# 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:__**(a)**

(i)

v

^{2}= u^{2}+ 2as
0 = 1.2

^{2}+ 2a(1.9)
a = - 0.38ms

^{-2}
So, deceleration = 0.38ms

^{-2}
(ii)

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

**(b)**

Power required = Fv = 16 x 1.2 = 19W

**(c)**

(i)

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

(ii)

(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 + ½at

^{2}
3.5 = 0 + 0.5 (0.098) t

^{2}
Time t = 8.5s

**(d)**

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 S

^{2}/ 2E B S^{2}/ E C E / 2S^{2}D 2E / S^{2}**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 =

**S**^{2}/ 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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