Tuesday, November 18, 2014

Physics 9702 Doubts | Help Page 15

  • Physics 9702 Doubts | Help Page 15

Question 79: [Kinematics > Linear motion]
Shopping trolley and its contents have total mass of 42 kg. Trolley is being pushed along horizontal surface at speed of 1.2 m s–1. When trolley is released, it travels distance of 1.9 m before coming to rest.
(a) Assuming that total force opposing motion of the trolley is constant,
(i) Calculate deceleration of trolley
(ii) Show that total force opposing motion of trolley is 16 N.

(b) Using answer in (a)(ii), calculate power required to overcome total force opposing motion of trolley at a speed of 1.2 m s–1.

(c) Trolley now moves down straight slope that is inclined at angle of 2.8° to horizontal, as shown.

Constant force that opposes motion of trolley is 16 N.
Calculate, for trolley moving down slope,
(i) Component down slope of trolley’s weight,
(ii) Time for trolley to travel from rest a distance of 3.5 m along length of slope.

(d) Use answer to (c)(ii) to explain why, for safety reasons, slope is not made any steeper.

Reference: Past Exam Paper – June 2008 Paper 2 Q3

Solution 79:
v2 = u2 + 2as
0 = 1.22 + 2a(1.9)
a = - 0.38ms-2
So, deceleration = 0.38ms-2

Force, F = ma = 42 x 0.38 = 16N

Power required = Fv = 16 x 1.2 = 19W

Component of weight down slope = (42 x 9.8) sin2.8 = 20.1N

(Resultant) accelerating force = 20.1 – 16 = 4.1N
Acceleration of trolley (= F / m) = 4.1 / 42 = 0.098ms-2

{Unlike the answer for part (a)(i) where the acceleration is given to 2 significant figures, here the acceleration should be taken to at least 3 significant figures. This is because the acceleration here is not the final answer, and rounding to 2 significant figures (here giving 0.10ms-2) would cause the final answer for the time to change (it would become 8.36 8.4).

So, final answers may be given to 2 significant figures (unless asked otherwise by the question) and intermediate calculations should NOT be rounded off to the same number of significant figures as for the final answers. Actually, using the exact value from the calculator (in the intermediate steps) is recommended. As for writing down the intermediate steps, it should be done in more significant figures than the final answers. This applies for all questions, not only this one.}

s = ut + ½at2
3.5 = 0 + 0.5 (0.098) t2
Time t = 8.5s

EITHER It allows plenty of time to stop runaway trolley
OR The speed of the trolley increases gradually
OR The trolley will travel faster

Question 80: [Pressure]
Liquid Q has twice density of liquid R. At depth x in liquid R, pressure due to the liquid is 4 kPa.
At what depth in liquid Q is pressure due to the liquid 7 kPa?
A 2x / 7                       B 7x / 8                       C 8x / 7                       D 7x / 2

Reference: Past Exam Paper – June 2014 Paper 12 Q18

Solution 80:
Answer: B.
[Hint: here ρ is the quantity that links R and Q. So, we need to identify ρ (in terms of the other known variables) first]
For liquid R, Pressure P = hρg = xρg = 4
So, density of liquid R, ρ = 4/xg

The density of liquid Q is twice that of liquid R.
For liquid Q, Pressure P = 7 = h(2ρ)g
So, depth in liquid Q, h = 7/(2ρg)
ρg = (4/xg) g = 4/x
Depth, h = (7/2) (4/x)-1 = 7x/8

Question 81: [Young modulus]
Elastic material with Young modulus E is subjected to tensile stress S. Hooke’s Law is obeyed.
What is the expression for elastic energy stored per unit volume of material?
A S2 / 2E                     B S2 / E                       C E / 2S2                     D 2E / S2

Reference: Past Exam Paper – June 2014 Paper 13 Q23

Solution 81:
Answer: A.
Elastic energy = ½ Fx
Elastic energy per unit volume = (½ Fx) / V

[Volume V = AL] Elastic energy per unit volume = (½ Fx) / AL.
[F / A = S] Elastic energy per unit volume = Sx / 2L

[Strain = x / L] Elastic energy per unit volume = S {strain} / 2
[Young modulus, E = stress S / strain. So, strain = S / E]

Elastic energy per unit volume = SS / 2E = S2 / 2E

Question 82: [Vectors]
Two physical quantities P and Q are added. Sum of P and Q is R, as shown.
Which quantity could be represented by P and by Q?
A kinetic energy                      B power                      C speed                       D velocity

Reference: Past Exam Paper – November 2012 Paper 13 Q2

Solution 82:
Answer: D.
Physical quantities may either be scalars or vectors. Vectors have both a magnitude and a direction, while scalars only have a magnitude.
Since P and Q have directions, they are vectors. Addition of 2 vectors result into another vector.
Kinetic energy is a scalar. Power (rate of doing work) is also a scalar. Speed is a scalar.
Velocity is a vector.

Question 83: [Forces > Equilibrium]
Diagram shows a rope bridge that student makes on adventure training course. The student has weight W.
Which formula gives tension T in rope?
A T = W / 2cosθ          B T = W / 2sinθ           C T = W / cosθ            D T = W / sinθ

Reference: Past Exam Paper – November 2010 Paper 12 Q11

Solution 83:
Answer: B.

There is a tension T in the rope on each side of the student (as indicated).
The sum of vertical components of the tensions T should be equal to the weight, W which acts vertically down.

Vertical component of tension T = T sinθ
2T sinθ = W
Tension, T = W / 2sinθ  

Question 84: [Forces > Moments]
A trailer of weight 30kN is hitched to a cab at X, as shown in the diagram.

What is the upward force exerted by cab on trailer at X?
A 3kN             B 15kN                       C 30kN                       D 60kN

Reference: Past Exam Paper – November 2011 Paper 12 Q14

Solution 84:
Answer: B.
For equilibrium (for the cab and the trailer to stay connected), clockwise moment = anticlockwise moment about a pivot.

Consider the free body diagram of the trailer alone. The forces should act such that the trailer is horizontal.
The back wheel rests on the ground (so, it can be used as the pivot). 

The trailer causes a (downward) force at point X (so that the trailer stays horizontal). So, the trailer causes a clockwise moment about the back wheel (this acts as the pivot). Its weight of 30kN acts at the centre of the trailer (at a distance of 10m from the back wheel).
(Clockwise) moment of trailer about the back wheel = 30 (10) = 300 kNm

To keep the trailer horizontal, the cab should exert an upward force, F on the trailer, causing an anticlockwise moment on the trailer. The cab is in contact with the trailer at point X, which is a distance of 20m from the back wheel.
(Anticlockwise) moment of cab on trailer about the back wheel = F (20) = 20F Nm

Considering the moments about point of contact (i.e, back wheel),
F x 20 = 300
Upward force of cab on trailer, F = 300 / 20 = 15kN


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