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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:

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Anna Litical drops a ball from rest from the top of 78.4 m high cliff.

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:

ΔE_p = mgΔh    or W Δh = 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 = 1.8x10²Nm^-1 (1cm à1.8N. So, 100cm = 1m à 1.8x10²)}

{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}

ΔE_s = ½ (1.8x10²) (0.036² – 0.021²) = 0.077J

3.

Work done by force F:

{Work done by force F = ΔE_s – ΔE_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 = (Δg / g) x 100% = 4.2% and g = 9.751ms^-2}

Δg = ([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. DECELERATION only occurs when the direction of acceleration is opposite to the direction of motion. 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].

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 T₁ and in string AC it is T₂. On Fig, draw vector triangle to determine magnitudes of T₁ and T₂:

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 T₂ 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 T₂. It should be parallel to the T₂ 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 T₁ (already available) and T₂ (drawn) should be about 7cm and 5.5 respectively. By proportion (9.2cm represents 7.0N), T₁ and T₂ can be obtained from their corresponding lengths.}

T₁ = 5.4 ± 0.2 N

T₂ = 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 α and β respectively with rod and AC = (2/3)AB, as shown. Equations, in terms of F, W, T, α and β, 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.

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)}

Fcosα = Tcosβ

(ii)

Resolution of forces vertically:

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

W = Fsinα + Tsinβ

(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] = Tsinβ[AB] which is simplified into the equation (since answer is not required in terms of AB):}

2W = 3Tsinβ

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 Ω and 1.0 km of steel core has resistance of 100 Ω.

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 Ω

Resistance of the steel core is 100Ω.

So, total approximate resistance of power cable = [(1/{10/6}) + (1/100)]^-1 = 1.639Ω = 1.6Ω

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 λ}

• 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) π 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.