Wednesday, June 10, 2015

Physics 9702 Doubts | Help Page 169

  • Physics 9702 Doubts | Help Page 169

Question 847: [Current of Electricity]
In circuit shown, the ammeters have negligible resistance and the voltmeters have infinite resistance.

The readings on the meters are I1, I2, V1 and V2, as labelled on diagram.
Which statement is correct?
A I1 > I2 and V1 > V2
B I1 > I2 and V1 < V2
C I1 < I2 and V1 > V2
D I1 < I2 and V1 < V2

Reference: Past Exam Paper – November 2014 Paper 13 Q37

Solution 847:
Answer: A.
This question required careful working.
Combined resistance of the left-hand section = [1/3 + 1/6]-1 = 2 Ω
Combined resistance of the right-hand section = [1/2 + 1/2]-1 = 1 Ω

From Kirchhoff’s law, the sum of V1 and V2 should be equal to the e.m.f. in the circuit.
So, potential difference V1 > potential differenceV2. [B and D are incorrect]

In the right-hand section, each parallel branch has half of the total current since the resistances are both 2 Ω, while in the left-hand section there is more current in the 3 Ω resistor than in the 6 Ω resistor (from Ohm’s law, I = V / R – the greater the resistance, the smaller the current).

This gives I1 > I2, so the correct answer is A.

In terms of calculations, the total current leaving the right-hand section is I2 + I2 = 2I2. As for the left-hand section, the current I1 is twice the current flowing through the 6 Ω resistor. Thus, the total current entering the left-hand section is I1 + ½ I1 = 1.5I1.

But the total current leaving the supply should be equal to the total current entering the supply.
1.5I1 = 2I2
I1 = 2I2 / 1.5 = 4I2 / 3 = 1.33I2
So, I1 > I2

Question 848: [Measurement > Uncertainty]
The speed of a car is calculated from measurements of distance travelled and the time taken.
Distance is measured as 200 m, with an uncertainty of ± 2 m.
Time is measured as 10.0 s, with an uncertainty of ± 0.2 s.
What is the percentage uncertainty in the calculated speed?
A ± 0.5 %                    B ± 1 %                       C ± 2 %                       D ± 3 %

Reference: Past Exam Paper – November 2012 Paper 12 Q7

Solution 848:
Answer: D.
Speed v = Distance d / Time t

Percentage uncertainty in the calculated speed = (Δv / v) × 100%
Δv / v = (Δd/d) + (Δt/t)
Δv / v = (2/200) + (0.2/10) = 0.03 = ±3%

Question 849: [Matter > Hooke’s law]
A beam, the weight of which may be neglected, is supported by three identical springs. When weight W is hung from the middle of the beam, the extension of each spring is x.

The middle spring and the weight are removed.
What is the extension when weight of 2W is hung from the middle of the beam?
A 3x / 2                       B 4x / 3                       C 2x                            D 3x

Reference: Past Exam Paper – June 2003 Paper 1 Q22

Solution 849:
Answer: D.
This is quite a tricky question which needs careful working, done on paper and not mentally.

Hooke’s law: F = kx

Since the system consist of more than one springs, we need to consider the effective spring constant of the springs, keff.

Each of the 3 springs exerts an upward vertical force (tension – from Newton’s 3rd law) on the beam such that these forces balances the downward weight W after extension.

When weight = W, extension = x
W = keff x
Effective spring constant of the 3 springs, keff = W / x

Let the spring constant of 1 spring be k. For 3 springs in parallel,
keff = k + k + k
Spring constant k of 1 spring = keff / 3 = (W/x) / 3 = W / 3x

Now, the middle spring is removed and a weight of 2W is used.
Effective spring constant of 2 springs, keff = (W/3x) + (W/3x) = 2W / 3x

Let the new extension be e.
From Hooke’s law,
2W = (2W/3x) e
New extension, e = 3x

Question 850: [Dynamics > Collisions]
Two equal masses travel towards each other on frictionless air track at speeds of 60 cm s–1 and 40 cm s–1

They stick together on impact.
What is the speed of the masses after impact?
A 10 cm s–1                 B 20 cm s–1                  C 40 cm s–1                  D 50 cm s–1

Reference: Past Exam Paper – June 2005 Paper 1 Q11 & June 2010 Paper 11 Q10 & Paper 12 Q11 & Paper 13 Q12

Solution 850:
Answer: A.
Let the mass of 1 trolley = M

From the conservation of momentum, the sum of momentum before collision should be equal to the sum of momentum after collision.

Sum of momentum before collision = M(60) + M(-40) = 20M
After collision, momentum = 2M(v)

2Mv = 20M
Final speed, v = 10cms-1


  1. in q 850 i didnt understand how we got 2mv

    1. After collision, they stick together. So, the mass becomes M + M = 2M. They move as a single body of velocity v.

    2. wont it be Mv--mv b4 collision?

  2. HI
    I have a question , when U are traveling in opp directions isn't one going to be negative so wont it be :

    1. When taking the 'sum', we need to 'add'. But since they are moving in opposite direction, one of them needs to be negative.

    2. Salam,

      why do we have to add? I get that one of them needs to be negative, but then why do u take the sum?

    3. ohk Sorry I get your point thanks!!

  3. Shouldn't we use relative velocity multiplied my mass to calculate total momentum before collision in 850.
    like (60-(-40))M?

    1. no as you would tend to make mistakes just like here.

      since you already inserted a -ve sign in front of 40, you have already accounted for the direction. You should not use another -ve sign between the 2 terms

  4. can relative velocity multiplied by mass be used instead in q 850?

    1. no, it should be done in terms of momentum because after the collision, both the velocity and the mass are different than they were initially


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