# Physics 9702 Doubts | Help Page 2

__Question 8: [Kinematics > Linear motion]__
Anna Litical drops a ball from rest
from the top of 78.4-meter high cliff. How much time will it take for the ball
to reach the ground and at what height will the ball be after each second of
motion?

**Reference:**

*Physics Classroom.com*

__Solution 8:__
Initial velocity of ball = 0ms

^{-1}
Total distance to travel = 78.4m

Take acceleration due to gravity
(downward), a = 9.8ms

^{-2}
Time taken to reach ground = t

Consider equation of uniformly
acceleration motion: s = ut + ½ at

^{2}
78.4 = 0 + 0.5(9.8)t

^{2}
Time taken to reach ground, t =
[78.4 / (0.5x9.8)]

^{0.5}= 4.0s
We also need to find the height
reached after each second (i.e. t = 1, 2, 3, 4)

Height reached at time t = 78.4 –
Distance travelled (downward) by ball for time t

Let distance travelled (downward) by
ball for time t = s

s = ut + ½at

^{2}= 0 + 0.5(9.8)t^{2}
s = 0.5(9.8)t

^{2}
When t = 1s, s = 4.9m, height = 78.4
m – 4.9 m = 73.5m

When t = 2s, s = 19.6 m, height =
78.4 m – 19.6 m = 58.8m

When t = 3s, s = 44.1 m, height
=78.4 m – 44.1 m = 34.3m

When t = 4s, s = 78.4 m, height =
78.4 m – 78.4 m = 0m (reached ground)

__Question 9: [Hooke’s law > Energy]__
Light spring of unextended length
14.2cm suspended vertically from fixed point, as illustrated. Mass of weight
3.8N hung from end of spring, as shown. Length of spring is now 16.3cm.
Additional force F then extends spring so that its length becomes 17.8cm, as
shown. Spring obeys Hooke’s law and elastic limit of spring not exceeded.

(i) Show spring constant of spring
is 1.8Ncm

^{-1}
(ii) For extension of spring from
length of 16.3cm to length of 17.8cm,

1. Change in gravitational potential
energy of mass on spring

2. Show that change in elastic
potential energy of spring is 0.077J

3. Work done by force F

**Reference:**

*Past Exam Paper – November 2009 Paper 22 Q4(c)*

__Solution 9:__
(i)

Show spring constant of spring is
1.8Ncm

^{-1}:
{F = kx. So,
k = F/x}

k = (F/x =) 3.8 / 2.1 = 1.8Ncm

^{-1}
(ii)

For extension of spring from length
of 16.3cm to length of 17.8cm,

1.

Change in gravitational potential
energy of mass on spring:

Î”E

_{p}= mgÎ”h or W Î”h = 3.8 x (1.5x10^{-2}) = 0.057J
2.

Show that change in elastic
potential energy of spring is 0.077J:

{k = 1.8Ncm

^{-1}= 1.8x10^{2}Nm^{-1}(1cm Ã 1.8N. So, 100cm = 1m Ã 1.8x10^{2})}
{At length
16.3cm, extension, x = 16.3 – 14.2 = 2.1cm = 0.021m. At length 17.8cm,
extension, x = 17.8 – 14.2 = 3.6cm = 0.036m}

Î”E

_{s}= ½ (1.8x10^{2}) (0.036^{2}– 0.021^{2}) = 0.077J
3.

Work done by force F:

{Work done
by force F = Î”E

_{s}– Î”E_{p}}
Work done = 0.077 – 0.057 = 0.020J

__Question 10: [Measurements > Uncertainties]__
Use answer in (b) to determine
absolute uncertainty in g. State value of g, with its uncertainty, to
appropriate number of significant figures:

Note: from (b), uncertainty in g = 4.2%

**Reference:**

*Past Exam Paper – November 2009 Paper 22 Q1(c)(ii)*

__Solution 10:__
{%
uncertainty in g = (Î”g / g) x 100% = 4.2% and g = 9.751ms

^{-2}}
Î”g = ([4.2/100] x 9.751 =) ± 0.41 / ±
0.42 to any number of s.f

g = (9.8 ± 0.4) ms

^{-2}**{**This uncertainty should be quoted to only 1 significant figure (to 1 s.f.). It is this significant figure that then determines the number of allowable decimal places in value for g.

**}**

__Question 11: [Kinematics > Linear motion]__
Asalamo alaikum,

can you please help me out..

I want to know if an object was
accelerating at 10 ms

^{-2}now it is accelerating at 6ms^{-2}is this called DECELERATION?
2, when a object was previously
accelerating in forward direction at 10ms

^{-2}but now it is accelerating at 12ms^{-2}but in opposite direction i.e backwards, is it still called Deceleration?
Thank you

__Solution 11:__
Wslm.

Deceleration is a decrease in the
numerical value of

**velocity**[Velocity is defined as the rate of change of displacement]. Acceleration is an increase in speed [Acceleration is defined as the rate of change of velocity].
For the first case, an object was
initially accelerating at 10ms

^{-2}, then the acceleration changes to 6ms^{-2}. This can be understood as follows. Initially, the increase in speed in one second is 10ms^{-1}. Then, the increase in speed in one second becomes 6ms^{-1}. In both cases, speed is increasing. So in both cases, the object is__accelerating__.
For the second case, the object was
initially accelerating at 10ms

^{-2}in the forward direction, then the acceleration changes to 12ms^{-2}in the opposite direction. This can be understood as follows. Define the forward direction as the positive direction. Initially, the increase in speed in one second is 10ms^{-1}in the forward direction. Then, the increase in speed in one second becomes 12ms^{-1}in the opposite direction.
For the acceleration in the opposite
direction, 2 cases are possible.

__1__

^{st}case:
The object is still moving in the forward
direction but it undergoes an acceleration in the opposite direction. In this
case, the forward speed would actually be decreasing until it becomes zero. A
decrease in speed is called DECELERATION. Now, if the acceleration continues in
the opposite direction, then the object would start moving in the opposite
direction. Its speed in the opposite direction would increase. So, for the
latter part, the object would be accelerating in the opposite direction (this
is not called deceleration).

{Acceleration is a vector
quantity, so it can be in any direction. Velocity is another vector quantity (which
is different than acceleration) and can also be in any direction. These 2
quantities (velocity and acceleration) should not be confused to be the same}

__2__

^{nd}case:
In this case, the object would be
moving in the opposite direction and the acceleration of 12ms

^{-2}is also in the opposite direction. As explained before, the object would be accelerating (this is not called deceleration) in the opposite direction.
To summarize, acceleration and
velocity are 2

__different__vector quantities describing motion. Since they are vectors, they are a magnitude and a direction. For linear motion, a positive magnitude would indicate a specific direction, while a negative value would indicate the opposite direction. For non-linear motion, they (the vectors) can be resolved into 2 perpendicular components and the directions can be specified, as for the linear motion case, by considering each component separately.
Thus, the sign associated with these
quantities usually indicate the direction. Just like a negative velocity
indicates that the object IS MOVING in a specific direction, a negative
acceleration also describes motion in a direction, while increasing the speed
in that direction.

**The velocity also decreases in this situation. Acceleration can be +ve or –ve but deceleration is not usually taken as negative since deceleration is a decrease in velocity while acceleration is the rate of change of velocity [which can be an increase or a decrease].**__DECELERATION__only occurs when the direction of acceleration is opposite to the direction of motion.

__Question 12: [Vectors]__
Weight of 7.0 N hangs vertically by
2 strings AB and AC, as shown. For weight to be in equilibrium, tension in
string AB is T

_{1}and in string AC it is T_{2}. On Fig, draw vector triangle to determine magnitudes of T_{1}and T_{2}:**Reference:**

*Past Exam Paper – June 2010 Paper 23 Q2(c)*

__Solution 12:__
Triangle should be drawn with the
correct shape

{To easily
draw the triangle, follow the instructions below [there may be other method to
draw it] (lines should be drawn lightly first):

Extend the dotted vertical
line shown by a significant amount. Then draw another dotted vertical line
which now passes through point B.

Now, consider T

_{2}which is 50^{o}to the vertical. Draw a dotted line, starting a point B, at an angle of 50^{o}to the second dotted vertical line drawn and which goes (upwards) towards the first dotted vertical line.
Join the point of
intersection formed to point B to obtain a proper line and include an arrow
showing to direction to be towards the point of intersection formed. This line
represents T

_{2}. It should be parallel to the T_{2}on AC and the direction should also be the same.
Finally, to complete the
diagram, join point A to the point of intersection to obtain a vertical line
pointing upwards. Measure the length of this line. This length would correspond
to 7.0N [length should be

**about**9.2cm]. Similarly, the length of T_{1}(already available) and T_{2}(drawn) should be**about**7cm and 5.5 respectively. By proportion (9.2cm represents 7.0N), T_{1}and T_{2}can be obtained from their corresponding lengths.}
T

_{1}= 5.4 ± 0.2 N
T

_{2}= 4.0 ± 0.2 N

__Question 13: [Moment > Equilibrium]__
Rod AB is hinged to wall at A. Rod
is held horizontally by means of cord BD, attached to rod at end B and to wall
at D, as shown. Rod has weight W and centre of gravity of rod is at C. Rod is
held in equilibrium by force T in cord and force F produced at hinge.

(a) Explain what is meant by

(i) centre of gravity of a body

(ii) equilibrium of a body

(b) Line of action of weight W of
rod passes through cord at point P. Why, for rod to be in equilibrium, force F
produced at hinge must also pass through point P:

(c) Forces F and T make angles Î± and
Î² respectively with rod and AC = (2/3)AB, as shown. Equations, in terms of F,
W, T, Î± and Î², to represent

(i) Resolution of forces
horizontally

(ii) Resolution of forces vertically

(iii) Taking of moments about A:

**Reference:**

*Past Exam Paper – June 2006 Paper 2 Q2*

__Solution 13:__**(a)**

(i)

The centre of gravity of a body is
the point at which the whole weight of the body may be considered to act.

(ii)

For the equilibrium of a body, the
sum of forces in any direction is zero and the sum of moments about any point
is zero.

**(b)**

{Forces passing through a point will have zero moment about that point. This is obvious because moment is defined as the product of the force and the perpendicular distance of the force from that point. When the force is acting on the point, its perpendicular distance from that point is zero. So, the moment of that force about the point is zero. For equilibrium, the sum of moments about a point should be zero. Since the other forces pass through point P, their moments about P are zero. So, for the sum of moment about P to be zero (condition for equilibrium), force F should also have zero moment about P. This occurs when force F passes through point P}

EITHER The forces T and W have zero moment about P. So, the force F must have zero moment (for equilibrium), i.e it must pass through P.

EITHER The forces T and W have zero moment about P. So, the force F must have zero moment (for equilibrium), i.e it must pass through P.

OR If all the forces pass through P,
the distance from P is zero for all forces. So, the sum of moments about P is
zero.

**(c)**

(i)

Resolution of forces horizontally:

{W does not have a horizontal
component since it is vertical. Resolving F and T horizontally gives the
equation: (F has a component to the right while T has a component to the left)}

FcosÎ± = TcosÎ²

(ii)

Resolution of forces vertically:

{F and T have components vertically
upwards while W is vertically downwards}

W = FsinÎ± + TsinÎ²

(iii)

Taking of moments about A:

{Taking moments about A. the weight
C is at a distance (2/3)AB from point A (clockwise moment) while the vertical
component of T is at distance AB from point A (anticlockwise moment). F has no
moment about point A since it acts at the point (its distance from the point is
zero). So, W[(2/3)AB] = TsinÎ²[AB] which is simplified into the equation (since
answer is not required in terms of AB):}

2W = 3TsinÎ²

__Question 14: [Resistance > Wires]__
Electric power cable consists of six
copper wires c surrounding a steel core s.

Length of 1.0 km of one of copper
wires has resistance of 10 Î© and 1.0 km of steel core has resistance of 100 Î©.

What is the approximate resistance
of a 1.0 km length of the power cable?

**Reference:**

*Past Exam Paper – November 2008 Paper 1 Q32*&

*June 2013 Paper 13 Q34*

__Solution 14:__**Answer: B.**

All the 7 wires in the cable are
connected in parallel (these are not separate cables, but one cable which
contains several wires).

{The

__cable__contains several__wires__as shown. Consider a current flowing in the cable – will the current flow in only 1 wire or all the wires at the same time? It is obvious that it will flow through all the wires at any instant. So, the current should be divided among the wires.
This is similar to the
usual electric circuits we usually deal with. Current is divided among the
components when they are connected in parallel. When the components are
connected in series, the same current flows through each of them.

So, in a cable, the wires
are connected in parallel. However, if 2

__cables__were present, then the same current would be made to flow through each of them. Connections of the__wires__(which are found in 1 cable) and connections of__cables__are considered differently.}
Consider the copper wires.

Combined resistance of copper wires,

R

_{C}= [(1/10) + (1/10) + (1/10) + (1/10) + (1/10) + (1/10)]^{-1}= [6/10]^{-1}= 10/6 Î©
Resistance of the steel core is 100Î©.

So, total approximate resistance of
power cable = [(1/{10/6}) + (1/100)]

^{-1}= 1.639Î© = 1.6Î©

__Question 15: [Stationary waves]__
What are the conditions for a
formation of a stationary wave: e.g same speed and frequency but what else?

__Solution 15:__
Conditions for the formation of
stationary waves:

- The waves must have the
__same speed__. - The waves must have the
__same frequency__.

{From the above 2
conditions, it can be concluded that the wavelengths of the waves must be the
same since speed v = f Î»}

- The waves must have the
__same (or almost the same) amplitude__. - The 2 (progressive) waves must be
travelling in in
__opposite directions__along the__same line of travel__and in the__same plane__. (Assuming the boundary conditions are met [a node is formed at the boundaries].)

As for the phase difference of the 2
waves, it should be

**constant**(coherent waves). Stationary waves contain notes (where displacement is minimum) and antinodes (where displacement is maximum).
For the formation of nodes, the
phase difference should be (out of phase) Ï€ rad (180

^{o}) and for the formation of antinodes, the phase difference should be (in phase) zero.
The particles in the same segment (between
2 adjacent nodes) are in phase while the articles in adjacent segments are in
anti-phase.

(Y)

ReplyDeleteYou have an amazing blog :)

ReplyDeleteI have a question though

In question number 9 (ii), why was the value of additional force not found out?

What I did was: I calculated the additional force using F=ke.

F=(1.8)(17.8-16.3)

F = 2.7 N.

Next, I calculated the change in gravitational potential energy like this:

(2.7+3.8)(17.8) - (3.8)(16.3)

= 53.76 J

Why is this wrong?

Please elaborate :)

Thanks.

DeleteBut why have you calculated change in GPE in this way:

(2.7+3.8)(17.8) - (3.8)(16.3)????????

GPE = mgh

Yeah, mg = weight which is a force, but we cannot use any value of force to represent mg (as you have done here).

In this case, the weight is constant. only the height is changing.