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Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy

Mathematics

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy

Overview

Define momentum and its change, and state Newton’s three laws of motion.
Apply the equation of motion to constant and variable mass, lifts and bodies connected over a pulley.
Define impulse and relate it to the change of momentum.
Define work, kinetic energy, potential energy and power, and use the work-energy principle and conservation of energy.

🌳 Concept tree

Lesson 1→Momentum, Newtons Three Laws of Motion, Impulse and Energy

├── 1 Momentum and its change

│ ├── Momentum

│ └── Change of momentum

├── 2 Newton's first law, inertia and force

│ ├── The law

│ └── Inertia and force

├── 3 Newton's second law
the equation of motion and variable mass

│ ├── The law

│ ├── Other forms when $F$ depends on time or position

│ ├── Units of force; weight and mass

│ └── Variable mass

├── 4 Newton's third law, lifts and scales

│ ├── The law

│ ├── Person in a lift

│ ├── The lift itself and the whole system

│ └── Scale readings

├── 5 Bodies connected by a string over a smooth pulley

│ ├── Two vertical masses

│ └── Special cases

├── 6 Impulse

│ ├── Definition

│ ├── Impulse and momentum

│ └── Impulsive force

├── 7 Work

│ ├── Work of a constant force

│ ├── Units of work

│ └── Work of a variable force

├── 8 Kinetic energy and the work-energy principle

│ ├── Kinetic energy

│ └── The work-energy principle

├── 9 Potential energy and conservation of energy

│ ├── Potential energy

│ ├── Conservation of energy

│ └── Motion with resistance

└── 10 Power

├── Definition

└── Units of power

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 1 / 12

1 Momentum and its change

Momentum

Momentum

\vec{H}

of a moving body is a vector with the direction of its velocity, equal to mass times velocity.

\vec{H} = m \vec{v}

In rectilinear motion use algebraic measuresH=mv.

Unit in SIkg⋅m/s. For constant mass, H is proportional to v, so it is called linear momentum.

A ball flying on a curved path shows vector functions and Newtons laws of motion: r(1) = (10, 5) and maximum height 5 m.

Change of momentum

If the velocity changes from v⃗1 to v⃗2 in a given interval ΔH⃗=m(v⃗2−v⃗1)
If the acceleration is given as a function of time, $Δ H = m \int_{t_{1}}^{t_{2}} a d t$ .

✍️ Worked examples

Example 1 A bike of mass $35$ kg moves east at $12$ m/s. Find its momentum.
Solution:
$H = m v = 35 \times 12 = 420$ kg}·m/s towards the east.
Example 2 A car of mass $2000$ kg has position $r = (3 t^{2} - 4 t + 1) \overset{</mo>}{c}$ metres. Find the magnitude of its momentum at the start and at $t = 3$ s.
Solution:
$v = \frac{d r}{d t} = (6 t - 4) \overset{</mo>}{c}$ .
At t=0 v=−4, so H=2000×(−4)=−8000; the magnitude is 8000 kg·m/s.
At t=3 v=14, so H=2000×14=28000 kg·m/s.
Example 3:
A $200$ g rubber ball falls from $90$ cm onto a horizontal surface and rebounds to $40$ cm.
Find the magnitude of the change in its momentum.
Solution:
Take upward as positive and $g = 9.8$ m/s $^{2}$ .
Before impact v1=−2(9.8)(0.9)=−4.2 m/s.
After impact v2=+2(9.8)(0.4)=2.8 m/s.
$Δ H = 0.2 (2.8 - (- 4.2)) = 0.2 \times 7 = 1.4$ kg·m/s.
Example 4 The acceleration of a body of mass $8$ kg is $a = 2 t - 6$ m/s $^{2}$ . Find its change of momentum for $3 \leq t \leq 5$ .
Solution:
$Δ H = 8 \int_{3}^{5} (2 t - 6) d t = 8 [t^{2} - 6 t]_{3}^{5}$ .
$= 8 [(25 - 30) - (9 - 18)] = 8 (- 5 + 9) = 32$ kg·m/s.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 2 / 12

2 Newton’s first law, inertia and force

The law

Newton’s first law every body stays at rest or in uniform motion in a straight line unless an unbalanced external force acts on it.
quot;Force quot; in the law means the resultant of all forces on the body.
Rest and uniform straight-line motion are both the quot;natural state quot; when the resultant force is zero.
Because of this the first law is called the law of inertia.

Inertiathe tendency of a body to resist any change in its state of rest or of uniform motion.

Forcethe effect that changes, or tries to change, the state of rest or uniform motion of a body.

A body moving with constant velocity has balanced forcesthe sums of the forces in opposite directions are equal.

✍️ Worked examples

Example 5:
A train of mass $200$ tonnes moves against a resistance proportional to $v^{2}$ .
The resistance is $9.6$ kg.wt per tonne at $72$ km/h.
The engine pulls with a uniform force of $4.32$ tonne.wt.
Find the maximum speed.
Solution:
At v1=72 km/h R1=9.6×200=1920 kg.wt.
Maximum speed is reached when the resistance equals the pull R2=4320 kg.wt.
$\frac{R_{2}}{R_{1}} = \frac{v_{2}^{2}}{v_{1}^{2}}$ , so $\frac{v_{2}^{2}}{7 2^{2}} = \frac{4320}{1920} = 2.25$ .
$v_{2} = 1.5 \times 72 = 108$ km/h.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 3 / 12

3 Newton’s second law, the equation of motion and variable mass

The law

Comparison aspect	Main idea	Keywords
Newton’s second law	the rate of change of momentum with respect to time is proportional to the force and takes place in the direction of the force. $\vec{F} = \frac{d}{d t} (m \vec{v})$	ratechangemomentumrespect
For constant mass this gives the equation of motion	∑F⃗=ma⃗or in algebraic measures ∑F=ma	algebraicmeasures
•	The acceleration has the direction of the resultant force, so fix the positive direction of $a$ first.	accelerationdirectionresultantforce

If F=f(t)F dt=m dv, so ∫t1t2F dt=m(v2−v1).

If F=f(s)F ds=m v dv, so ∫s1s2F ds=m∫v1v2v dv.

The unit of force is chosen so that the constant of proportionality is $1$ : $1 kg \times 1 {m/s}^{2} = 1$ newton.
$1 g \times 1 {cm/s}^{2} = 1$ dyne.
$1 kg.wt = 9.8$ N and $1 g.wt = 980$ dyne.
Weight is the force of the Earth’s attraction on the body: $w = m g$ . Mass is a constant property of the body.

✍️ Worked examples

Example 6 A force of $10$ N acts on a body of mass $8$ kg at rest. Find the distance travelled in $12$ s and the velocity then.
Solution:
$8 a = 10$ , so $a = 1.25$ m/s $^{2}$ .
$v = 0 + 1.25 \times 12 = 15$ m/s.
$s = \frac{1}{2} (1.25) (12)^{2} = 90$ m.
Example 7 A body of mass $3$ kg falls from $10$ m onto sand and sinks $5$ cm. Find the (constant) sand resistance.
Solution:
Free fall v2=0+2(9.8)(10)=196, so v=14 m/s.
In the sand 0=142+2a(0.05), so a=−1960 m/s2.
Take downward as positive ma=mg−R, so 3(−1960)=3(9.8)−R.
$R = 29.4 + 5880 = 5909.4$ N $= 603$ kg.wt.

Variable mass

When the mass changes, use $F = \frac{d}{d t} (m v)$ , and for a body gaining mass at a rate $\frac{d m}{d t}$ with constant velocity, $F = v \frac{d m}{d t}$ .

✍️ Worked examples

Example 8 A body of variable mass $m = 2 t + 1$ has displacement $S = (\frac{1}{2} t^{2} + t) \overset{</mo>}{i}$ . Find its momentum and the force acting on it.
Solution:
$v = \frac{d S}{d t} = (t + 1) \overset{</mo>}{i}$ .
$H = m v = (2 t + 1) (t + 1) \overset{</mo>}{i} = (2 t^{2} + 3 t + 1) \overset{</mo>}{i}$ .
$F = \frac{d H}{d t} = (4 t + 3) \overset{</mo>}{i}$ .
Example 9 A metal ball of mass $100$ g moves at a uniform $10$ m/s in dusty air; dust sticks to it at $0.6$ g/s. Find the force on the ball.
Solution:
$F = v \frac{d m}{d t} = (1000 cm/s) (0.6 g/s) = 600$ dyne.
The mass at time $t$ is $m = 100 + 0.6 t$ grams.
Example 10 A force $F = 5 r + 6$ newtons acts on a body of mass $1$ kg that starts at rest at $r = 0$ . Find its velocity at $r = 4$ m.
Solution:
$F = m v \frac{d v}{d r}$ , so $v d v = (5 r + 6) d r$ .
$\int_{0}^{v} v d v = \int_{0}^{4} (5 r + 6) d r$ .
$\frac{v^{2}}{2} = [\frac{5 r^{2}}{2} + 6 r]_{0}^{4} = 40 + 24 = 64$ .
$v^{2} = 128$ , so $v = 8 \sqrt{2}$ m/s.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 4 / 12

4 Newton’s third law, lifts and scales

The law

Newton’s third law to every action there is a reaction equal in magnitude and opposite in direction (acting on the other body).
A body at rest on a plane presses on the plane with a force equal to its weight, and the plane reacts with an equal upward force.

Comparison aspect	Main idea	Keywords
Two forces act on the person	the weight mg (down) and the reaction N of the floor (up)	weightdownreactionfloor
Lift at rest or moving uniformly	N=mg
Lift accelerating upward (or decelerating while moving down)	ma=N−mg so N=m(g+a)
Lift accelerating downward (or decelerating while moving up)	ma=mg−N so N=m(g−a)
•	The pressure of the person on the floor equals $N$ ; this is the apparent weight.	pressurepersonfloorequals
•	If the lift falls freely ( $a = g$ ) then $N = 0$ (weightlessness).	liftfallsfreelythen

Forces on the empty-lift partits weight m′g, the pressure P of the person (down) and the cable tension T (up).

Lift moving up with acceleration am′a=T−P−m′g. Lift moving down: m′a=m′g+P−T.

For lift and person together(m+m′)a=T−(m+m′)g going up, and (m+m′)a=(m+m′)g−T going down.

A spring scale hung from the ceiling of the lift reads the tension $T$ in the spring.
A pressure scale on the floor reads the pressure $N$ of the body.
Reading greater than mg the lift accelerates upward or decelerates while moving downward.
Reading less than mg the lift accelerates downward or decelerates while moving upward.
Reading equal to mg the lift is at rest or moves uniformly.
The readings going up and going down with the same acceleration add up to $2 m g$ .
A balance (comparing masses) is the only instrument that gives the real weight in all conditions.

✍️ Worked examples

Example 11:
A man of mass $80$ kg is in a lift.
Find his pressure on the floor when the lift (a) goes up with $a = 0.49$ m/s $^{2}$ , (b) moves uniformly, (c) goes down with $a = 0.98$ m/s $^{2}$ .
Solution:
(a) $N = m (g + a) = 80 (9.8 + 0.49) = 823.2$ N $= 84$ kg.wt.
(b) $N = m g = 784$ N $= 80$ kg.wt.
(c) $N = m (g - a) = 80 (9.8 - 0.98) = 705.6$ N $= 72$ kg.wt.
Example 12:
The spring scale in a lift going up with $a = 2.45$ m/s $^{2}$ reads $50$ kg.wt.
Find the mass of the body and the reading when the lift goes down with the same acceleration.
Solution:
Going up ma=T−mg, so m(2.45)=50(9.8)−m(9.8).
$m (12.25) = 490$ , so $m = 40$ kg.
Going down T=m(g−a)=40(9.8−2.45)=294 N =30 kg.wt.
Example 13 A lift cable can bear $1200$ kg.wt. The empty lift has mass $600$ kg and rises with $a = 1.4$ m/s $^{2}$ . Find the maximum load.
Solution:
Let the load be m kg (m+600) a=T−(m+600)g.
$(m + 600) (1.4 + 9.8) = 1200 \times 9.8$ .
$m + 600 = \frac{11760}{11.2} = 1050$ .
$m = 450$ kg.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 5 / 12

5 Bodies connected by a string over a smooth pulley

Two vertical masses

When the pulley is small and smooth and the string is light and inextensible, the tension is the same on both sides.
Let $m_{1} gt; m_{2}$ . Equations of motion: $m_{1} a = m_{1} g - T$ and $m_{2} a = T - m_{2} g$ . $a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g, T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g$
The pressure on the pulley axis is $P = 2 T$ .
If the masses start level, after time $t$ each has moved $s = \frac{1}{2} a t^{2}$ and the vertical distance between them is $2 s$ .

Comparison aspect	Main idea	Keywords
Equal masses	$a = 0$ and the system stays at rest or moves with uniform velocity.	systemstaysrestmoves
Extra mass m3 added to one side	(m1+m3)a=(m1+m3)g−T and m2a=T−m2g
Extra mass removed after t seconds	the system keeps moving in the same direction with the velocity gained but with the new (smaller) acceleration which may be a deceleration	systemkeepsmovingsame
String cut after t seconds	each body continues in its direction with the velocity gained and then moves under gravity alone (a=±g)	bodycontinuesdirectionvelocity
•	If the heavier side is unknown and the system returns to its start, $m_{1} lt; m_{2}$ ; if it moves uniformly, $m_{1} = m_{2}$ .	heaviersideunknownsystem

✍️ Worked examples

Example 14 Two masses start level and after $1$ s the vertical distance between them is $20$ cm. Find $m_{1} : m_{2}$ ( $m_{1} gt; m_{2}$ ).
Solution:
Each moved 10 cm 10=12a(1)2, so a=20 cm/s2.
$a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$ , so $\frac{20}{980} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}$ .
$m_{1} + m_{2} = 49 m_{1} - 49 m_{2}$ , so $50 m_{2} = 48 m_{1}$ .
m1 m2=25:24.
Example 15:
Masses $5$ kg and $3$ kg hang level at $245$ cm above the ground.
Find $a$ , $T$ and the pressure on the pulley.
The string is cut after $1$ s; find the speed of each mass on reaching the ground.
Solution:
$a = \frac{5 - 3}{5 + 3} (9.8) = 2.45$ m/s $^{2}$ .
$T = 3 (9.8 + 2.45) = 36.75$ N, so the pressure on the pulley is $2 T = 73.5$ N.
At the cut v=2.45 m/s and each has moved 12(2.45)(1)=1.225 m.
The $5$ kg mass is $1.225$ m above the ground moving down: $v^{2} = 2.4 5^{2} + 2 (9.8) (1.225) = 30.0125$ , so $v \approx 5.48$ m/s.
The $3$ kg mass is $3.675$ m above the ground moving up: $v^{2} = 2.4 5^{2} + 2 (9.8) (3.675) = 78.0325$ , so $v \approx 8.83$ m/s on reaching the ground.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 6 / 12

6 Impulse

Definition

Comparison aspect	Main idea	Keywords
Impulse	of a constant force $\vec{F}$ acting for time $t$ is the vector $\vec{I} = \vec{F} t$	constantforceactingtime
•	Its direction is that of the force	directionforce
Unit	newton·second (N·s) which equals kg·m/s. In the CGS system it is dyne·s = g·cm/s	newtonsecondequalssystem
Other units	kg.wt·s and g.wt·s

Comparison aspect	Main idea	Keywords
Using F=ma and v−v0=at	I=Ft=m(v−v0)
Impulse equals the change of momentum.	For a variable force: $I = \int_{0}^{t} F d t = m (v - v_{0})$ .	variableforce
•	On a force-time graph the impulse is the area under the curve	forcetimegraphimpulse

Comparison aspect	Main idea	Keywords
•	An impulsive force is a very large force acting for a very short time causing a large change in momentum without a change in position	impulsiveforceverylarge
Equation	mv1+F t=mv2 with t extremely small (examples: a bat hitting a ball a collision)	extremelysmallexampleshitting
Falling onto the ground	the force on the ground is F+w. Rising into a ceiling: F−w. Hitting a vertical wall horizontally: F	forceRisingceilingHitting

✍️ Worked examples

Example 16 A force of $25$ kg.wt acts on an object for $\frac{1}{10}$ s. Find the impulse in N·s.
Solution:
$I = F t = 25 \times 9.8 \times \frac{1}{10} = 24.5$ N·s.
Example 17 A force $F = t^{2} - 4 t + 5$ newtons acts on a body. Find the impulse in the first $3$ s.
Solution:
$I = \int_{0}^{3} (t^{2} - 4 t + 5) d t = [\frac{t^{3}}{3} - 2 t^{2} + 5 t]_{0}^{3}$ .
$= 9 - 18 + 15 = 6$ N·s.
Example 18 A force of $5$ N acts on a $4$ kg body at rest on a smooth plane for $8$ s. Find the impulse and the final velocity.
Solution:
$I = F t = 5 \times 8 = 40$ N·s.
$I = m (v - 0)$ , so $40 = 4 v$ and $v = 10$ m/s.
Example 19 A rubber ball of mass $\frac{1}{4}$ kg falls $10$ m and rebounds to $2.5$ m. The contact lasts $\frac{1}{10}$ s. Find the impulse and the impulsive force.
Solution:
Falling v1=−2(9.8)(10)=−14 m/s. Rebound: v2=+2(9.8)(2.5)=7 m/s.
$I = m (v_{2} - v_{1}) = \frac{1}{4} (7 + 14) = 5.25$ N·s.
$F = \frac{I}{t} = \frac{5.25}{0.1} = 52.5$ N.
Example 20:
A $100$ g ball hits a wall at $9$ m/s and rebounds at $7.2$ km/h for $0.1$ s of contact.
Find the impulse and the pressure of the ball on the wall.
Solution:
$7.2$ km/h $= 2$ m/s. Take the rebound direction as positive: $v_{1} = - 9$ , $v_{2} = 2$ .
$I = 0.1 (2 + 9) = 1.1$ N·s.
$F = \frac{1.1}{0.1} = 11$ N, which is the pressure of the ball on the wall.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 7 / 12

7 Work

Work of a constant force

The work done by a constant force $\vec{F}$ moving a particle through displacement $\vec{S}$ is the scalar product: $W = \vec{F} \cdot \vec{S} = ∥ \vec{F} ∥ ∥ \vec{S} ∥ \cos θ$
Work is a scalar that can be positive, negative or zero. For $\vec{F} = (F_{x}, F_{y})$ and $\vec{S} = (S_{x}, S_{y})$ : $W = F_{x} S_{x} + F_{y} S_{y}$ .
θ=0 (force along displacement) W=FS.
θ lt;90∘ W gt;0 (the force helps the motion).
θ=90∘ W=0 (for example the weight of a body moving horizontally, or the normal reaction).
θ gt;90∘ W lt;0; this is resistance work (friction, resistance).
Work depends only on the displacement $\vec{A B}$ , not on the path. The total work for two displacements equals the work for the resultant displacement.

Joulework of a

1

N force moving a body

1

1 J = 1 N \cdot m

(SI unit).

Ergwork of a

1

dyne force over

1

cm.

1 J = 1 0^{7}

erg.

kg.wt·mwork of a

1

kg.wt force over

1

1 kg.wt \cdot m = 9.8

✍️ Worked examples

Example 21 The force $\vec{F} = 6 \overset{</mo>}{i} + 8 \overset{</mo>}{j}$ moves a particle in a straight line from $A (3, - 4)$ to $B (7, 2)$ . Find the work.
Solution:
$\vec{S} = \vec{A B} = (7 - 3, 2 + 4) = (4, 6)$ .
$W = 6 (4) + 8 (6) = 24 + 48 = 72$ work units.
Example 22 $\vec{F} = 3 \overset{</mo>}{i} + 5 \overset{</mo>}{j}$ moves a body from $A (2, 4)$ to $B (5, 3)$ and then to $C (8, - 2)$ . Check that the work along the two displacements equals the work along $A C$ .
Solution:
$\vec{A B} = (3, - 1)$ , so $W_{1} = 3 (3) + 5 (- 1) = 4$ .
$\vec{B C} = (3, - 5)$ , so $W_{2} = 3 (3) + 5 (- 5) = - 16$ .
Total 4−16=−12.
$\vec{A C} = (6, - 6)$ , so $W = 3 (6) + 5 (- 6) = - 12$ . They are equal.
Example 23 A resistance of $100$ N acts on a body displaced $300$ m. Find the work done by the resistance.
Solution:
The resistance opposes the displacement, so $W = - F S = - 100 \times 300 = - 3 \times 1 0^{4}$ J.
Example 24:
A body of mass $10$ kg slides $6$ m down a rough plane inclined at $3 0^{\circ}$ with $μ_{k} = 0.2$ .
Find (in kg.wt·m) the work done by the weight, the normal reaction and the friction.
Solution:
Weight Ww=(10)sin⁡30∘×6=10×12×6=30 kg.wt·m.
Normal reaction perpendicular to the displacement, so the work is 0.
Friction Fk=μkN=0.2×10cos⁡30∘=3 kg.wt, and the work is −FkS=−63≈−10.4 kg.wt·m.

Work of a variable force

Divide the displacement into tiny parts $Δ s$ in which $F$ is nearly constant; adding gives $W = \int_{A}^{B} F d s (F is the component of the force along the displacement)$
On a force-displacement graph, work is the area under the curve; area below the axis counts as negative work.

✍️ Worked examples

Example 25 A force $F = 3 s^{2} - 4$ newtons acts along a line. Find the work from $s = 2$ m to $s = 5$ m.
Solution:
$W = \int_{2}^{5} (3 s^{2} - 4) d s = [s^{3} - 4 s]_{2}^{5}$ .
$= (125 - 20) - (8 - 8) = 105$ J.
Example 26 $\vec{F} = 2 \overset{</mo>}{i} + 3 \overset{</mo>}{j}$ acts on a particle with $\vec{r} (t) = (t + 5) \overset{</mo>}{i} + (t^{2} + 4) \overset{</mo>}{j}$ . Find the work from $t = 1$ to $t = 5$ .
Solution:
$\vec{S} = \vec{r} (5) - \vec{r} (1) = (10, 29) - (6, 5) = (4, 24)$ .
$W = 2 (4) + 3 (24) = 8 + 72 = 80$ work units.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 8 / 12

8 Kinetic energy and the work-energy principle

Kinetic energy

Kinetic energy $T$ is the energy a particle has because of its velocity: $T = \frac{1}{2} m ∥ \vec{v} ∥^{2} = \frac{1}{2} m v^{2}$
It is a non-negative scalar and equals zero when the velocity is zero. Its units are the same as the units of work (joule, erg, kg.wt·m).

✍️ Worked examples

Example 27 A $100$ g body moves with $\vec{v} = 5 \overset{</mo>}{i} + 12 \overset{</mo>}{j}$ cm/s. Find its kinetic energy in erg and in joules.
Solution:
$∥ \vec{v} ∥ = \sqrt{25 + 144} = 13$ cm/s.
$T = \frac{1}{2} (100) (169) = 8450$ erg.
$= \frac{8450}{1 0^{7}} = 8.45 \times 1 0^{- 4}$ J.

Comparison aspect	Main idea	Keywords
For a constant force	12m(v22−v12)=F⋅S
For a variable force	dTdt=F v so ∫dT=∫F ds. T2−T1=W	dTdt
Work-energy principle	the change in kinetic energy of a particle between two positions equals the work done by the forces acting on it between them	changekineticparticlebetween
•	If a particle returns to its starting point under gravity its final kinetic energy equals its initial kinetic energy so a vertical projectile has the same speed at a point going up and coming down	particlereturnsstartingpoint

✍️ Worked examples

Example 28 A $200$ g bullet moving at $400$ m/s embeds $20$ cm in a barrier. Find the constant resistance.
Solution:
$T_{1} = \frac{1}{2} (0.2) (400)^{2} = 16000$ J and $T_{2} = 0$ .
$W = - R (0.2)$ .
$T_{2} - T_{1} = W$ , so $- 16000 = - 0.2 R$ .
$R = 80000$ N $= 8 \times 1 0^{9}$ dyne.
Example 29:
A $300$ g body slides from the top of an inclined plane of height $1$ m.
The work done against resistance is $1.59$ J.
Find its speed at the bottom.
Solution:
Work of the weight $= m g h = 0.3 (9.8) (1) = 2.94$ J.
Net work $W = 2.94 - 1.59 = 1.35$ J.
$\frac{1}{2} (0.3) v^{2} - 0 = 1.35$ , so $v^{2} = 9$ and $v = 3$ m/s.
Example 30:
A body of mass $1$ kg moves at $12$ m/s against a resistance $6 x^{2}$ newtons, where $x$ is the distance travelled.
Find the work done by the resistance and the speed at $x = 2$ m and $x = 4$ m.
Solution:
Work done between 0 and 4 W=−∫046x2 dx=−[2x3]04=−128 J.
At x=2 W=−[2x3]02=−16 J.
$\frac{1}{2} (1) (v^{2} - 144) = - 16$ , so $v^{2} = 112$ and $v = 4 \sqrt{7}$ m/s.
$T = \frac{1}{2} (112) = 56$ J.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 9 / 12

9 Potential energy and conservation of energy

Potential energy

The potential energy of a particle at a position is the work done by the force on it as it moves from that position to a fixed reference position $O$ (where $P = 0$ ).
Change in potential energy equals minus the work of the force: $P_{B} - P_{A} = - W$
For a body of mass $m$ at height $h$ above the reference level (weight is the force): $P = m g h$ .
Units are the same as for work and kinetic energy.

If the body moves without resistance, $T_{B} - T_{A} = W$ and $P_{B} - P_{A} = - W$ , so $T_{A} + P_{A} = T_{B} + P_{B}$
The sum of kinetic and potential energy stays constant during the motion.

When a body descends a rough plane (or moves in a resisting medium): $change in potential energy = change in kinetic energy + work done against resistance$

✍️ Worked examples

Example 31:
A body of mass $300$ g is $10$ m above the ground and is released.
Find its potential energy, the constant sum of its energies, and its kinetic energy at $3$ m above the ground.
Solution:
$P_{A} = m g h = 0.3 (9.8) (10) = 29.4$ J, and $T_{A} = 0$ , so $T + P = 29.4$ J throughout.
At 3 m PB=0.3(9.8)(3)=8.82 J.
$T_{B} = 29.4 - 8.82 = 20.58$ J.
Example 32 A $3$ kg body slides from rest from the top of a smooth plane $20$ m long, inclined at $3 0^{\circ}$ . Find its speed at the bottom.
Solution:
Height $h = 20 \sin 3 0^{\circ} = 10$ m, so $P_{A} = 3 (9.8) (10) = 294$ J.
At the bottom $P_{B} = 0$ , so $\frac{1}{2} (3) v^{2} = 294$ .
$v^{2} = 196$ and $v = 14$ m/s.
Example 33:
A $2$ kg block slides from rest from $4$ m high down a smooth curve $A B$ , then along a rough horizontal plane $B C$ with $μ_{k} = \frac{1}{5}$ .
How far along $B C$ does it travel before stopping?
Solution:
Smooth part TB=PA=2(9.8)(4)=78.4 J.
On $B C$ the potential energy does not change, so $0 = (0 - 78.4) + μ_{k} N S$ .
$N = m g = 19.6$ N, so $μ_{k} N = 3.92$ N.
$3.92 S = 78.4$ , so $S = 20$ m.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 10 / 12

10 Power

Definition

Power is the time rate of doing work: $Power = \frac{d W}{d t}$ .
Since $d W = \vec{F} \cdot d \vec{S}$ , $Power = \vec{F} \cdot \vec{v}$ ; when $\vec{v}$ has the direction of $\vec{F}$ : $Power = F v$
Power is a scalar defined at an instant. With constant force, power is proportional to speed, so the maximum power is reached at maximum speed.
Average power $= \frac{W}{Δ t} = \frac{W}{t_{2} - t_{1}}$ .
If the power is given as a function of time, work $= \int_{t_{1}}^{t_{2}} Power d t$ .

Units of power

Watt

= 1

joule/second

= 1

N·m/s

= 1 0^{7}

erg/s.

kg.wt·m/s

1 kg.wt \cdot m/s = 9.8

watt.

Kilowatt

1 kW = 1000

Horse power(hp): the power of a machine doing

75

kg.wt·m per second.

1 hp = 75 \times 9.8 = 735

= 0.735

kW.

✍️ Worked examples

Example 34 A person of mass $50$ kg climbs a tower $441$ m high in $15$ minutes. Find the average power in watts.
Solution:
$F = m g = 50 \times 9.8 = 490$ N.
Average speed $v = \frac{441}{15 \times 60} = 0.49$ m/s.
Power $= F v = 490 \times 0.49 = 240.1$ W.
Example 35:
A car of mass $2$ tonnes moves uniformly at $108$ km/h on a level road with resistance $150$ kg.wt per tonne.
Find the engine power in horse power.
Solution:
Uniform motion, so the engine force equals the resistance F=150×2=300 kg.wt.
$v = 108 \times \frac{5}{18} = 30$ m/s.
Power $= 300 \times 30 = 9000$ kg.wt·m/s.
In horse power 900075=120 hp.
Example 36:
A train of mass $200$ tonnes climbs a slope with $\sin θ = \frac{1}{200}$ at a uniform $27$ km/h against resistance of $18$ kg.wt per tonne.
Find the engine power going up, and going down the same slope at the same speed.
Solution:
$v = 27 \times \frac{5}{18} = 7.5$ m/s; resistance $= 200 \times 18 = 3600$ kg.wt.
Component of the weight along the slope $= 200000 \times \frac{1}{200} = 1000$ kg.wt.
Going up F=3600+1000=4600 kg.wt, power =4600×7.5=34500 kg.wt·\text{m/s =460 hp.
Going down F+1000=3600, so F=2600 kg.wt, power =2600×7.5÷75=260 hp.
Example 37 The power of an engine is $9 t^{2} + 4 t$ . Find the work done in the first $3$ s and in the fourth second.
Solution:
First 3 s W=∫03(9t2+4t) dt=[3t3+2t2]03=81+18=99.
Fourth second W=∫34(9t2+4t) dt=[3t3+2t2]34=224−99=125.
Example 38 A force $\vec{F} = 3 \hat{i} + 4 \hat{j}$ newtons acts on a particle whose displacement is $\vec{S} = 3 t^{2} \hat{i} + 6 t \hat{j}$ . Find the work and the power at $t = 2$ s.
Solution:
$W = \vec{F} \cdot \vec{S} = 9 t^{2} + 24 t$ .
Power $= \frac{d W}{d t} = 18 t + 24$ .
At t=2 power =60 W.

Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy · 11 / 12

Comparison aspect	Main idea	Keywords
RULE 1	In Newton's second law: If mass varies with time m(t) Force F = d/dt(m · v) and CANNOT be simplified to m · a!	Newtonsecondmassvaries
RULE 2	Apparent weight on elevator scale: Scale reads more than real weight (N > mg) only when acceleration is UPWARDS	Apparentweightelevatorscale
RULE 3	Impulse of a force: I = F · Δt = m · (v2 – v1). Area under the Force-Time (F – t) curve equals Impulse	ImpulseforceAreaunder
RULE 4	1 Horsepower = 75 kg·wt·m/s = 75 × 9.8 Watts = 735 Watts	HorsepowerWatts
📊 Matrix	Apparent Weight States of a Body in an Elevator	ApparentWeightStatesBody

Elevator Motion State	Scale Reaction Force Formula (N)	Comparison with Real Weight (W = mg)
Stationary or Moving with Constant Velocity	N = m · g	N = W (Apparent weight equals real weight)
Accelerating Upwards (or Decelerating Downwards)	N = m · (g + a)	N > W (Apparent weight is GREATER)
Accelerating Downwards (or Decelerating Upwards)	N = m · (g – a)	N < W (Apparent weight is LESS)
Free Fall (Cable Snaps: a = g)	N = m · (g – g) = 0	N = 0 (Apparent weightlessness)

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Lesson 1: Momentum, Newtons Three Laws of Motion, Impulse and Energy

🌳 Concept tree

├── 1 Momentum and its change

│ ├── Momentum

│ └── Change of momentum

├── 2 Newton's first law, inertia and force

│ ├── The law

│ └── Inertia and force

│ ├── The law

│ ├── Units of force; weight and mass

│ └── Variable mass

├── 4 Newton's third law, lifts and scales

│ ├── The law

│ ├── Person in a lift

│ ├── The lift itself and the whole system

│ └── Scale readings

│ ├── Two vertical masses

│ └── Special cases

├── 6 Impulse

│ ├── Definition

│ ├── Impulse and momentum

│ └── Impulsive force

├── 7 Work

│ ├── Work of a constant force

│ ├── Units of work

│ └── Work of a variable force

│ ├── Kinetic energy

│ └── The work-energy principle

│ ├── Potential energy

│ ├── Conservation of energy

│ └── Motion with resistance

└── 10 Power

├── Definition

└── Units of power

Momentum

Change of momentum

✍️ Worked examples

The law

✍️ Worked examples

The law

✍️ Worked examples

Variable mass

✍️ Worked examples

The law

✍️ Worked examples

Two vertical masses

✍️ Worked examples

Definition

✍️ Worked examples

Work of a constant force

✍️ Worked examples

Work of a variable force

✍️ Worked examples

Kinetic energy

✍️ Worked examples

✍️ Worked examples

Potential energy

✍️ Worked examples

Definition

Units of power

✍️ Worked examples

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Mathematics Thanaweya Amma (2026/2027)

🌳 Conceptual Hierarchy Tree

Newtonian Dynamics and Work-Energy

├── 2. Motion in Elevators (Apparent Weight N)

📋 Master Teacher's Blackboard Secrets

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