Download smjk yu hua kajang marking scheme peperiksaan

SMJK YU HUA KAJANG
MARKING SCHEME
PEPERIKSAAN PERTENGAHAN TAHUN 2014
PHYSICS PAPER 2 FORM 4
NO
MARKING CRITERIA
No 1
1 (a)
(b) (i)
(b) (ii)
(b) (ii)
Systematic
+ 0.03 cm
To measure the internal diameter
Actual Reading: 4.38 cm – 0.03cm = 4.35cm
MARKS
SUB
TOTAL
1
1
1
1
TOTAL
4
No 2
(a) (i) Momentum is product of mass and velocity or
Momentum = mass x velocity
(b) (i) Momentum before collision: 2.5 + 0 = 2.5 kgms-1
(ii) Momentum after collision: 0.4 + 2.1 = 2.5 kgm-1
(c) Compare:
Total momentum before collision equals to total momentum after
collision.
SUB
1
(d) (i) Principle of conservation of momentum
1
1
1
1
TOTAL
No 3
(a)(i)
(ii)
(iii)
(b)
(c)
(d)
No 4
(a)
(b)(i)
(ii)
(iii)
-1
Initial Velocity u = s/t = 0.5/0.02 = 25 cms
Final velocity: v = s/t = 0.8/0.02 = 40 cms-1
a = (v-u)/t = (40-25)/0.02x3 = 250 cms-2 = 2.50 ms-2
Uniform acceleration or constant acceleration
Net Force F=ma  1.5 x 2.5 = 3.75 N
New distance of ST = 1.6 cm
Force is defined as the rate of change of momentum
Horizontal Component: F cosθ = 20 x cos 40 = 15.32 N
Verticle component : F sinθ = 20 x sin 40 = 12.86 N
Reaction force acting on stroller
R = mg + Fy = 8x10 + 12.86 = 92.86 N
(No need to draw Free Body Diagram)
(c) Acceleration : F = ma  a = Fx/m = 15.32 /8 = 1.92 ms-2
(d) Decreases
MARKS
5
SUB
TOTAL
1
1
1
1
1
1
TOTAL
6
SUB
1
1
1
1
TOTAL
2
1
TOTAL
7
No 5
(a)(i) A to B : Constant velocity //uniform velocity // acceleration = 0
(ii) B to C: Uniform acceleration //constant acceleration
(iii) C to D : Uniform deceleration // constant deceleration
(b) Acceleration in 1st 3 seconds
a = 5/5 = 1 ms-2
(c) Total distance travelled
s = 12.5m + 35m + 22.5m + 15m
= 85m
(d) Average velocity = total displacement / total time
v = 85/18 = 4.72 ms-1
SUB
1
1
1
2
1
1
1
TOTAL
No 6
(a)(i) Fw = FR
If in words also accept answer
(ii) Fw > FR
If in words also accept answer
(iii) Fw < FR
(b) F = ma
FW – FR = ma
20000 – FR = 1500 x 5
FR = 20000 – 7500 = 12,500N
(c) Impulsive Force = (mv – mu)/t
Note: 110 kmh-1 is converted to 30.556 ms-1
Impulsive Force = (0 – 1500 x 30.556) / 0.5
= 91668 N or 91667 N
TOTAL
SUB
1
8
TOTAL
1
1
2
1
1
(d) 2 ways to reduce the impulsive force
Way 1: Soft front bumpers
Way 2: Crumple zone at the front bonnet
1
1
TOTAL
9
No 7
(a) Forces in equilibrium means
Net force or resultant force acting is zero //
Force vectors when added can form a closed polygon.
(b)
T1
String
T2
o
SUB
2
TOTAL
1
o
45
45
W
T1 and T2 can be interchanged.
If direction of T1 and T2 is not correct, less 1 mark
No labels : no marks
(c)
1
2
T2
W
T1
Triangle is correctly drawn. Angle not important
Labels are all correctly placed
(d) Since forces are in equilibrium
Resultant Force = 0
1
(e) Tension of the string
T sin 45 + T sin 45 = 2 x 10
2T sin 45 = 20
 T = 14.14 N
(f) String will snap/break
1
1
1
TOTAL
10
No 8
(a)
X: Meter ruler
Y: Vernier Caliper
Z: Micrometer Screw gauge
(b) Average
X : 1.3 cm ( If 1.26 or 1.27 is wrong)
Y: 1.23 cm
Z: 1.204 cm
(c) Accurate instrument is : Z
(d)(i) Passengers are thrown forward
(ii) Inertia
(iii) 1. Anti inertia Seatbelts
2. Air bags
3. ABS Brakes
4. Crumple zone
5. Head rest
6. Bumpers
7. Shatter proof windscreen
(Select any 3)
(iv) Higher speed, higher frictional force
Value: > 30,000 N
(Any value greater than 30000N accept)
Both answers right, then award 1 mark
SUB
1
1
1
TOTAL
3
2
1
1
1
3
1
TOTAL
12
No 9
(a) (i) Momentum is the product of mass and velocity or
p=mv where m: mass (kg) and v is the velocity (ms-1)
(ii)
o Total momentum in diagram 9.1 is zero
o Total in Diagram 9.2
m1v1 + m2v2 = 60x5 + 50(-6) = 0
o Total momentum before and after boy jumped to jetty are equal
o Principle of conservation of momentum
(b) (i)
SUB
1
TOTAL
1
1
1
4
1
1
3
36
48
t/s
Straight line through the (0,0)
Values are marked correctly on axis
labels on axis and units of physical quantity
1
1
1
No units in axis label – less 1 mark
(ii) Acceleration of train: Gradient of graph
a = 36/48
= 0.75 ms-2
(c)(i)
2
1
1
2
V / ms-1
36
48
120
t/s
(ii) Total distance travelled is the area under the v-t graph
Area = ½ x (120 + 72) x 36 = 3456 m
(d) (i)
(i) Area of shaded part is 432m
V / ms-1
432 = ½ (t-120) x 36

(t-120)
= 24 sec
36
(ii)
120 t
t/s
6
2
2
Area= 432m
48
2
(ii) Deceleration = gradient
a = 36/24 = - 1.5 ms-2
(iii) Acceleration is the rate of change of velocity
2
TOTAL
20
No 10
(a) (i) Impulsive Force
(ii) 1. The surface of metal block is hard
2. The time of impact is shorter
3. The impulsive force is bigger
(iii) Drop on a soft surface // drop on a pile of sand/mud
SUB
1
1
1
1
TOTAL
1
3
1
(b) (i)
10
Specification
Mass of pile
driver
Height
of
pile driver
Pile Material
Characteristic
Bigger mass of
load
Higher position of
pile drive
Steel Pile is used
Shape of the Shape must
base pile
pointed
Choose
Q is chosen
Explanation
Produce a bigger force during
impact
Produce a higher velocity when
striking the pile
Steel is stronger and stiffer. Is
not brittle and does not crack on
impact
be Produce a higher pressure //
easy to push the pile into the
ground
Because piling system Q has big
mass of pile driver, position of
pile driver is high and steel piles
are use with sharp pointed base
2
2
2
2
2
(c)(i) W = mg
= (50) (10) = 500 N
(ii) Velocity of metal pile just before it hits the pile
1
2
v2 = u2 + 2gh
= 0 + 2 (10) (20)
= 400
 v = 20 ms-1
(iii) Impulsive Force
F = mv – mu
t
= 50 (0-20)
0.5
1
= 2000 N
1
1
2
1
TOTAL
Prepared by Pradeep Kumar Chakrabarty
SMJK Yu Hua Kajang
2014
20