Friday, December 5, 2014

Physics 9702 Doubts | Help Page 32

  • Physics 9702 Doubts | Help Page 32

Question 197: [Forces > Moments]

A uniform metre rule of mass 100 g is supported by a pivot at the 40 cm mark and a string at the 100 cm mark. The string passes round a frictionless pulley and carries a mass of 20 g as shown in the diagram.

At which mark on the rule must a 50 g mass be suspended so that the rule balances?
November 2002:
A 4cm                         B 36 cm                       C 44 cm                       D 96 cm

June 2011 & November 2014:
A 4 cm                        B 36 cm                       C 44 cm                       D 64 cm

Reference: Past Exam Paper – November 2002 Paper 1 Q14 & June 2011 Paper 11 Q13 & Paper 13 Q12 & November 2014 Paper 13 Q14

Solution 197:
Answer: C.
Since the metre rule is uniform, its centre of mass is at the centre (at the 50cm mark).
Let the acceleration of free fall = a

The weight due to the centre of mass causes a clockwise moment on the rule about the pivot (at a distance of [50cm – 40cm =] 10cm from the pivot).
{Note that the distance may be left in cm (provided that all distances are taken in cm) as the final answer required (among the choices) is in cm}
Clockwise moment due to centre of mass = 100a x 10 = 1000a

The weight of the 20g mass causes an anticlockwise moment on the rule.
Anticlockwise moment due to 20g mass = 20a x (100 – 40) = 1200a

For equilibrium, the clockwise moment should be equal to the anticlockwise moment. Since the anticlockwise moment is currently greater than the clockwise moment, the 50g mass must be suspended such that it produces the remaining clockwise moment for equilibrium. That is, the 50g mass should be on the right side of the pivot. [A is incorrect]

Let the distance of the 50g mass from the pivot be x.
For equilibrium,
1000a + 50a (x) = 1200a
Distance, x = (1200 – 1000) / 50 = 4cm

Therefore, the 50g should be suspended at the (40 + 4 =) 44cm mark.

Question 198: [Dynamics]
Two blocks X and Y, of masses m and 3m respectively, are accelerated along smooth horizontal surface by force F applied to block X as shown.

What is the magnitude of force exerted by block X on block Y during this acceleration?
A F / 4                         B F / 3                         C F / 2                         D 3F / 4

Reference: Past Exam Paper – June 2003 Paper 1 Q10

Solution 198:
Answer: D.
The force F, applied to block X, causes both blocks to accelerate.
Total mass being accelerated = m + 3m = 4m

From Newton’s second law of motion: F = ma
Acceleration, a = F / 4m

From this acceleration, the force acting on block Y (FY) may be calculated using Newton’s second law of motion now on block Y only. This force is exerted by block X on block Y.
FY = (3m)a = (3m) (F / 4m) = 3F / 4

Question 199: [Measurements > CRO]
Time-base on cathode-ray oscilloscope is set at 6 ms / cm.
Trace consisting of two pulses is recorded as shown in diagram.

What is time interval between the two pulses?
A 0.42 ms                    B 0.75 ms                    C 1.33 ms                    D 27 ms

Reference: Past Exam Paper – November 2003 Paper 1 Q5 & June 2011 Paper 12 Q5

Solution 199:
Answer: D.
Time-base setting of the c.r.o. is 6ms / cm.
1cm represents 6ms

The 2 pulses are separated by a distance of 4.5cm.
4.5cm corresponds to a time interval of (4.5 x 6 =) 27ms

Question 200: [Kinematics > Projectile motion]
Motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above ground, landing 10 m away as shown.

What was speed at take-off?
A 5 m s–1                     B 10 m s–1                   C 15 m s–1                   D 20 m s–1

Reference: Past Exam Paper – June 2004 Paper 1 Q9

Solution 200:
Answer: D.
This problem deals with projectile motion. So, the vertical and horizontal motion may be considered separately.

Consider the vertical motion:
Since the rider takes off horizontally, its initial vertical velocity, u is zero.
(Downward) Acceleration of free fall, g = 9.81ms-2
Vertical distance, s = 1.25m
Consider the equation for uniformly accelerated motion: s = ut + ½ at2
Let t be the time taken by the motorcycle to reach the ground

1.25 = 0(t) + 0.5 (9.81) t2
Time, t = √[1.25 / (0.5 x 9.81)] = 0.505s

Now, consider the horizontal motion:
There is no acceleration for the horizontal motion. That is, the motorcycle moves at a constant speed.
The distance travelled in a time t = 0.505s is 10m.
Speed = Distance / Time = 10 / 0.505 = 20ms-1  

Question 201: [Forces > Moment]
Rigid uniform bar of length 2.4 m is pivoted horizontally at its mid-point.

Weights are hung from two points of the bar as shown in diagram. To maintain horizontal equilibrium, a couple is applied to the bar.
What is the torque and direction of this couple?
A 40 N m clockwise
B 40 N m anticlockwise
C 80 N m clockwise
D 80 N m anticlockwise

Reference: Past Exam Paper – June 2006 Paper 1 Q15

Solution 201:
Answer: A.
The pivot is placed at the 1.2m from each end of the bar.

Before applying the couple to the bar,
Clockwise moment on bar = 300 x (1.2 – 0.8) = 120Nm
Anticlockwise moment on bar = 200 x 0.8 = 160Nm

For horizontal equilibrium, a clockwise moment should be provided.
Required clockwise moment = 160 – 120 = 40Nm

So, the torque and direction of the couple should be 40Nm clockwise.

Question 202: [Hooke’s law]
Fig shows variation with force F of extension x of spring as the force is increased to F3 and then decreased to zero.

(a) State, with a reason, whether spring is undergoing an elastic change.

(b) Extension of spring is increased from x1 to x2.
Show that work W done in extending spring is given by W = ½ k (x22 – x12), where k is spring constant.

(c) Trolley of mass 850 g is held between two fixed points by means of identical springs, as shown.

When trolley is in equilibrium, springs are each extended by 4.5 cm. Each spring has spring constant 16Ncm–1.
Trolley is moved a distance of 1.5 cm along direction of the springs. This causes extension of one spring to be increased and the extension of the other spring to be decreased. Trolley is then released. Trolley accelerates and reaches its maximum speed at the equilibrium position.
Assuming that springs obey Hooke’s law, use expression in (b) to determine maximum speed of the trolley.

Reference: Past Exam Paper – June 2006 Paper 2 Q5

Solution 202:
There is no hysteresis loop / no permanent deformation occurs. So, the spring undergoes an elastic change.

Work done = Area under graph line = ½ Fx
Force, F = kx,  so Work done = ½ kx2
Work done in extending spring = ½ k(x22 – x12)

Work done = average force x distance = [½ (F2 + F1) (x2 – x1)
F = kx, so Work done = ½ k (x2 + x1) (x2 – x1) = ½ k (x22 – x12)

{When the trolley is moved in one direction, the extension of one spring increases while that of the other decreases. Thus, the elastic energy stored in the former spring increases and that in the latter spring decreases.
When the trolley is moved a distance of 0.015m (= 1.5cm), one of the spring is (4.5cm + 1.5cm =) 0.060m and the other spring is (4.5cm – 1.5cm =) 0.030m. At the equilibrium position, both springs are 0.045m.
At the equilibrium position, the elastic energy stored in the springs is converted in all the kinetic energy of the trolley. Spring constant, k = 16Ncm-1. }
Gain in energy of trolley = ½ k(0.0602 – 0.0452) + ½ k(0.0302 – 0.0452) = 0.36J
Kinetic energy of trolley = ½ mv2 = ½ (0.85) v2 = 0.36J
Speed, v = 0.92ms-1

Question 203: [Kinematics > Linear motion]
Experiment is performed to measure acceleration of free fall g. Body falls between two fixed points. Four measurements shown below are taken.
Which measurement is not required for calculation of g ?
A distance fallen by the body
B initial velocity of the body
C mass of the body
D time taken for the body to fall

Reference: Past Exam Paper – June 2014 Paper 13 Q7

Solution 203:
Answer: C.
The equations of uniformly accelerated motion are:
s = ut + ½at2    and v2 = u2 + 2as

It can be seen that the distance, initial velocity and time taken for the body to fall are needed to calculate a, the acceleration of free fall. But mass is not required. The acceleration of free fall is constant (in a specific location), no matter what the mass is (However, the gravitational mass on the body will depend on its mass).


  1. Please clear my doubts
    02/O/N/05 Q.5(c)(ii),(d)(ii)
    01/M/J/06 Q.15,22,24,25
    02/M/J/06 Q.2(c)(iii),Q.5(c)

    1. All questions for November 2005 Paper 2 are solved at

      The others will be added soon

    2. Q15 of June 2006 P1 is explained above

    3. Q2 of June 2006 P2 is explained as solution 13 at

    4. June 2006 P2 Q5 is explained above

    5. June 2006 P1 Q22 is explained as solution 205 at

    6. June 206 P1 Q24 is explained as solution 214 at

    7. June 2006 P1 Q25 has been explained at

  2. Replies
    1. Q4 explained as solution 225 at

    2. Q15 is explained as solution 232 at

  3. This comment has been removed by the author.

  4. please send the worked solution for Q9 O/N 2010 paper 11

    1. check at

  5. 9702_w04_qp_1

    qs 28


    1. See solution 835 at

  6. Does elastic change mean the same as plastic deformation? In question 202

    1. No. For elastic change, the object regains its shape when the deforming force is removed. For plastic deformation, the shape of the object is permanently changed.

  7. in q 203 why cant we use the formula f=ma?

    1. F, which would be the weight here, is not known

  8. Can you please provide an explanation for 9702 w02 paper 1 Q10?
    please I really need it.

    1. see solution 470 at


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