# 9702 November 2007 Paper 2 Worked Solutions | A-Level Physics

## Paper 2

__Question 1__**(a)**

Distinguish between systematic and
random errors:

For systematic error: Choose any 1:
e.g.

There is a constant error (in all
the readings)

It cannot be eliminated by averaging

The error is in the measuring
instrument

For random error: Choose any 1: e.g.

The readings are scattered (equally)
about the true value

The error is due to the observer

The error can be eliminated by
averaging

**(b)**

Cylinder of length L has circular
cross-section of radius R. Volume V of cylinder given by V = πR

^{2}L
Volume and length of cylinder are
measured as V = 15.0 ± 0.5 cm

^{3}; L = 20.0 ± 0.1 cm.
Radius of cylinder, with
uncertainty:

{R

{R

^{2}= πL / V. So 2(ΔR / R) = (ΔL / L) + (ΔV / V)}
15 = πR

^{2}(20)
R = 0.4886cm (accept any number of sf)

% uncertainty in V = 3.3% (or
=0.5/15)

% uncertainty in L = 0.5% (or
=0.1/20)

% uncertainty in R = 1.9% (half of
the sum of % uncertainties)

R = 0.489 ± 0.009cm

__Question 2__
Girl G is riding bicycle at constant
velocity of 3.5 ms

^{–1}. At time t = 0, she passes boy B sitting on bicycle that is stationary. At time t = 0, boy sets off to catch up with girl. He accelerates uniformly from time t = 0 until he reaches speed of 5.6 ms^{–1}in a time of 5.0 s. He then continues at constant speed of 5.6 ms^{–1}. At time t = T, boy catches up with girl. T measured in seconds.**(a)**

In terms of T, distance moved by
girl before boy catches up with her:

Distance = (Speed x Time =) 3.5T

**(b)**

For boy,

(i)

Distance moved during his
acceleration:

Distance = Average Speed x time

(a = (5.6 – 0) / 5)

Distance, s = ut + ½ at

^{2}= 0 + ½ [(5.6 – 0) / 5] x 5^{2}= 14m
(ii)

Distance moved during time that he
is moving at constant speed. Give answer in terms of T:

(Time he is moving with constant
speed = T - 5)

Distance = 5.6 (T – 5) (or 3.5T – 14)

**(c)**

Use answers in (a) and (b) to
determine time T taken for boy to catch up with girl:

3.5T = 14 + 5.6(T – 5)

Time, T = 6.7s

**(d)**

Boy and bicycle have combined mass
of 67 kg.

(i)

Force required to cause acceleration
of boy:

Acceleration, a = (5/6 / 5 =) 1.12ms

^{-2}
Force = ma = (67 x 1.12 =) 75N

(ii)

At speed of 4.5 ms

^{–1}, total resistive force acting on boy and bicycle is 23 N. Output power of boy’s legs at this speed:
Power = (force x speed =) [75+23] x
4.5 = 440W

__Question 3__**(a)**

(i)

Potential energy is defined as the stored
energy available to do work

(ii)

Difference between gravitational
potential energy and elastic potential energy:

Gravitational potential energy is
due to height / the position of a mass OR distance from a mass OR moving a mass
from one point to another.

Elastic potential energy is due to
the deformation / stretching / compressing

**(b)**

Small sphere of mass 51 g suspended
by light inextensible string from fixed point P. Centre of sphere is 61 cm
vertically below point P, as shown. Sphere is moved to one side, keeping string
taut, so that string makes an angle of 18° with vertical.

(i)

Gain in gravitational potential
energy of sphere:

The height raised, h = [61 –
61cos(18) =] 3.0cm

Gain in energy = [mgh =
0.051(9.8)(0.030) =] 1.5x10

^{-2}J
(ii)

Moment of weight of sphere about
point P:

Moment = Force x perpendicular
distance

Moment = (0.051 x 9.8) (0.61sin(18))
= 0.094Nm

__Question 4__**{Detailed explanations for this question is available as Solution 1025 at Physics 9702 Doubts | Help Page 214 -**

*http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-214.html*}

__Question 5__**{Detailed explanations for this question is available as Solution 240 at Physics 9702 Doubts | Help Page 38 -**

*http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-38.html*}

__Question 6__**{Detailed explanations for this question is available as Solution 1020 at Physics 9702 Doubts | Help Page 213 -**

*http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html*}

__Question 7__**(a)**

Evidence for nuclear atom provided
by α-particle scattering experiment. Results of this experiment:

Most of the α-particles are deviated through small angles

A few α-particles are deviated
through angles greater than 90

^{o}.**(b)**

Estimates for diameter for

(i)

Atom: allow 10

^{-9}m to 10^{-11}m
(ii)

Nucleus: allow 10

^{-13}m to 10^{-15}m
Why cant I use P=VI in no. 6 c) i ?

ReplyDeleteThe explanation has been updated. Check again.

Deleteis it because minimum area of the MATERIAL has to be at least 2 minimum, so bubble can be anywhere between 0 and 1.2 as it fits inside the wire??

ReplyDeleteAlso, what does the difference in extensions on a thicker rod matter for its breaking?

The explanation has now been updated. Check again.

Deletemcq 29 oct/nov 2007??

ReplyDeleteSee solution 1097 at

Deletehttp://physics-ref.blogspot.com/2016/02/physics-9702-doubts-help-page-234.html