# 9702 June 2012 Paper 22 Worked Solutions | A-Level Physics

__Question 1__**{Detailed explanations for this question is available as Solution 1022 at Physics 9702 Doubts | Help Page 213 -**

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

__Question 2__**(a)**

(i)

Speed of ball as it hits the ground:

v

^{2}= u^{2}+2as = (8.4)^{2}+ 2(9.81)(5)
v = 12.99ms

^{-1}
(ii)

Show that time taken for ball to
reach ground =0.47s:

s = ut + ½ at

^{2}or t = (v-u) / a
5 = 8.4t + ½ (9.81)t

^{2}or t = (12.99 – 8.4) / 9.81
t = 0.468s

**(b)**

Reasonable shape

Suitable scale

Correctly plotted 1

{Graph to show variation with time t of velocity v of ball: First point is at (0,8.4) since velocity of ball at t=0 is 8.4ms

^{st}and last points at (0,8.4) and (0.88 – 0.96,0) with non-vertical line at 0.47s{Graph to show variation with time t of velocity v of ball: First point is at (0,8.4) since velocity of ball at t=0 is 8.4ms

^{-1}. Due to gravity, velocity of ball increases until v = 12.99ms^{-1}where the ball reaches the ground. This is at t=0.47s. The gradient between these 2 points [(0,8.4) and (0.47,12.99)] which represents the acceleration is constant and equal to 9.81ms^{-2}. The speed after the ball rebounds is 4.2ms^{-1}but since the ball is now moving upwards, v = - 4.2ms^{-1}. Additionally, since contact time is 20ms = 0.02s [non-zero], the line representing the change from the maximum +ve speed to this –ve one is not vertical. As for the final stage, the velocity of the ball at B is zero. The time taken for the velocity to be zero is t = (0 - -4.2) / (9.81) = 0.428s. So, velocity is zero at t = 0.428 + 0.47 = 0.898s. Gradient is similar to first section (=9.81). Graph drawn: on x-axis, smallest unit = 0.02s. on y-axis, smallest unit = 1ms^{-1}.}**(c)**

Mass of ball = 0.050kg. Ball moves
from A and reaches B after rebounding.

(i)

1.

Change in kinetic energy:

Kinetic energy at B is zero, so Î”KE = ½ mv

^{2}or Î”KE = ½ mu^{2}– ½ mv^{2}
Î”KE = ½ (0.05)(8.4)

^{2}= (-)1.8J
2.

Change in gravitational potential
energy:

Final maximum height, h = (4.2)

^{2}/ 2(9.81) = (0.9m)
Change in PE = mgh – mgh

_{1}
Change in PE = 0.05 x 9.81 x (0.9 –
5) = (-)2.0J

(ii)

Explain total change in energy of
ball for this motion:

The total change in energy is -3.8J.
Energy is lost to ground (on impact) / energy of deformation of the ball /
thermal energy in ball.

__Question 3__**{Detailed explanations for this question is available as Solution 614 at Physics 9702 Doubts | Help Page 121 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html*}

__Question 4__**(a)**

Pd across XY:

Total resistance in circuit = 8 + 12
= 20kÎ©

Current = 12 / (20x10

^{3}) or potential divider formula
p.d = [12/(20x10

^{3})] x 12x10^{3}= 7.2V**(b)**

Resistance LDR = 4.0kÎ©. Current in
ammeter:

Parallel resistance (across XY) = [(1/4)
+ (1/12)]

^{-1}= 3kÎ©
Total resistance in circuit = 3 + 8
= 113kÎ©

Current = 12 / (11x10

^{3}) = 1.09x10^{-3}A or 1.1x10^{-3}A**(c)**

Intensity of light on LDR increased.

(i)

Change to ammeter reading:

The resistance of the LDR decreases.
So, the total resistance (of the circuit) is less. Hence, current increases.

(ii)

Change to p.d across XY:

The resistance across XY is now
less. So, there is less proportion of the 12V across
XY. Hence, p.d. is less.

__Question 5__**{Detailed explanations for this question is available as Solution 626 at Physics 9702 Doubts | Help Page 124 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html*}

__Question 6__**{Detailed explanations for this question is available as Solution 631 at Physics 9702 Doubts | Help Page 125 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-125.html*}

__Question 7__**{Detailed explanations for this question is available as Solution 641 at Physics 9702 Doubts | Help Page 127 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-127.html*}
could you explain 3(B2) in detail please.It would be really appreciated! ^_^ in desperate need of help

ReplyDeleteDetails have been added for the question. See again

DeleteIn question 5(c), why must the line pass through 1.5 on the y axis? i get that it has to pass through 0.25 on the x axis but how did you find that it has to pass through 1.5 on the y axis?

ReplyDeleteDetails have been updated for question 5

Deletethank you :)

Deletecan you also upload the diagram for question 6(b)?

can you please draw question 6(b)?

ReplyDeleteQuestion 6 has been updated

DeleteHi, I am just wondering if the uncertainty is 0.09 x 10^-3,shouldnt the actual measurement be multiples of the uncertainty? or follow the decimal place of uncertainty?Confused about this part...I would really appreciate if you can answer me in time.Thanks!!!

ReplyDeleteIf you are referring to a specific question, give me the reference of the question. It's difficult to explain otherwise.

DeletePlease tell me the difference between 9702 June 2012 Paper 22 Q 1c and 9702 October 14 P23 Q2b),your help is much appreciated ^^

DeleteThe explanation has been updated. See if it helps.

DeleteCan you please solve 0ctober November 2001 Paper 2 Question 7?

ReplyDeleteAnyways great blog. Keep up the good work!

Check solution 1093 at

Deletehttp://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-232.html

these have been extremely helpful,by any chance do you have one for chemistry papers 9701?

ReplyDeletetried to make one at

Deletehttp://chemistry-ref.blogspot.com/

but it requires too time work/time. if anyone's willing to contribute it would help

Dear Admin, thanks for sharing the workings. Referring to Question 2(b), the velocity vs. time plot, shouldn't the graph start at v = -8.4 m/s at t = 0 s instead, since the velocity is downward (negative)? Also, shouldn't the gradient of the graph be -9.8 m/s^2, since the gravitational acceleration is downward? Looking forward your kind advice. Thank you.

ReplyDeleteit depends. if we take the downward direction as positive, then the graph is as above.

DeleteThank you for the advice!

Delete