Wednesday, November 15, 2017

A ball is released from rest above a horizontal surface and bounces several times. The graph shows how, for this ball, a quantity y varies with time.




Question 4
A ball is released from rest above a horizontal surface and bounces several times.
The graph shows how, for this ball, a quantity y varies with time.


What is the quantity y?
A acceleration
B displacement
C kinetic energy
D velocity





Reference: Past Exam Paper – June 2013 Paper 12 Q8





Solution:
Answer: B.
The ball is thrown from rest above a horizontal surface and it bounces several times. 

We know that the ball falls due to the acceleration due to gravity ‘g’. However, the value of g is constant (= 9.81 m s-2) and does not vary when there is no air resistance. [A is incorrect]


Consider choice B.
Let the displacement be taken from the position above the surface and take the positive direction to be downwards. The ball falls until it reaches the surface (this is the maximum displacement).

Now, the ball falls under gravity. It accelerates; that is, its speed increases. In a displacement-time graph, this is indicated by an increasing gradient. This is indeed indicated by the graph.

As it hits the surface, the ball rebounds. However, due to energy loss (in form of heat, sound …), the ball does not reach its original height. As the ball moves up, its displacement (taken from its original position above before the release) decreases but does not come to zero as the ball does not reach its original height. [B is correct]


Under careful consideration, it can be shown that ‘y’ cannot be kinetic energy nor velocity.

Tuesday, November 14, 2017

A transverse wave travels along a rope. The graph shows the variation of the displacement of the particles in the rope with distance along it at a particular instant.




Question 2
A transverse wave travels along a rope. The graph shows the variation of the displacement of the particles in the rope with distance along it at a particular instant.

 

At which distance along the rope do the particles have maximum upwards velocity?
A 0.5 m                       B 1.0 m                       C 1.5 m                       D 2.0 m





Reference: Past Exam Paper – June 2017 Paper 13 Q22





Solution:
Answer: A.
This question is easily tackled by sketching on the diagram given in the question.

Since the wave is moving to the right, we can draw the position of the wave at the next instant by simply replicating the wave, but shifted to the right.


In this way, we can know how each point has moved in the next instant. It can be seen that particles at both 0.5 m and 1.0 m will move upward at the next instant but the largest upward change in displacement occurs at a distance of 0.5 m.

At 1.5 m and 2.0 m, the particles would move downward in the next instant, and the maximum downward velocity is at 1.5 m.

Sunday, November 12, 2017

A network of resistors, each of resistance 1 Ω, is connected as shown. The current passing through the end resistor is 1 A.




Question 3
A network of resistors, each of resistance 1 Ω, is connected as shown.
The current passing through the end resistor is 1 A.
What is the potential difference (p.d.) V across the input terminals?
A 2 V               B 5 V               C 8 V               D 13 V





Reference: Past Exam Paper – November 2015 Paper 13 Q37





Solution:
Answer: D.


We know that
- When simplifying a circuit into a series connections of resistors, the e.m.f. in the circuit is equal to the sum of p.d. across the components.
- For components connected in parallel to each other, the p.d. across them is equal.
- At any junction, the sum of current entering the junction is equal to current leaving the junction.

To understand the flow of current in the circuit above, the junctions have been marked as A, B, C and D, and the currents have been indicated. The way the circuit above is drawn is a bit tricky and confusing for some. It is easier to draw all the resistors in a horizontal way so that it becomes easier to follow the flow of current.


I will now explain how I arrived at this circuit, which is an equivalent to the circuit given in the question.

Consider junction A. To its left is resistor R1 and at the junction, the current splits to go into resistor R2 and R3. This is why we have 2 branches at A.

R2, which is the lower branch of the two, is connected to junction C. R3 is connected between A and B.

At junction B, the current splits into 2 branches; one with a single resistor R4 and another one with the 2 resistors R5 and R6 connected in series. The 2 branches join at junction D.

There is no resistor between C and D. So, the circuit drawn is correct.


Consider the current I5. It passes through the 2 resistors on the right on BD since the 2 resistors are connected in series. The total resistance in that part is 2 Ω.
p.d. in that branch = IV = 1 × (1+1) = 2V

Now, consider junction B: I3 = I4 + I5
We know that the greater the resistance, the smaller the current flowing. When the total resistance is 2 Ω (as above), the current is 1A. Current I4 passes through only a 1 Ω resistor, so the current would be twice that above. So, I4 = 2A.
p.d. across R4 = IV = 2 × 1 = 2 V
This is expected since it is in parallel with the above 2 resistors.

Also, I3 = I4 + I5 = 2 + 1 = 3 A
p.d. across R3 = IV = 3 × 1 = 3 V


The sum of p.d. across the upper branch AC is 3V + 2V = 5V
For a parallel connection, the sum of p.d. across the junctions are equal. So, the p.d. across R2 is also 5 V.

Current I2 flowing through R2 = V / R = 5 / 1 = 5 A

At junction A: I1 = I2 + I3 = 5 + 3 = 8 A
p.d. across R1 = IR = 8 × 1 = 8 V



e.m.f. in circuit = sum of p.d. across any loop
e.m.f. in circuit = 8 + 3 + 2 = 13 V                 OR = 8 + 5 = 13 V
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