• Physics 9702 Doubts | Help Page 50

Question 295: [Forces > Hooke’s law]
The force F required to extend a sample of rubber by a length x was found to vary as shown in the diagram.

The energy stored in the rubber for an extension of 5m was
For J78 / II / 36 & J83 / II / 40 & J89 / I / 30:
A more than 200J
B 200J
C between 100J and 200J
D 100J
E less than 100J

For November 2004 Paper 1 Q23:
A less than 100 J
B 100 J
C between 100 J and 200 J
D more than 200 J

Reference: Past Exam Paper – J78 / II / 36 & J83 / II / 40 & J89 / I / 30 & November 2004 Paper 1 Q23



Solution 295:
For J78 / II / 36 & J83 / II / 40 & J89 / I / 30: Answer: E.
For November 2004 Paper 1 Q23: Answer: A.
The energy stored in the rubber is given by the area under the force-extension graph.

From the graph, for an extension x of 5m, the force required is 40N.
Consider a straight line joining the points (0, 0) and (5, 40). The area under that line would be greater than the area under the curve shown (since the straight line is above the curve).

Area under graph for straight line = ½ (5) (40) = 100J

So, the energy stored in the rubber is less than 100J.









Question 296: [Dynamics > Newton’s laws of motion]
(a) Define the term acceleration.                    [2]
Use your definition to explain why it is that the velocity of a body may be in a different direction from its acceleration.                [2]

(b) Discuss whether the resultant force on a body may or may not be in the same direction as its direction                                    [2]

(c) The moving head of an electronic printer has a mass of 0.20kg and moves along the line of print in a jerky motion. After printing each character, the head accelerates sideways under a force of 10N and then immediately decelerates to rest under a force of 30N ready to print the next character. The characters are 2.5mm apart.
(i) Sketch a graph to show how the velocity of the head of the printer varies with time.      [2]
(ii) Calculate the time taken for the movement between characters.              [6]
(iii) An additional time of 8.0ms is required for the printing of each character. Find the maximum number of characters that can be printed each second.                [2]
(iv) What physical problems might require the printer to run at a slower speed than the value you have calculated? What changes could be made to increase the printing speed?            [6]

Reference: Past Exam Paper – J88 / II / 8



Solution 296:
(a)
Acceleration is defined as the rate of change of velocity.
(Acceleration a = (v – u) / t where v is the final velocity, u is the initial velocity and t is the time)
Both velocity and acceleration are vectors – they have a magnitude and a direction. So, a change in either the magnitude or the direction causes a change in a vector.

Consider a change in magnitude only.
If the magnitude of the velocity increases, then the acceleration is in the same direction as the velocity.
If the magnitude of the velocity decreases, then the acceleration is in a direction opposite to the velocity. So, the directions of the velocity and acceleration are different. [Here, ‘acceleration’ is called deceleration]

Consider a change in the direction of the velocity vector.
This causes a change in velocity with time [= acceleration]. The direction of the acceleration is the same as the direction of the resultant change in velocity and the acceleration is therefore in a direction different from the velocity.


(b)
From Newton’s second law of motion, it is found that the force vector may be given as

F = ma

where a is the acceleration (vector) and m is the mass (scalar).
Since the force and the acceleration are the only vectors involved in the equations, they can only be in the same direction.


(c)
(i)

(or the horizontal lines at zero velocity could be omitted)


(ii)
From the question, the head accelerates sideways (from rest) under a force of 10N after printing each character and then immediately decelerates to rest under a force of 30N ready to print the next character. The characters are 2.5mm apart.
So, the head prints a character, then moves sideways by the motion described until it comes to rest (at the position a new character would be printed) and prints another character.
Therefore, the distance travelled after being accelerated by the force of 10N and then decelerated to rest by the force of 30N is 2.5mm. This is the area under the graph in part (i) for the ‘triangles’ formed in the graph.

Mass of head = 0.20kg
Acceleration caused by 10N force, a₁ = 10 / 0.2 = 50ms^-2
Acceleration (deceleration) caused by 30N force, a₂ = (-) 30 / 0.2 = -150ms^-2

The forces each cause a constant acceleration. So, the equations of uniformly accelerated motion can be used.

Consider the motion caused by the 10N force.
Initial speed, u₁ = 0. Final speed = v₁. Distance travelled by head = s₁. Time to travel distance s₁ = t₁. Acceleration = a₁.
v₁² = u₁² + 2a₁s₁
v₁² = 0 + 2 (50) s₁ = 100s₁       …………………… (1)

Now consider the motion caused by the 30N force.
Initial speed, u₂ = v₁ (as can be seen from the graph). Final speed, v₂ = 0. Distance travelled by head = s₂. Time to travel distance s₂ = t₂. Acceleration = a₂.
v₂² = u₂² + 2a₂s₂
0 = v₁² + 2 (-150) s₂
0 = v₁² – 300s₂                         …………………… (2)

As stated above, the total distance travelled is the distance between each character. So,
s₁ + s₂ = 2.5x10^-3                     …………………… (3)

From equation (2),
v₁² = 300s₂                               …………………… (4)

Replacing equation (4) in (1) gives
300s₂ = 100s₁                          …………………… (5)

But from (3),
s₂ = 2.5x10^-3 – s₁                     …………………… (6)

Replacing equation (6) into (5) gives
300 (2.5x10^-3 – s₁) = 100s₁
s₁ = 300 (2.5x10^-3) / 400 = 0.75 / 400 = 1.875x10^-3m

s₂ = 2.5x10^-3 – s₁ = (2.5 – 1.875) x10^-3 = 0.625x10^-3m

Replacing the value of s₂ in equation (4) gives
Speed v₁ = √(300 x 0.625x10^-3) = 0.4330ms^-1

For the motion caused by the force of 10N,
a₁ = (v₁ – u₁) / t₁
Time t₁ = (0.4330 – 0) / 50 = 8.66x10^-3s

For the motion caused by the force of 30N,
a₂ = (v₂ – u₂) / t₂
Time t₂ = (0 – 0.4330) / (-150) = 2.887x10^-3s

Total time for movement between characters = t₁ + t₂ = (8.66 + 2.887) x10^-3 = 11.5x10^-3 = 11.5ms


(iii)
To print a character, the head must first be moved to its position (this takes a total time of 11.5ms as calculated above) and the it takes 8ms for the printing itself.
Total time to print a character = 11.5 + 8 = 19.5ms = 19.5x10^-3s
Number of characters that can be printed in 1 second = 1 / (19.5x10^-3) = 51.3s^-1


(iv)
Possible physical problems that might require the printer to run at a slower speed:
Heat buildup, Vibration, Noise, Friction, Improper parts of the printer could cause mechanical advantage loss
Time may be required to start / measure / stop the ink flow to each dot.
Supposing the drive system is not very ‘stiff’, the head might overshoot the required position – that is, the head could keep on moving beyond the required position even if the drive has stopped, due to the inertia of the head.
The head might oscillate about the required position before coming to rest. So, a ‘setting’ time might be required after the deceleration so that the oscillations die away.


Changes that could be made to increase the printing speed:
{F = ma. Acceleration, a = F / m}
The forces used to accelerate and decelerate the head could be increased.
Making the printer head lighter would cause it to accelerate faster.









Question 297: [Matter > Young modulus]
Young modulus of steel is determined using length of steel wire and is found to have the value E.
Another experiment is carried out using a wire of same steel, but of twice the length and half the diameter.
What value is obtained for Young modulus in the second experiment?
For June 2008:
A ¼ E                          B ½ E                          C E                              D 2E

For November 2011:
A ½ E                          B E                              C 2E                            D 4E

Reference: Past Exam Paper – June 2008 Paper 1 Q24 & November 2011 Paper 11 Q23 & Paper 13 Q21



Solution 297:
June 2008 - Answer: C.
November 2011 – Answer: B.
For a specific material, the Young modulus is constant.
Since the SAME steel is used in both experiments, the Young modulus is the same.








Question 298: [Matter > Strain energy]
Rubber band is stretched by hanging weights on it and force-extension graph is plotted from the results.

What is best estimate of the strain energy stored in the rubber band when it is extended 30 cm?
For June 2009:
A 2.0 J                         B 2.6 J                         C 5.1 J                         D 200 J

For November 2014:
A 1.8 J                         B 2.6 J                         C 5.1 J                         D 200 J

Reference: Past Exam Paper – June 2009 Paper 1 Q21 & November 2014 Paper 11 & 12 Q21



Solution 298:

Answer: A.

Strain energy stored in the rubber band is given by the area under the force-extension graph.

EITHER
This question could be tackled by drawing a straight line on the graph (that would approximately correspond to the mean force). A horizontal line at about 6 N shows approximately the same area as under the whole graph. This gives 6 N × 0.3 m = 1.8 J.

OR
Take the area from extension = 0cm to extension = 7cm to be that of a triangle.
Area 1 = ½ (0.07 – 0) (5) = 0.175J

Consider the section from extension = 7cm to 20cm to correspond to a constant force of 5.5N. The area is that of a rectangle.
Area 2 = (0.20 – 0.07) (5.5) = 0.715J

From extension = 20cm to 22cm, a parallelogram is formed with parallel sides being 5.5N and 6N. Area of parallelogram = ½ (sum of parallel side) (height)
Area 3 = 0.5 (5.5 + 6) (0.02) = 0.115J

The last section from extension = 22cm to 30cm can also be considered to be a parallelogram of parallel side 6N and 17N.
Area 4 = 0.5 (6 + 17) (0.30 – 0.22) = 0.92J

Total energy stored = 0.175 + 0.715 + 0.115 + 0.92 = 1.925J
So, choice A is the best estimate.