• Physics 9702 Doubts | Help Page 119

Question 602: [Dynamics]

A 1.2 kg mass is supported by person’s hand and two newton-meters as shown.

When person’s hand is removed, what is the initial vertical acceleration of the mass?

A 0.6 m s^–2                  B 2 m s^–2                     C 4 m s^–2                     D 6 m s^–2

Reference: Past Exam Paper – November 2012 Paper 12 Q14

Solution 602:

Answer: D.

The newton-meters gives the magnitude of the forces in the direction shown.

We are asked for the initial vertical acceleration of the mass, so we do no need to consider the horizontal components of forces involved.

Resultant upward force = 4cos(37) + 3cos(53) = 4.999.. = 5.0N

Resultant vertical force = ma = 12 – 5 = 7N (downwards)

Acceleration a = 7 / 1.2 = 5.8 = 6ms^-2

Question 603: [Operational Amplifier]

(a) State three properties of ideal operational amplifier (op-amp).

(b) Circuit incorporating an ideal op-amp is to be used to indicate whether a door is open or closed.

Resistors, each of resistance R, are connected to the inputs of the op-amp, as shown in Fig.1.

Switch S is attached to the door so that, when the door is open, the switch is open. The switch closes when door is closed.

(i) Explain why polarity of the output of the op-amp changes when the switch closes.

(ii) A red light-emitting diode (LED) is to be used to indicate when door is open. A green LED is to indicate when the door is closed.

On Fig.1,

1. draw symbols for the LEDs to show how they are connected to output of the op-amp,

2. identify the green LED with letter G.

Reference: Past Exam Paper – November 2010 Paper 43 Q10

Solution 603:

(a) Examples: (choose any 3)

Infinite input impedance / resistance

Zero output impedance / resistance

Infinite gain

Infinite bandwidth

Infinite slew rate

(b)

(i)

With the switch S opened, V^– is less (positive) than V⁺, so the output is positive.

{When the switch S is opened, current flows through the resistor R at the top. Note that switch S is connected only to the vertical connection on the left, not to that on the right too. From the potential divider equation, for V^–, the p.d. is the same across each resistor R, and so, V^– is at +1.5V.

As for V⁺, there are 3 resistors R and the p.d. across each of them is the same (= 1.0V each). V⁺ is connected after 2 resistors R, so the potential at V⁺ is 2(1.0) = 2.0V.

V_OUT = A₀ (V⁺ – V^–)

So, V_OUT is positive.}

With the switch S closed, V^– is more (positive) than V⁺, so the output is negative.

{When the switch S is closed, current flows through the switch S instead of the top resistor R on the left vertical connection. So, V^– is at a potential of 3.0V. The potential at V⁺ is still 2.0V. From the equation above, V_OUT is now negative.}

(ii)

1. The diode should be connected correctly between the output and earth.

{When the door is opened, the switch is opened. When the switch is opened (as explained above), the output is positive. So, the red LED should be directed from the output towards the earth. Similarly, the green LED is in the opposite direction}

2. The green LED should be identified correctly

Question 604: [Kinematics]

The velocity-time graph below is for a stone thrown vertically up into air. Air resistance is negligible.

The stone is thrown up at time zero.

Area X represents a distance of 5 m. Area Y represents a distance of 3 m.

What is the displacement of the stone from its initial position at time t?

A 2 m                          B 3 m                          C 5 m                          D 8 m

Reference: Past Exam Paper – November 2010 Paper 11 Q8 & Paper 13 Q6

Solution 604:

Answer: A.

At time zero, the stone is thrown up. When the graph intercepts the time axis (x-axis), the velocity of the stone is zero. This means that the stone has reached its maximum height which is given by the area X = 5m.

Then, the stone falls down due to the force of gravity. In the graph, this is represented by the velocity being negative. So, area Y represents the distance that the stone has fallen after reaching its maximum. This distance is 3m from the maximum height, not 3m from the initial position.

Displacement from initial position = 5 – 3 = 2m

Question 605: [Current of Electricity]

Graph shows how current I varies with voltage V for a filament lamp.

Since the graph is not a straight line, resistance of the lamp varies with V.

Which row gives the correct resistance at stated value of V?

V / V   R / Ω

A         2.0       1.5

B         4.0       3.2

C         6.0       1.9

D         8.0       0.9

Reference: Past Exam Paper – June 2013 Paper 13 Q33

Solution 605:

Answer: C.

The I-V graph of a filament lamp is not a straight line.

A filament lamp does not follows Ohm’s law in that the variation of I with V is not linear (a straight line is not obtained), but the resistance is still obtained as follows.

Resistance R = V / I

{The values of current I read from the graph may not be exact for all the points, but they are close to the correct value. The resistance would be approximately the same though.}

Choice A:

At V = 2.0 V, current I = 1.8 A

Resistance R = 2.0 / 1.8 = 1.11 Ω

Choice B:

At V = 4.0 V, current I = 2.5 A

Resistance R = 4.0 / 2.5 = 1.6 Ω

Choice C: [Correct choice]

At V = 6.0 V, current I = 3.2 A

Resistance R = 6.0 / 3.2 = 1.875 = 1.9 Ω

Choice D:

At V = 8.0 V, current I = 4.0 A

Resistance R = 8.0 / 4.0 = 2.0 Ω

Question 606: [Work, energy and Power > Gravitational Potential Energy]

Mass m is situated in a uniform gravitational field.

When the mass moves through displacement x, from P to Q, it loses an amount of potential energy E.

Which row correctly specifies magnitude and direction of the acceleration due to gravity in this field?

magnitude       direction

A         E / mx              →

B         E / mx              ←

C         E / x                 →

D         E / x                 ←

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q14

Solution 606:

Answer: A.

Gravitational Potential Energy E = mgh

To easily answer this question, compare this situation (where the gravitational field lines are horizontal) with our usual case on Earth (where the gravitational field lines are vertical). Both are gravitational fields, so the same principles apply.

  

At a specific location, the force of gravity acts downwards (towards the centre if we look at the Earth as a whole). The acceleration due to gravity is in the same direction as the force of gravity. The gravitational field lines point downwards towards the surface of the Earth (towards its centre). The direction of the field lines show the direction that an object would move when placed in that specific field.

We know that as the height of an object on Earth increases, its gravitational potential energy increases. In other words, as the object moves against the gravitational field, it gains potential energy. Conversely, if the object moves in the direction of the field, its potential energy decreases.

The question states that when the mass moves through displacement x, from P to Q, it loses an amount of potential energy E. So, the direction of the field is from P to Q – that is to the right. [B and D are incorrect]

Gravitational Potential Energy E = mgh

[h = x here]

E = mgx

Acceleration due to gravity, g = E / mx