 𝑚𝑔 = 𝑔
⇒ 𝑚 = 9 𝑘𝑔
𝑔
Hence, upward acceleration: 𝑎 = 9 𝑚𝑠 −2.
20. A vertical shield is made of two plates of wood and iron respectively, the iron being 2 cm
and the wood 4 cm thick. A bullet fired horizontally goes through the iron first and the
penetrates 2 cm into the wood. A similar bullet fired with the same velocity from the
opposite direction goes through the wood first and then penetrates 1 cm into the iron.
Compare the average resistance exerted by the iron and the wood.
SOLUTION:
wood
iron
1st bullet
2cm
2nd bullet
1cm
2 cm
4 cm
Let 𝑎𝑖 = retardation due to the iron.
𝑎𝑤 = retardation due to the wood.
If 𝑉𝑖 = the initial velocity of the first bullet that strikes the iron first,
𝑉𝑤 = final velocity of the bullet at the end of the iron, then:
Motion 1st bullet through the iron:
2
1
𝑉𝑤 2 = 𝑉𝑖 2 − 2𝑎𝑖 𝑆𝑖 where 𝑆𝑖 = 100 = 50
1
∴ 𝑉𝑤 2 = 𝑉𝑖 2 − 2𝑎𝑖 ×
50
Prepared by Elsam Senjobe @ GHS 2021.
22
𝑎
∴ 𝑉𝑤 2 = 𝑉𝑖 2 − 25𝑖 …………………………..(i)
Motion of 1st bullet through the wood:
2
0 = 𝑉𝑤 2 − 2𝑎𝑤 ×
100
𝑎𝑤
2
∴ 0 = 𝑉𝑤 −
25
𝑎
∴ 𝑉𝑤 2 = 25𝑤 ……………………………..(ii)
1
(𝑖) − (𝑖𝑖): 0 = 𝑉𝑖 2 − (𝑎𝑤 + 𝑎𝑖 )
25
1
2
∴ 𝑉𝑖 = 25 (𝑎𝑤 + 𝑎𝑖 ) …………………….(iii)
When the bullet is now travelling in the opposite direction i.e. from wood to iron:
Motion of bullet through the wood:
4
𝑉𝑡 2 = 𝑉𝑖 2 − 2𝑎𝑤 ×
100
2𝑎𝑤
2
2
∴ 𝑉𝑡 = 𝑉𝑖 − 25 ………………….(iv)
Motion of the bullet through the iron:
1
0 = 𝑉𝑡 2 − 2𝑎𝑖 ×
100
𝑎𝑖
2
∴ 0 = 𝑉𝑡 −
50
𝑎
∴ 𝑉𝑡 2 = 50𝑖 ………………………(v)
1
𝑎𝑖
(𝑖𝑣) − (𝑣): 0 = 𝑉𝑖 2 − (2𝑎𝑤 + )
25
2
1
𝑎𝑖
2
∴ 𝑉𝑖 = 25 (2𝑎𝑤 + 2 ) ……………….(vi)
Comparing (iii) to (vi):
𝑎𝑖
𝑎𝑤 + 𝑎𝑖 = 2𝑎𝑤 +
2
1
∴ 𝑎𝑖 = 𝑎𝑤
2
∴
𝑎𝑖
2
=
𝑎𝑤 1
But the resistances are proportional to the retardations they produce, therefore
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑖𝑟𝑜𝑛: 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑜𝑜𝑑 = 𝑎𝑖 : 𝑎𝑤 = 2: 1
21. A 100mg bullet, travelling at 150 𝑚𝑠 −1, will penetrate into a fixed block of wood before
coming to rest. Find the velocity with which it would emerge if fired through a fixed
Prepared by Elsam Senjobe @ GHS 2021.
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board 4 cm thick, the resistance being supposed uniform and to have the same value in
each case.
SOLUTION:
Let 𝑑 = retardation and using 𝑉 2 = 𝑈 2 − 2𝑑𝑆
8
0 = 1502 − 2𝑑 ×
100
8𝑑
2
∴ 150 =
25
1502 ×25
∴𝑑=
𝑚𝑠 −2 this is the common retardation.
4
When the wood is 4 cm thick, the velocity of emergence is obtained as follows:
𝑉 2 = 𝑈 2 − 2𝑑𝑆
∴ 𝑉 2 = 1502 − 2 ×
1502
∴ 𝑉 =
2
1502 × 25
4
×
4
100
2
∴𝑉=
150
√2
=
150√2
= 75√2 𝑚𝑠 −1
2
22. A bullet of mass 30 g is fired into a fixed block of wood with a velocity of 294 𝑚𝑠 −1 , and
1
is brought to rest in 150 𝑠. Find the resistance exerted by the wood, supposing it to be
uniform.
SOLUTION
30
1
Mass of bullet, 𝑚 = 100 𝑘𝑔, 𝑢 = 294 𝑚𝑠 −1 , 𝑡 = 150 𝑠, 𝑣 = 0 𝑚𝑠 −1
From : 𝑣 = 𝑢 − 𝑎𝑡
𝑢
294
=
= 294 × 150 = 44100 𝑚𝑠 −1
𝑡 1⁄
150
The resistance exerted by the wood, supposing it to be uniform is:
30
𝐹 = 𝑚𝑑 =
× 44100
1000
∴ 0 = 𝑢 − 𝑑𝑡 ⇒ 𝑑 =
∴ 𝐹 = 1323 𝑁
Prepared by Elsam Senjobe @ GHS 2021.
24
23. The figure below shows a truck A of mass 1000 kg pulling a trailer B of 3000 kg. The
frictional force on A is 1,000 N and on B is 2,000 N, and the trucks engine exerts a
force of 8000 N. Calculate
(i)
the acceleration of the truck,
(ii)
the tension T in the tow-bar connecting A and B.
A ( TRUCK)
B ( TRAILER)
3000 𝑘𝑔
2000 𝑁
T
1000 𝑘𝑔
8000 𝑁
1000 𝑁
SOLUTION:
(i)
Using 𝐹 = 𝑚𝑎 where 𝐹 is the resultant force;
For A: 8000 − 1000 − 𝑇 = 1000 𝑎
∴ 7000 − 𝑇 = 𝑚𝑎 ………………………………(i)
For B: 𝑇 − 2000 = 3000𝑎 …………………..(ii)
Adding (i) and (ii)
5000 = 4000𝑎
5000
∴ 𝑎 = 4000 = 1.25 𝑚𝑠 −1 is the acceleration of the truck.
(ii)
From (ii)
𝑇 = 2000 + 3000 × 1.25
Tension in tow-bar, 𝑇 = 5750 𝑁
Prepared by Elsam Senjobe @ GHS 2021.
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