FM 4 PHYSICS MID YEAR EXAMS 2015
PAPER 1
SMJK YU HUA KAJANG
SOLVING PROBLEMS THAT REQUIRE HOTS (HIGHER ORDER THINKING SKILLS)
Questions No 15,16 and 17
Diagram 5 show a displacement-time graph of a cyclist for a duration of 30 seconds. Questions 15 to 17 will be based on the graph
Diagram 5
15.What can you deduce regarding the final position of the cyclist?
A.The cyclist is 1200 m away from the starting point
B.The cyclist is 1500 m away from the starting point
C.The cylclist is 60 m behind the starting point[PC1]
D.The cyclist is 180 m in front of the starting point
16.Based on the displacement-time graph in question 15, determine the velocity of the cyclist from A to B
A. 60 ms-1B. – 12 ms-1[PC2]C. 0 ms-1D. - 20 ms-1
17.At location C, explain the state of motion of the cyclist
A.StationaryB.cycling with a uniform velocity[PC3]
C.decelerating uniformlyD.accelerating uniformly
Analysis of question
A displacement-time graph is shown above. The gradient of the graph will represent velocity. The Area under a displacement-time graph does not represent anything at all.
OA: The cyclist travels in a straight line a distance of 120m for 10 sec with a uniform velocity.
AB: Cyclist turns back immediately and heads back to its starting position.
He does not stop at 0 but continues cycling past his starting position for another 60m.
Q15: The answer is C
Velocity from A to B is a uniform velocity but in the reverse direction. Since velocity is a vector, it will have a negative value. The gradient of the graph is -120/10 = -12 ms-1.
Q16: The answer is B
At location C, since the gradient of the graph is not zero, cyclist is moving with a uniform velocity
Q17: B
Additional questions that can be given
If you are asked to convert this s-t graph into a v-t graph, it will look like this
OA: Uniform velocity of 12 m/s. AB: Uniform velocity of -12 m/s and BC: Uniform velocity of – 6 m/s
Q18:
Requires a knowledge of momentum as a vector quantity.
Velocity towards the right is positive
Velocity towards the left is negative
Change in momentum is mv – mu = 0.5 (20) – 0.5(-30)
= 10 – (-15) = 25 kg ms-1 or Ns
Q20
On an s-t graph, a object moving with a constant acceleration should show a quadratic curve
As in C.
D show uniform deceleration
B show uniform velocity ( acceleration =0)
A shows zero velocity or stationary object
For question 23, Netwons 2nd Law of motion is applied
Net Force F = ma Forward Thrust – Fricitional Force = ma
4000 – Fr = 1200(2.5) Fr = 4000 – 3000 = 1000N
Ans: B
Q24
This question also uses the Newton’s 2nd law of motion. It requires analytical and logical thinking. Higher Order Thinking skills is required to solve this problem.
When an object moves with a uniform velocity, the net force acting of the object is zero.
Since both Car A and Car B move with a uniform velocity, the net force acting on both cars is zero. But that does not mean that they are subjected to the same fricitional force. A car moving at a higher uniform velocity will be subjected to a higher frictional force. A car moving at a higher speed will have a greater engine thrust compared to a car moving with lower speed.
For Car A: Net Force F = 0 F1 - Fr1 = 0 F1 = Fr1 ( F1 = Engine Thrust for Car A, and Fr1 = Frictional Force acting on the tires of car A)
For Car B: Net Force F = 0 F2 - Fr2 = 0 F2 = Fr2 ( F2 = Engine Thrust for Car B, and Fr2 = Frictional Force acting on the tires of car B)
Therefore; since F2 > F1 , that means that Fr2 > Fr1.
This question shows us that when a car is driving as a higher uniform velocity on a road, the enjine thrust is higher and thus ,frictional force acting on the tyres of the car is higher. At higher speeds, the car tyres wear off faster and consumes more petrol (since engine thrust is higher).
Ans: C
Q25
This question requires you to have an understanding and application of the formula
a = (v-u)/t .
Since the ticker tape shows an unform increase in length, the acceleration produced by the object is uniform. Since each contains 10 ticks, the time taken for each tape is 0.02 x 10 = 0.2s
We assume that the ticker timer is operating with a 50Hz AC source
Initial Velocity u = s/t = 4/0.2 = 20 cms-1
Final velocity v = s/t = 28/0.2 = 140 cms-1
Time taken is calculated using less one ticker tape ,ie (6-1) x 0.2 = 5 x 0.2 = 1 s
a = (140 – 20)/1 = 120 cms-2 or 1.2 ms-2
Ans: B
Q26
This question requires an understanding and application of the law of conservation of momentum.
The total momentum before collision = total momentum after collision. Since no numeric values are given, students will have to use logical formulas to solve this problem
mu + 0 = 2mv u = 2v or v = ½ u
Ans: C
Q27:
Understanding and application of Law of conservation of momentum. Understand that velocity is a vector quantity and object moving from right to left as in the 0.5kg ball will have a negative velocity
Use thinking skills that the final direction of the combined objects will follow the initial direction of the higher momentum object
Total momentum before collision = Total momentum after collision
m1u1 + m2mu2 = (m1 + m2) v
1 x 2 + 0.5 (-6) = (1+0.5)v 2-3 = 1.5v
V = -1/1.5 = -0.67 ms-1
Therefore, when the 2 balls stick together , they will move towards the left with a combined velocity of 0.67 ms-1 . The final direction will follow the initial direction of the higher momentum object.
Ans: D
Q28:
Fixing a hammer head into its handle involves inertia. Options B, C and D requires an understanding on impulsive force.
Ans: A
Q29:
In vacuum, there is no frictional force acting on the objects. In this case, both the marbles will undergo free fall. Both the marbles will reach the base at the same time.
Gravitational force acting on marble A is higher than that of marble B since F = mg
Ans: C
Something to think about
If this cylinder is in a micro-gravity situation (like inside of a space ship , ISS orbiting the earth), then both marbles will float and may not fall to the base. It gives a sense of weightlessness to the objects. Since orbital space stations like the ISS is revolving around the earth, they are also subjected to the earth’s gravitational acceleration, g but this force of gravity is balanced by the centripetal force acting on the space ship. Centripetal force acts on an object in constant circular motion.
Centripetal force is given by the formula F= mv2/ r where v is the linear velocity and r is the radius of rotation ( height above the earth plus the radius of the earth).
Q31
In experiment 1; Force is the manipulative variable and acceleration is the responding variable.
In experiment 2; mass the manipulative variable and acceleration is the responding variable
Ans: A
Q32
What can you deduce from the experiment F=ma?
F and a are both vector quantities - true
F is the net force – true
For F=ma to be true, both the F and a vectors must be parallel – true
F=ma is Newton’s 2nd Law of motion
Q33
Since from experiment 1 : a is directly proportional to F, Graph III is true
Since from experiment 2; a is inversely proportional to m, graph I is true
Ans C
fm 4 paper 1 ppt analysis / pradeep kumar chakrabarty (smjk yu hua kajang)[PC1]15 C
[PC2]16 B
[PC3]17 B