Force and Motion JPN Pahang Ans

8/9/2019 Force and Motion JPN Pahang Ans

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FORCE AND MOTION

Distance and displacement

1. i.has only a magnitude and directionii. has both magnitude and direction

2. i. length of the path taken

ii. distance of an object from a point in a certain direction3. a. by car = 41+53 = 94km

b. by plane = 60kmThe path traveled by the plane is shorter than travelled by the car.So, distance = 94km. Displacement = 60km to East

S peed and velocity

1. Speed is the distance traveled per unit time or rate of change of distance2. Velocity is the speed in a given direction or rate of change of displacement

3. Average of speed; ):(,)(, _ )(, _ tan _ 1! msunit t sV st takentime m straveled cedistotal

4. Average of velocity; )(,)(, _

)(, 1! mst

sV

st takentimem snt displaceme

5. i.1175

4400300

tan

!!

!

k h

Ti ece Dis

Speed

ii. 1125

4

500!!! k h

ti

e

nt Displace evelocity (in the direction of 053)

Acceleration and deceleration

1. The velocity of the car increases2. Acceleration is the rate of change of velocity. Then

t uv

geti eofchanocityinitialvel ity Finalveloc

a !!

3. i. From A to B 2102

020!! m sa AB

ii. From B to C 21022040

!! m sa BC

4. When the velocity of an object decreases, in calculation a will be negative.5. st msumsv 5,30,0 11 !!!

Then, 265

300 !!! m st

uva


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FORCE AND MOTION

Analyzing of motion

1. i. The frequency of the ticker timer = 50Hz ( 50 ticks in 1 second)

So, 1 tick ondsond

sec02.050

sec1!!

ii. xy = displacement over time tt=7 ticks = 0.14s

iii. U niform velocity ; Acceleration ; Acceleration , then deceleration

iv. 19.8714.0

3.12,

14.002.07,3.12

!!

!!!

cm sv

elocity

s s X tim ecmnt displace m e

v. 20.25

)2.0(5

0.150.40!!

!

m s

t uv

a

The equation of motion

1. The important symbols: s:displacement, v: final velocity, u: initial velocity, t:time,a:acceleration.

2. a. t vu s )(21

! b.t

uva !

c. at uv!

d.2

21

at ut s !

d. a suv 222 !

3. G iven 110! m su , 23! m sa , t=20s, s??

m

a t ut s

800

)20)(3(21

)20)(10(

21

2

2

!

!

!

4. G iven 116! m su ,v=0(rest), s=8m, a=?

2

22

22

16

)8(2160

2

!

!

!

m s

a

a suv


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FORCE AND MOTION

e) G raph is quadratic form.

Displacement increases with time.

G raph gradient decreases uniformly.

The object moves with decreasing velocity, with uniformdeceleration.

f) OA = uniform velocity (positive move ahead)

AB = velocity is zero (rest)

BC = uniform velocity (negative reverse)

The velocity-time Graph

a) No change in velocity

Zero gradient the object moves with a constant velocity orthe acceleration is zero.

The area under the graph is equal to the displacement of themoving object

s = v x t

b) Its velocity increases uniformly

The graph has a constant gradient

The object moves with a uniform acceleration

The area under the graph is equal to the displacement,s of the moving object:

)(21

t v s v!

c) The object moves with a uniform acceleration for t 1 s

After t 1 s, the object decelerates uniformly (negative gradient) until it comes to rest.

The area under the graph is equal to the displacement of the moving object:

2

21

vt s !

d) The shape of the graph is a curve

Its velocity increases with time.


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The gradient of the graph increases.

The object moves with increasing acceleration.

The area under the graph is equal to the total displacement ofthe moving object.

e) The shape of graph is a curve

Its velocity increases with time.

The gradient of the graph decreases uniformly.

The object moves with a decreasing acceleration.

The area under the graph is the total displacement of themoving object.

Examples

1) G iven : S OP =20 m S OQ = 20 m S OR = 0 mSOS = -10 m

tOP=2 s t PQ = 3 s t QR = 2 stRS= 1 s

i. 1102

20!! m s OP

1102

200!! m s QR

1101

010!! m s RS

ii. mS 101020!!

2) G iven: 10! msV , 110! msV , 110! msV

, 10! msV

i. 25.24

010!! m sa QP ,

204

1010!! m sa PQ ,

252100

!! m sa QR

ii. mS 0.70)10)(104(21

!!

Exercise 2.2

1) a) The body remains in rest 5 m at the back of initial point.

b) The body start move at 10 m infront of the initial point, then back to initial pointin2s. The body continue it motion backward 10 m. The body move with uniformvelocity.

c) The body move with increasing it velocity.2) a) The body move with uniform velocity , 5 m s-1 backward.

b) The body start it motion with 10 m s-1 backward and stop at initial point in 2 s,then continue its motion forward with increasing the velocity until 110 ms in 2 s.


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FORCE AND MOTION

2.3 Understanding inertia

1. A pillion rider is hurled backwards when the motorcycle starts to move

2. Bus passengers are thrust forward when the bus stop immediately

3. Large vehicle are made to move or stopped with greater difficulty

4. Inertia of an object is the tendency of the object to remain at rest or if moving, to

continue its uniform motion in a straight line.

Mass and inertia

1 i. An adult

ii. An adult

2. The larger the mass, the larger its inertia

3. The larger mass have the tendency to remain its situation either at rest of in moving

Effects of inertia

1. Application of inertiai. Drying off an umbrella by moving and stopping it quickly.

ii. Building a floating drilling rig that has a big mass in order to be stable andsafe.

iii. To tight the loose hammer 2. We should take precaution to avoid the effect

i. During a road accident, passengers are thrust forward when their car issuddenly stopped.

ii. Passengers are hurled backwards when the vehicle starts to move and arehurled forwardwhen it stops immediately.

iii. A person with a heavier/larger body will find it move difficult to stop hismovement.

iv. A heavier vehicle will take a long time to stop.

Exercise 2.3

1. Inertia is the tendency of the object to remain at rest or, if moving, to continue itsuniformmotion in a straight line . Y es, the inertia increase with the mass increased.

2. a) A wooden block move up of a wooden dowel.A wooden block has inertia toremains at rest.


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FORCE AND MOTION

b) The wooden block move downward of a wooden dowel.A wooden block hasinertia to continue it motion.

2.4 Analyzing momentum

1. it has momentum

2. depends on its mass and velocity

3. As the product of its mass and velocity, that is

Momentum, vm p v! 1! kgmsUnit

The principle of conservation of momentum

In the absence of an external force, the total momentum of a system remains unchanged.

1. Elastic collision: The colliding objects move separately after collision.Momentum: 22112211 vmvmumum !

2. Inelastic collision: The co lliding objects move together after collision.Momentum: vmmumum )( 212211 !

3. Explosion: The objects involved are in contact with each other before explosion andare separated after collision

Momentum: 221121 0)( vmvmmm !

Example 1:

G iven: kg m 100! , 130! msu , 125! msv , kg m 90! , 120! msu , ?! Bv

B B A A B B A A vmvmumum !

Bv9025100209030100 !

1

56.25!

msv B

Example 2:

G iven: kg m A 100! ,130! msu A , kg m B 90! ,

120! msu B , ?! B Av

B A B A B B A A vmmumum !


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FORCE AND MOTION

B Av! 90100209030100

126.25! msv B A

Example 3:

G iven: kg g mb 002.02 !! , kg m g 1! , 0!b g u ,1150! msvb , ?! g v

bb g g umvm!0

150002.010 ! g v 13.0! msv g

Exercise 2.4

1. Arrow: g ma 150! ,115! msva

Wooden block: g mwb 450! , 0!wbv

Wooden block + Arrow: g m ba 600! , ?!w bav

wbawbawbwbaa vmvmvm !

w bav!vv 6.0045.01515.0

175.3! msv wba

2. Riffle: ?,0.5 !! r r vkg m

Bullet: 180,50 !! msv g m bb

bbr r vmvm !

8005.00.5 !r v 18.0! m svr

Understanding the effect of a forceIdea of force

1. Force can make an object;a. Move

b. Stop the movingc. Change the shape of the object


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FORCE AND MOTION

d. Hold the object at rest

Idea of balanced forces

1. i. In a stationary state

ii. Moving at uniform velocity

2. Magnitude R = W but R acts in an opposite direction to the weight.(object is inequilibrium)

3. Force, F = FrictionResultant = F-Friction

= 0 (objects is in equilibrium)Examples:

a) A car move at constant velocity

b) A plane flying at constant velocity

Idea of unbalanced forces

1. Because the force act are not balanced2. The ball move in acceleration when ball is kicked by a resultant force,F.

So, the ball move in the direction as the applied force.

Relationship between forces, mass and acceleration (F = ma)

1. It is found that, F a w when m is constant and ma 1w when F is constant

Therefore, m F a w

From m F a w ,

ma F w Therefore, kma F ! , k = constant

2. ma F !


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FORCE AND MOTION

t

m um v F ! , unit impulsive force: newton(N)

F is defined as impulsive force which is the rate of change of momentum over theshort period of time.

Example1:

Impulsive, N s

mumv F t

100

)10(5)10(5!

!

!

and impulsive force, N 1001

100!

Example 2:

Impulsive, N s

m um v Ft

100)10(5)10(5

!!

!

and impulsive force, N

502100

!

Impulsive force, t F 1w

Therefore, F decreases when the time of collision increases.

Exercise 2.6

1.

N s

Ft F im pulse

0.168.020

!

v!

!

2. i). A large canvas bag will increase the time of collision.ii) When the time of collision increase the impulsive force will decrease.

2.7 BEING AWARE OF THE NEED FOR S AFETY FEATURE S IN VEHICLE S


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FORCE AND MOTION

S afety features ImportancePadded dashboard I ncreases the time interval of collision so the

impulsive force produced during an impact is thereby reduced

Rubber bumper A bsorb impact in minor accidents, thus prevents damage to the car

Shatter-proof windscreen P revents the w indscreen from shattering Air bag A cts as a cushion for the head and body in an

accident and thus prevents injuries to the driver and passengers.

Safety seat belt P

revents the passengers from being throw

n out of the car. Slo w sdo w n the for w ard movement of the passengersw hen the car stopsabruptly.

Side bar in doors P revents the collapse of the front and back of the car into the

passenger compartment. A lso gives good protection from a side-oncollision.

Exercise 2.7

1. - The absorber made by the elastic material: To absorb the effect of impact (hentaman) during it moving- Made by the soft material of bumper : To increase the time during collision, then the impulsiveforce will be decreased.


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FORCE AND MOTION

- The passengers space made by the strength materials.: To decrease the risk trap to the passenger during accident.- Keep an air bag at the in front of dash board and infront of passengers: Acts as a cushion for the head and body in an accident andthus prevents injuries to the driver and passengers.- Shatter-proof windscreen: Prevents the windscreen from shattering.

2.8 Understanding gravity1. It pulled by the force of gravity 2. as earths gravitational force 3. the object is said to be free falling 4. is known as acceleration due to gravity

5. on the strength of the gravitational field

Gravitational field

1. the gravitational field of the earth2. is on the force of gravity3. as the gravitational force acting on a 1 kg mass

4. m

F g ! where F : gravitational force; m : mass of an object

5. 18.9! N kg g

6. That an object of mass 1 kg will experience a gravitational force of 9.8N.7. Example 1:

N

m g F

5888.960 !v!

!Example 2: 18

604800

!!! N kg m

F g

Example 3: a)

176.11

2.18.90!

!!

ms

at uvb)

m

a t ut s

1.7

2.18.9212.10

21

2

2

!

!

!

Weight

1. As the gravitational force acting on the object2. m g wweight !, ; where g= acceleration due to gravity

Example: N mg W 5868078.96000 !v!!

Exercise 2.8


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FORCE AND MOTION

1.

a) It is at stationary state, S=0m b) It is moving with increasing velocityc) It is moving with uniform acceleration

2. 2

2

2

6.9

79.0217.000.3

21

!

!

!

m s g

g

gt ut s

The answer less than the constant because of the air frictional force

2.9 Idea of equilibrium forces

1. An object is stationary

2. moving with uniform velocity

i) Magnitude of R=W and it acts in opposite directionSo the resultant force, W-R=0(objects in equilibrium)


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FORCE AND MOTION

ii) Magnitude of Ucosmg

! and acts in opposite direction

So the resultant force = 0cos !Rmg U (objects in equilibrium)iii) Force , F=Frictional force; Resultant force = F-Frictional force = 0

(object in equilibrium)

Addition of force

1. A resultant force is a single force that represents in magnitude and directiontwo or more forces acting on an object. F resultant = the total of forces(including the directions of the forces)

N F F F force sultamt 15510, _ Re 21 !!! N F F F force sultamt 5510, _ Re 21 !!!

Example 1: Resultant force, N F 70053006000 !! They were not in equilibrium

Example 2:Resultant force, N F 525505.10 !v!

Resolution of a force

1.

reverse process of finding the resultant force

F

F

F F y x

!! UU sin;cos

UU sin;cos F F F F y x !!


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FORCE AND MOTION

Example 1:

N F F x

255.050

60cos50cos

!!

!! U N

F F y

3.43

866.05060sin50

sin

!

!!

! U

F is the resultant force of F x and F y

Therefore, F can be resolved into

F = mg sin 400 + 200

= 800(0.6427) + 200

= 514.2 + 200

= 714.2 N

P roblem S olving

1. the resultant force is equal to zero2. the sum of each component is equal to zero


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FORCE AND MOTION

3. kg mlamp 5.1! , N W lamp 7.14!

c)

N T

T W

T T

lamp

82.770sin281.95.1

70sin2

70sin2'

!v

!

!

!

r

r

r

4.

F maximum when both of forces act in same direction,

N F imum 24618max !! ;

F minimum when both of forces act in opposite direction,

N F im um 12618min !!


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FORCE AND MOTION

Example 4:

J

F sW

4808.0600 !v!

!

Energy

1. It is the potential to do the work

2. created nor be destroyed

3. potential energy, kinetic energy, electrical energy, sound energy, nuclear energy andchemical energy.

4. When we are running up a staircase the work done consists of energy change fromChemical energy >Kinetic energy > Potential energy. The energy quantity consumed isequal to the work done.

5. If 100J of work is done, it means 100 J of energy is consumed.

Work done and the change in kinetic energy

1. energy of an object due to its motion


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FORCE AND MOTION

2.)2

1( 2vmmas

F sWork

!!

!

The formula of kinetic energy, 221 m v E k !

3. G iven : N F kg m 200;100 !!

a) Kinetic energy, J Fs E k 200010200 !v!!

b) Velocity,1

2

32.6

200021

!

!p

m sv

m vv

Work done and gravitational potential energy

1. energy of an object due to its position.(possessed by an object due to its position in a

gravitational field)

2. Maximum potential energy, mg F w heremgh F sW !!! :,

3. J

W

147

)5.1)(8.9(10!

!

; Therefore work done = 147J and Ep = 147J

P rinciple of conservation of energy

1. created or destroyed but can be changed from one form to another form.

2.

Kinetic energy decrease andpotential energy increase

KKinetic energy increaseand potential energy

decrease


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FORCE AND MOTION

3. G iven, ;8.9;0;20 2 !!!! vm s g umh

12

2

2

8.19,39221)20)(8.9(

21

!!

!

!

!

m svv

m vm

m vm gh

E E k p

P ower

1. The rate of doing work.

Therefore, power,t

W P so

tim et a ken

workdone P !! ,,

Where, P : power in watt/W

W : work in joule/J

t : time to do work in seconds/s

2. ?,8.9,,8.0,2,180: 2 !!!!! P ms g and st mhkg mGiven

W t

mgh

t

W P

44108.0

28.9180!

vv!

!!

Efficiency

1. As the percentage of the energy input that is transformed into useful energy.

2. %100 _

_ _ X

input energyinput energyuseful

E fficiency !

3.


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FORCE AND MOTION

4. a) Solution : G iven : m = 0.12 kg, s= 0.4 m, t = 5 s, E input = 0.8 J

J

s F F output 48.0

4.01012.0

!

vv!v!

b)

%60%10080.048.0

%100

!v!

v!

E fficiency

E E E fficiency

input

output

Exercise 2.10

1. J

FsW

9001090 !v!

!

The energy transferred to the force =900J

2. a) kg m 150500.3 !v! mh 5.1!

J

m ghw

2205

5.18.9150

!

vv!!

b) W t

W P 82.8

2502205

!!!

2.11 Appreciating the importance of maximizing the efficiency of devices

1. some of energy transformed into unwanted forms of energy.

2. The efficiency of energy converters is always less than 100%.

3. The unwanted energy produced in the devices goes to waste.


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FORCE AND MOTION

5. the best possible use of the input energy.

Ways of increasing the efficiency of devices

1. Engine must be designed with the capability to produce greater amount of mechanical work.

2. Light Fittings

- replace filament light bulb with fluorescent lamps which have higher efficiency.

- use a lamp with a reflector so that the illumination can be directed to specific areas of the

user.

Air-conditioners

- choose a model with a high efficiency.

- accommodate the power of air-conditioner and the size of the room

- Ensure that the room totally close so that the temperature in the room can be maintained.

Refrigerators

- choose the capacity according to the size of the family.

- installed away from source of heat and direct sunlight.

- the door must always be shut tight.

- more economical use a large capacity refrigerator.

- use manual defrost consumption.

Washing machines

- use a front loading as such more economical on water and electricity.

- front loading use less detergent as compared to a top loading machine.

Operation of electrical devices

1. when they are in good operating

2. condition. Will increase the life span of device.


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FORCE AND MOTION

3. Example : the filter in an air-conditioner and fins of the cooling coil of a refrigerator must be periodically cleaned.


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Reinforcement Chapter 2

1. C 6. D

2. C 7. D

3. A 8. D

4. C 9. A

5. A 10. D

P art B : S tructure Questions

1. a) i. Method A

ii. The forces given parallel with the surface of motion. So all the forces given are used to

move the breakdown car.

b) i. N

F F F friction given300200500 !!

!ii.

N

F F F friction given

5020060cos500

50cos

!!

!r

r

iii.205.0

10000.50

!

!

!

msa

a

ma F

c) The acceleration of car A 23.0! ms

To move Car B with the same acceleration of Car A, increase the force given to 1000N.

2. a) The difficulty to move the tin depends to its mass

b) From position M the velocity of tin is more than the velocity compare when it is from N.

Ek increase then the force to stop it will be increased.

c) Mass and velocity

d) Inertia of tin Q will decrease because the mass of tin decreased.

3. a) i. J

mv E k

0.4)2)(2(21

21

2

2

!!

!


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FORCE AND MOTION

ii. J

m gh E p0.60)0.3)(10)(2( !!

!

iii. J

E E E pk

0.640.600.4 !!

!

c) i. 64.0J(the conservation of energy)

ii. J

s F W f 0.4

0.44

!

v!v!

d) J

E E E f s0.600.40.64 !!

!

J

sa t E E a tS E p sk 0.20)0.2)(10)(2(0.60

_

!!

!

e)))(2(2

12

1

0.20 _ 22 vm v

J T a t E k

!!

!

1

2

5.4

20!

!

msv

v

P art C Essay Question

Answer

1a) (i) momentum is product of mass and velocity

(ii) - The shape of car changed but the shape of wall remained.

- The shape of ball remained but the shape of the racquet string was changed. (The racquet

string is elastic but the wall is harder)

- The time taken of collision between the ball and racquet string more than the time taken

when the car hit the wall.

- The impulsive force will decrease when the time of collision increased.

- The concept is the impulsive force.


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b) - To decrease the time of collision between the ball and the racquet string.

- Impulsive force will be increased.

- The force act to the ball will be increased.

- The velocity of ball will be increased.

c) - Make a gradually narrower at the front shape (tapering)

: To decrease air friction

- Made by the high strength and high rigidity of materials

: To decrease the probability to become dented (kemik).

- Made by the low density of material.

: To reduce the mass/weight

- The structure is fractional engine

: The mass will be decreased and the velocity will increase.

- Made by the high of heat capacity of materials

: It will be high heat resistance.

2.(a) Increase the velocity

(b)

(c) - time reaction mast be short : fast to detect the signal to start its move

- has a small of mass : to decrease the inertia, then easier to start move and to

stop its moving.

- thrust force is high : has more power during its moving / increase the

acceleration


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FORCE AND MOTION

- friction force is low : decrease the lost of force

- the best car is A : because it has short of time reaction, small of mass, high of

thrust force and low friction of force.

(d) (i)

J E B .505)50(4.25.12 !!

J

mghupmoveto suitable E

0.45030sin50108.1

) _ _ _ (

!

vv!

!

r

E E B " (Car B can mov up the plane)

(ii) ;ma F !

261.5

8.14.25.12!

!

m sa

a

Force and Motion JPN Pahang Ans

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