# 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 I

_{1}, I_{2}, V_{1}and V_{2}, as labelled on diagram.
Which statement is correct?

A I

_{1}> I_{2}and V_{1}> V_{2}
B I

_{1}> I_{2}and V_{1}< V_{2}
C I

_{1}< I_{2}and V_{1}> V_{2}
D I

_{1}< I_{2}and V_{1}< V_{2}**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 V

_{1}and V_{2}should be equal to the e.m.f. in the circuit.
So, potential difference V

_{1}> potential differenceV_{2}. [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 I

_{1}> I_{2}, so the correct answer is A.
In terms of calculations, the total
current leaving the right-hand section is I

_{2}+ I_{2}= 2I_{2}. As for the left-hand section, the current I_{1}is twice the current flowing through the 6 Ω resistor. Thus, the total current entering the left-hand section is I_{1}+ ½ I_{1}= 1.5I_{1}.
But the total current leaving the
supply should be equal to the total current entering the supply.

1.5I

_{1}= 2I_{2}
I1 = 2I

_{2}/ 1.5 = 4I_{2}/ 3 = 1.33I_{2}
So, I

_{1}> I_{2}

__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, k

_{eff}.
Each of the 3 springs exerts an
upward vertical force (tension – from Newton’s 3

^{rd}law) on the beam such that these forces balances the downward weight W after extension.
When weight = W, extension = x

W = k

_{eff}x
Effective spring constant of the 3
springs, k

_{eff}= W / x
Let the spring constant of 1 spring
be k. For 3 springs in parallel,

k

_{eff}= k + k + k
Spring constant k of 1 spring = k

_{eff}/ 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, k

_{eff}= (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}
in q 850 i didnt understand how we got 2mv

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

Deletewont it be Mv--mv b4 collision?

DeleteHI

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

MV--MU???

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

DeleteSalam,

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

ohk Sorry I get your point thanks!!

DeleteShouldn't we use relative velocity multiplied my mass to calculate total momentum before collision in q.no 850.

ReplyDeletelike (60-(-40))M?

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

Deletesince 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

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

ReplyDeleteno, it should be done in terms of momentum because after the collision, both the velocity and the mass are different than they were initially

Delete