1

time for the motion of a car.

Diagram 1 (a) State the physical quantity represented by (i) the gradient of the graph.

................. [1 mark]

(ii) the area under the graph. ............................................[1 mark] (b) Complete the table below.

Section of the graph Type of motion of the car

OA

AB

[2 marks] Soalan N9-2008 2. The equations of motion provides a useful means of predicting information about an objects motion in a straight line. The equation which describe an objects motion is given by the equation : s = ut + at2 where s is displacement, u is initial velocity, a is acceleration and t is time. (a) (i) What is meant by base quantity ? .................................. [1 mark] (ii) What is the base unit for acceleration? ...................................... [1 mark] . (b) Based on the equation, give one example of

QUEST. MARKS

1/04

2/04

3/07

4/06

5/10

6/14

7/20

TOTAL/65

Name: ................................................................. Class:............................ LINEAR MOTION SBP 2008 1. Diagram 1 shows the graph of velocity against

2

Base quantity Derived quantity.

Scalar quantity Vector quantity

[2 mark]

Perlis 2009 3. Diagram 1.1 shows a photo picture of a stroboscope for a motion of an air puck. The

strosboscope is set to give 20 flashes per second.

(i) Explain the motion of the air puck?

............................................................................................................................... [1 mark]

(ii) How many images will be produced in 2 s?

............................................................................................................................... [1 mark]

(iii) Find the acceleration?

[2 mark]

(a) Diagram 1.2 shows the photo picture of a moving ping pong ball taken by a mechanical strosboscope which has two slits and rotate with 5 rotation per second.

(i) What is the time interval between two adjacent images?

............................................................................................................................... [1 mark] (ii) Find the average speed for the motion

3

............................................................................................................................... [1 mark]

(iii) If the number of slit of the strosboscope is increase, what will happen to the distance between the images? ............................................................................................................................... [1 mark]

Pahang 2008 4. Diagram 3.1 shows a motorcycle driven by a snatch thief overtaking a police car at 4.00 p.m.

The police car pursued the motorcycle. Graph from Diagram 3.2 shows velocity against time graph for both vehicles.

Diagram 3.1

Diagram 3.2

(a) Based on the graph in Diagram 3.2, state the type of motion for,

(i) Motorcycle

..............[1 mark]

(ii) Police car

..............[1 mark]

(b) (i) What is the net force acting on the motorcycle? .......................................................................................................[1mark ]

(ii) Give a reason for your answer in 3(b)(i).

..................................................................................................... [ 1 mark ]

Police car

Kereta Polis

Velocity / ms-1

Halaju / ms-1

Time

Waktu

0 4.00 pm

30

35

4.01 pm 4.02pm

Motorcycle

Motosikal

4

(c) Based on the graph in Diagram 3.2, calculate the distance travelled by the police

car from 4.00 pm to 4.02 pm. .

[2 marks]

Selangor 2009 5. Diagram 7.1 shows a tanker with a mass of 2900 kg. The trailer is carrying petrol in one big

tank.

(b) If the tanker starts from rest and achieves a velocity of 40 m s-1

in 20 s, calculate (i) its acceleration.

[2 marks] (ii) the force acting on the tanker.

[2 marks]

(b) The tanker in Diagram 7.1 is used for transporting large amounts of petrol. Suggest modifications that can be made based on the following aspects.

(i) number of tyres

............................................................................................................................... [1 mark]

Reason ............................................................................................................................... [1 mark]

(ii) number and size of tanks

............................................................................................................................... [1 mark] Reason. .............................................................................................................................. [1 mark]

5

(c) Diagram 7.2 shows the speed limit displayed on the back of the tanker.

(i) What is meant by speed?

............................................................................................................................... [1 mark]

(ii) Explain why the speed limit must be imposed on heavy vehicles.

................................................................................................................................ ................................................................................................................................[1 mark]

Ahir2010 6. Diagram 6 shows the velocity-time graph of a car

Diagram 6 Based on the graph,

(a) What is the maximum velocity of the car

......[1 mark]

(b) Explain the pattern of the movement of the car at :

JK

KL

LM

N

O

6

[3 marks] (c) Calculate the displacement travel by the car from J to O

[2 marks]

(d) Is the value of distance and displacement travel by the car from J to O is same? If not, what are the differences?

..... .....[2 marks]

(e) Calculate the average velocity travel by the car from J to O

[2 marks] (f) When does the car stop? Explain

. [2 marks] (g) Change the graph above to the graph acceleration against time from J to O

[2 marks]

8

SKEMA JAWAPANLATIH TUBI 3 LINEAR MOTION

1.(a)(i) (ii)

(b)

Acceleration Displacement OA : Uniform acceleration / velocity increases uniformly AB : Uniform velocity / acceleration is zero

1 1 1 1 Total 4

2(a) (i)

(ii) (b)

base quantity is a quantity that cannot be defined in any other physical quantity ms-2 Base quantity - time / displacement Derived quantity - velocity / acceleration Scalar quantity time Vector quantity - displacement / acceleration

1 1 1 1 Total 4 3(a)(i)

(ii) (iii) b(i) (ii) (iii)

Constant accelerations/increasing in velocity 40 images 4000cms-2/40ms-2

0.1s 50cms-1 decreases

1 1 2 1 1 1 Total 7

4(a) (i) Constant/ uniform velocity // acceleration is zero 1

(ii) Increasing velocity uniformly followed by constant velocity/ zero acceleration

1

(b) (i) zero // F = 0 1

(ii) constant / uniform velocity // acceleration is zero 1

(c) Distance = area under graph

= m315035601202

1

2 Total 6

5.a(i )

(ii)

b(i)

(ii)

c(i)

(ii)

a = 40 0 20

= 2 ms-2

1. F = 2900 x 2

2 = 5800 N//kgms-2

1. increase the number of tyres

2. reduces pressure

1. use many small tanks // divide large tanks to small tanks

2 reduces the effect of inertia

distance // rate of change of distance

time

momentum of heavy vehicle depends on the speed

directly proportional to the speed or greater momentum

causes more damage /

difficult to control direction /

greater impulsive force when accident occurs

2 2 1 1 1 1 1 1 Total 10

9

v / ms-1

5

t / s

0 20 50 65

6 a) 20 ms-1 b) Decreasing in velocity//uniform deceleration c) = (1/2x10x20)+(1/2x(10+20)x10)+(10x10)+(1/2x10x5)

(1/2x10x5) = 350 m

d) - Not - 50 m

e) = 350/40 = 8.75 ms-1

f) -At N // t = 35 s -Velocity is zero

g) Graph acceleration against time

1 1 1 1 1 1 1 1 1 1 1 1 Total: 12

7(a) the quantity of matter 1

1

(b) (i) (b) (ii)

PROPERTIES REASON

high mass the vehicles becomes more stable

high engine capacity -the power is higher -produce high acceleration

diesel types of engine the cost is low

bigger diameter of the tyre the pressure is low // more stable

suitable vehicle is P because it has high mass, high engine capacity, using diesel and bigger diameter of the tyre.

2 2 2 2 2

(c) (ii) y- axis ( with correct value only ) x axis (with correct value only )

1 1 1 1 1 1 1

(c) (ii) Distance = Area under the graph = x 5 x 20 + 30 x 5 + x 15 x 5 = 237.5 m

1 1

(c) (iii) a = gradient of the graph in the 1st part. = 5/ 20 = 0.25 ms-2

1 1

Total 20

1m

1m 1m

10 20 30 35 40

a/ms-2

t/s

2

-1

-2

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