# Physics 9702 Doubts | Help Page 12

__Question 64: [Energy > Potential energy]__
Diagram shows two identical vessels X and Y
connected by short pipe with a tap.

How much potential energy is lost by water during
this process? (g = acceleration of free fall)

**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q15*

__Solution 64:__**Answer: B.**

Consider the water in both vessels as

**one body with one mass**.
Since the mass of the
water is spread all over the volume it occupies, the

**centre of mass**should be considered when calculating the (gravitational) potential energy.
Before the tap is
opened, the centre of mass of the water in vessel X may be considered to be at
its centre (that is, at

__half the water level__[h/2]).
Potential energy of water before tap is opened =
m g (h/2) = mgh / 2

When the tap is opened, the

**total mass of water in the whole body is the same**, but the water__level__is now at a height of h/2 in both vessels.
The centre of mass
may be considered to be at the centre of the volume of water, that is, at a
height of half the water level [water level = h/2, so centre of gravity is now at
a height of h/4]

Potential energy of water after tap has been
opened = m g (h/4) = mgh / 4

Lost in potential energy = (mgh / 2) – (mgh / 4)
=

**mgh / 4**

__Question 65: [Energy > Potential energy]__**Reference:**

*???*

__Solution 65:__(Gravitational) Potential energy = mgh

Lost in (gravitational) potential energy = mgΔh = 80000 x 9.81 x (10000 – 1000)

Lost in (gravitational) potential energy = 7.1x10

^{9}J

This energy is lost as work done against drag forces (to keep the aircraft’s speed constant) and against air resistance (when decreasing in altitude).

__Question 66: [Dynamics > Resultant forces]__Mass of 2.0 kg rests on a frictionless surface. It is attached to 1.0 kg mass by a light, thin string which passes over frictionless pulley. 1.0 kg mass is released and accelerates downwards.

What is the speed of 2.0 kg mass as the 1.0 kg mass hits floor, having fallen a distance of 0.50 m?

A 1.8 m s

^{–1}B 2.2 m s

^{–1}C 3.1 m s

^{–1}D 9.8 m s

^{–1}

**Reference:**

*Past Exam Paper – November 2012 Paper 11 Q13*

__Solution 66:__**Answer: A.**

The gravitational force acts on the
1.0kg mass, causing an acceleration of 9.81ms

^{-2}.
Weight of 1.0kg box = 1 x 9.81 = 9.81N

But as the 1.0kg mass moves, the

__2.0kg also undergoes motion__. Therefore, considering the whole system, a**force of 9.81N is acting on a total mass of (2.0 + 1.0 =) 3.0kg**.
The resultant acceleration, a on
this total mass can be obtained as follows

3a = 9.81

Acceleration, a = 9.81 / 3 = 3.27ms

^{-2}
Hence, each of the 2 masses
undergoes a resultant acceleration of 3.27ms

^{-2}.
Initial speed of 2.0kg mass, u = 0

Distance travelled, s = 0.5m

v

^{2}= u^{2}+ 2as = 0 + 2 (3.27) (0.5)
Speed, v = √(2 x 3.27 x 0.5) =

**1.8ms**^{-1}

__Question 67: [Kinetic energy > Speed]__
Kinetic energy of particle is
increased by a factor of 4. By what factor does its speed increase?

A 2 B
4 C 8 D 16

**Reference:**

*Past Exam Paper – November 2004 Paper 1 Q15 & November 2012 Paper 11 Q18*

__Solution 67:__**Answer: A.**

Kinetic energy, KE is proportional
to v

^{2}(KE = ½ mv^{2}). So, speed v is proportional to √(KE).
Kinetic energy is increased by a
factor of 4. (KE becomes 4KE)

So, new speed is √(4KE) = (√4) (√(KE)) = 2√(KE)

Since, speed v is proportional to √(KE),

New speed = 2v

Therefore, speed increases by a
factor of (2v / v =) 2

__Question 68: [Dynamics > Resultant forces]__
A light spring has a mass of 0.20kg
suspended from its lower end. A second mass of 0.10kg is suspended from the
first by a thread. The arrangement is allowed to come into static equilibrium
and then the thread is burned through. At this instant, what is the upward
acceleration of the 0.20kg mass? (Take g as 10ms

^{-2}.)
A zero B 3.3ms

^{-2}C 5.0ms^{-2}D 6.7ms^{-2}E 10ms^{-2}**Reference:**

*Past Exam Paper – November 1987 / I / 5*

__Solution 68:__**Answer: C.**

For static equilibrium, the
resultant force is zero. Thus, the (upward) force in the spring is equal to the
total (downward) weight of the masses.

Spring force, T = 0.20 (10) + 0.1
(0.10) = 3N

When the thread is burnt,
the weight of the 0.10kg mass no longer contributes any forces to the system,
but at that

**instant**, the spring force is still 3N.
So, the weight of the 0.20kg acts
downwards while the spring force acts upwards. The resultant (upward) force on
the 0.20kg mass is given by

Resultant force = ma = T – mg = 3 –
0.20(10) = 1N

Acceleration, a = 1 / 0.20 = 5ms

^{-2}

__Question 69: [Dynamics > Resultant forces]__
A raindrop of mass m is falling
vertically through the air with a steady speed v. The raindrop experiences a
retarding force kv due to the air, where k is a constant. The acceleration of
free fall is g.

Which expression gives the kinetic
energy of the raindrop?

A mg / k B mg

^{2}/ 2k^{2}C m^{3}g^{2}/ k^{2}D m^{3}g^{2}/ 2k**Reference:**

*Past Exam Paper – June 1992 / I / 6 & November 1996 / I / 7 & November 2010 Paper 12 Q15*

__Solution 69:__**Answer: D.**

For raindrop to fall with a steady
speed v, terminal speed must be reached (net force and
thus, net acceleration is zero).

This occurs when the retarding force
kv equals to the weight mg.

So, mg = kv giving

**v = mg / k**.
Kinetic energy of raindrop, KE = ½
mv

^{2 }= ½ m(mg / k)^{2}=**m**.^{3}g^{2}/ 2k^{2}
salam! in q65, wont energy be converted to kinetic energy?

ReplyDeleteWslm.

DeleteThe question states that the aircraft's speed is constant. So, kinetic energy is also constant [since the aircraft was initially moving, it has some initial kinetic energy].

Kinetic energy = (1/2) mv^2. If v is not changing, the kinetic energy is not changing also.

so in q65 we assume that the constant speed with which it descends is the same as that with which it was travelling at the altitude of 10000 m?

DeleteYes, but we are only assuming this because it is said in the question.

DeleteJazakAllah

DeleteCould you please explain question 6 of October/November 2012 p11?

ReplyDeleteSee solution 433 at

Deletehttp://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-82.html

Thank you for your post =D

ReplyDeleteI don't understand why we need to consider centre of gravity in solution 65

ReplyDeletethis has always been the case, but people do not usually pay attention.

Deletethe centre of mass is where all of the mass of the object is assumed to be and it is where the weight would act.

Potential energy is due to height so shouldn't it be mgh at the start?

Deleteyest and 'height' is actually the height of the centre of mass.

Delete