# Physics 9702 Doubts | Help Page 8

__Question 45: [Resistance > Power]__
Electric heater consists of three
similar heating elements A, B and C, connected as shown.

Each heating element is rated as 1.5
kW, 240 V and assumed to have constant resistance. Circuit is connected to 240
V supply.

**(a)**Calculate resistance of one heating element.

**(b)**Switches S

_{1}, S

_{2}and S

_{3}may be either open or closed. Complete Fig to show total power dissipation of heater for switches in positions indicated.

S

_{1}S_{2}S_{3}Total power / kW
Open closed closed

Closed closed open

Closed closed closed

Closed open open

Closed open closed

**Reference:**

*Past Exam Paper – June 2008 Paper 2 Q6*

__Solution 45:__**(a)**

EITHER Power, P = VI and V = IR OR P = V

^{2}/ R
Resistance (= V

^{2}/ P = 240^{2}/ 1500) = 38.4Ω**(b)**

S

_{1}S_{2}S_{3}Total power / kW
Open closed closed

**zero**
Closed closed open

**1.5kW**
Closed closed closed

**3.0kW**
Closed open open

**0.75kW**
Closed open closed

**2.25kW**
{A and B are in series and
these 2 are in a parallel combination with C and the 240V supply. A, B and C
each has a resistance of 38.4Ω.
Power = V

^{2}/ R = 240^{2}/ R where R is the total resistance being considered in the circuit. In this case, the power can be related to the p.d. across the heater. [P is proportional to V^{2}– this method is actually quicker than always calculating the total resistance]
Consider the switches. If S

_{1}is opened, no current would flow in the circuit. So, power = zero. Closing S_{2}causes B to be short circuited while opening S_{2}allows the resistance of B to be included in the total resistance of the circuit. Opening S_{3}causes C to be short circuited.
For case 1, no current
flows since S

_{1}is open. Power = 0.
For case 2, B and C are
short circuited. So, p.d. across A = 240V. Total power = 1.5kW.

For case 3, only B is
short circuited. Since A and C are in parallel, the same p.d. of 240V is across
them. Each has power = 1.5kW. Total power = 1.5 + 1.5 = 3.0kW.

For case 4, only C is
short circuited. Since A and B are in series, the p.d. across each of them is
120V each. Since P is proportional to V

^{2}, when V is halved, P becomes one quarter. So, the power of each heater A and B is (1.5 / 4)kW. Total power = (1.5/4) + (1.5/4) = 0.75kW.
For case 5, none of the
heaters is short circuited. A p.d. of
240V would be across C since it is in parallel with the supply. So, power due
to C = 1.5kW. There would be a p.d. of 120V across each heater A and B and as
in case 4, total power due to A and B = 0.75kW. Total power = 1.5 + 0.75 =
2.25kW}

__Question 46: [Kinematics > Projectile motion]__
A marble rolls off a table 4m high
at a speed of 10ms

^{-1}.**(a)**How far from the edge of the table does it strike the floor?

**(b)**What is its velocity (magnitude and direction) as it strikes the floor?

**Reference:**???

__Solution 46:__**(a)**

Horizontal velocity = 10ms

^{-1}
Vertical height, s = 4m

Take acceleration of free fall, g =
9.81ms

^{-2}
Consider the vertical motion.

Time taken for the ball to reach the
ground = t

Initial vertical velocity, u = 0ms

^{-1}
s = ut + ½ at

^{2}
4 = 0 + 0.5(9.81)t

^{2}
Time t = [4.0 / (0.5 x 9.81)]

^{0.5}= 0.903s
Assuming air resistance is
negligible, horizontal acceleration = 0.

For horizontal motion,

Speed = Distance / Time

Distance from edge of table = Speed
x Time = 10 x 0.903 =

**9.03m****(b)**

When ball strikes the floor,

Horizontal component of velocity =
10ms

^{-1}
Vertical component of velocity
(downwards) when ball strikes floor = v

v = u + at = 0 + 9.81(0.903) =
8.858ms

^{-1}
Magnitude of velocity = √(10

^{2}+ 8.858^{2}) = 13.359 = 13.4ms^{-1}
Angle of velocity below horizontal =
tan

^{-1}(8.858 / 10) = 41.5^{o}
By convention, angles are read in an
anticlockwise direction from the horizontal. The above angle is also the acute
angle formed from the direction of the velocity and above the horizontal. But
it is not the angle that should be stated (due to the convention).

So, the ball strikes the floor with
a velocity of

**13.4ms**at an angle of {180 – 41.5 =}^{-1}**138.5**.^{o}from the horizontal

__Question 47: [Kinematics > Projectile Motion]__
Ball is thrown from S with initial
velocity of 15.0 m s

^{–1}at 60.0° to horizontal. Assume air resistance is negligible.**(a)**For ball at S, calculate

(i) Horizontal component of
velocity:

(ii) Vertical component of velocity:

**(b)**Horizontal distance from S to wall is 9.95 m. Ball hits wall at P with velocity that is at right angles to wall. Ball rebounds to point F that is 6.15 m from wall. Using answers in (a),

(i) Calculate vertical height gained
by ball when it travels from S to P

(ii) Show that time taken for ball
to travel from S to P is 1.33 s

(iii) Show that velocity of ball
immediately after rebounding from wall is about 4.6 m s

^{–1}**(c)**Mass of ball is 60 × 10

^{–3}kg.

(i) Calculate change in momentum of
ball as it rebounds from wall.

(ii) State and explain whether
collision is elastic or inelastic.

**Reference:**

*Past Exam Paper – November 2011 Paper 21 Q3*

__Solution 47:__**(a)**

(i) Horizontal component of velocity
= 15cos60

^{o}= 7.5ms^{-1}
(ii) Vertical component of velocity
= 15sin60

^{o}= 13ms^{-1}**(b)**

(i)

{Consider vertical
motion,}

v

^{2}= u^{2}+ 2as
{Since the ball hits the
wall

__horizontally__, it has no vertical component. So, we consider only the horizontal component of velocity. Also, since air resistance is negligible, the horizontal component of velocity remains unchanged.
So, (taking the upward
direction as +ve) initial velocity, u = (+) 13ms

^{-1}. When ball hits the wall, its velocity is zero. So, final velocity, v = 0. Acceleration due to gravity, g is downwards. So, a = g = (-) 9.81ms^{-2}.
[For Cambridge A-Level Physics, the magnitude of
acceleration due to gravity (on earth) is always taken to be 9.81ms

^{-2}, unless stated otherwise. As for the direction, it depends whether the upward or downward direction is taken to be positive. Acceleration due to gravity is always downwards (towards surface). Here the direction of the acceleration is negative as the positive direction has been defined to be upwards]
The equation becomes: 0 =
13

^{2}+ 2(-9.81) s}
Vertical height, s = 13

^{2}/ (2x9.81) = 8.6(1)m
(ii)

{For 1

^{st}case, consider vertical motion: v = u + at. 0 = 13 + (-9.81) t giving t = 13 / 9.81. For the OR case, consider horizontal motion. (There is no acceleration). Speed = Distance / Time, giving time t = distance / speed = 9.95 / 7.5}
Time, t = 13 / 9.81 = 1.326s OR Time, t = 9.95 / 7.5 = 1.327s

(iii)

{Consider horizontal
motion. The time for the ball to return to the ground is the same as in (ii),
the time for the upward journey of the ball. [At the

__instant__the ball hits the wall (horizontally), it has no vertical component. In the same way, at the**the ball rebounds; it still has only a horizontal velocity. So, the (resultant) velocity of the ball at that instant is actually equal to the horizontal component (since the vertical component is zero). There is no need to find the resultant here. We calculate the resultant only when the 2 components of the velocity are not zero.] Velocity = distance / time = 6.15 / 1.33.}**__instant__
Velocity of ball after rebound =
6.15 / 1.33 = 4.6ms

^{-1}**(c)**

(i)

{Take motion towards the
wall to be +ve and away from the wall to be –ve. [As
explained above, at the instant the ball rebounds, it only has a horizontal
component of velocity.] Change
in momentum, Δp = mΔv =
m(v

_{f}– v_{i}) where v_{f}is the final velocity (= - 4.6ms^{-1}) and v_{i}is the initial velocity (= (+) 7.5ms^{-1}).}
Change in momentum of ball = (60x10

^{-3}) [– 4.6 – 7.5] = (-) 0.73Ns
(ii)

The final velocity / kinetic energy
is less after the collision OR The relative speed of separation is less than
the relative speed of approach. Hence, the collision is inelastic.

__Question 48: [Momentum > Rocket]__
A rocket of mass M is travelling at
a constant speed v away from a stationary observer when it instantaneously
ejects one-third of its mass at a speed of 6v towards the observer. What is the
new speed of the rocket?

A. 1.5v B. 2.0v C.
4.5v D. 5.0v

**Reference:**

*???*

__Solution 48:__
Define the direction away from
observer be the positive direction.

__Before ejection of mass__,

Sum of momentum before ejection of
mass = Mv

__After ejection of mass__,

Since the one-third mass (M / 3) is
moving towards the observer, its speed = – 6v.

New mass of rocket = 2M / 3

Let new speed of rocket = v

_{f}
From the

**conservation of momentum**, sum of momentum before ejection of the mass should be equal to the sum of momentum after ejection of the mass.
Mv = (M/3)(– 6v) + (2M/3)v

_{f}
Mv = – 2Mv + 2Mv

_{f}/3
3Mv = 2Mv

_{f}/3
v

_{f}= (3 x 3)v /2 = 4.5v [Correct answer is C]

__Question 49: [Energy > Water flow]__**(a)**Distinguish between gravitational potential energy and electric potential energy.

**(b)**Body of mass m moves vertically through distance h near Earth’s surface. Use defining equation for work done to derive expression for gravitational potential energy change of body.

**(c)**Water flows down a stream from reservoir and then causes water wheel to rotate, as shown.

As water falls through vertical
height of 120 m, gravitational potential energy is converted to different forms
of energy, including kinetic energy of water. At the water wheel, kinetic
energy of water is only 10% of its gravitational potential energy at reservoir.

(i) Show that speed of water as it
reaches wheel is 15 m s

^{–1}.
(ii) Rotating water wheel is used to
produce 110 kW of electrical power. Calculate mass of water flowing per second
through wheel, assuming that production of electric energy from kinetic energy
of the water is 25% efficient.

**Reference:**

*Past Exam Paper – November 2011 Paper 21 Q4*

__Solution 49:__**(a)**

Electric potential energy is the
energy (stored) when a charge is moved and gravitational potential energy is
the energy (stored) when a mass is moved due to work done in an electric field
and work done in gravitational field respectively.

**(b)**

Work done = Force x distance moved
(in the direction of force)

and Force = mg

(Gravitational potential energy =)
mg × h OR mg × Δh

**(c)**

(i)

Show that speed of water as it
reaches wheel is 15 m s

^{–1}:
{0.1 x Gravitational
potential energy = Kinetic energy, ½mv

^{2}}
0.1 m (9.81) (120) = 0.5 m v

^{2}
Speed of water, v = 15.3ms

^{-1}
(ii)

{Power = energy / time.
(Kinetic) energy = ½mv

^{2}. Power = (½mv^{2}) / t. Mass of water flowing per second: m / t}
Power, P = 0.5 mv

^{2}/ t
{Only 25% of the kinetic
energy produces electric energy.}

Mass of water flowing per second, m
/ t = (110x10

^{3}) / [0.25 x 0.5 x 15.3^{2}] = 3740kgs^{-1}
for the first question I dont quite get the meaning of short circuited :s

ReplyDeleteWhen a component is short circuited, current does not flow through it.

DeleteFor example, component B is connected to switch B which is assumed to have zero resistance. When it is closed, all currents would flow through the switch and none through the component as the later has some resistance while the switch has zero resistance.

Note that previously, there was a mistake in part of the explanation. It was

'Closing S3 causes C to be short circuited.'

It should have been

'Opening S3 causes C to be short circuited.'

I already corrected it.

I didn't understood how u calculated mass of water per second?? I mean how u substituted the values for time

ReplyDeleteconsider the unit of 'mass of water per unit time'.

Deleteit is kg / s

since distance per unit time is m /s

and kg / s is mass/time (mass divided by time)

so, the quantity that we want to find is mass/time. In the equation, we do not replace values for mass or time and instead we make 'mass/time' the subject of formula

In solution 47, u r using vertical component of velocity but u r also stating that vertical component is zero plz explain

ReplyDeleteat S (initially), it has a vertical component. But when it hits the wall at right angles, it does not have any vertical component as the motion at that instant is horizontal (90 deg to the wall).

Delete