# Physics 9702 Doubts | Help Page 197

__Question 953: [Dynamics > Momentum]__**(a)**State the principle of conservation of momentum.

**(b)**A ball X and a ball Y are travelling along the same straight line in the same direction, as shown in Fig.1.

Ball X has mass 400 g and horizontal
velocity 0.65 m s

^{–1}.
Ball Y has mass 600 g and horizontal
velocity 0.45 m s

^{–1}.
Ball X catches up and collides with
ball Y. After the collision, X has horizontal velocity 0.41 m s

^{–1}and Y has horizontal velocity v, as shown in Fig.2.
Calculate

(i) total initial momentum of the
two balls,

(ii) velocity v,

(iii) total initial kinetic energy
of the two balls.

**(c)**Explain how you would check whether the collision is elastic.

**(d)**Use Newton’s third law to explain why, during collision, the change in momentum of X is equal and opposite to the change in momentum of Y.

**Reference:**

*Past Exam Paper – November 2014 Paper 23 Q4*

__Solution 953:__**(a)**The principle of conservation of momentum states that, for a system (of interacting bodies), the

__total__momentum remains constant provided there is no

__resultant__force acting (on the system)

**(b)**

(i)

{Total momentum = momentum
of X + momentum of Y.

Momentum p = mv}

Total initial momentum = m

_{1}v_{1}+ m_{2}v_{2}= (0.4×0.65) + (0.6×0.45) = 0.26 + 0.27 = 0.53 Ns
(ii)

{Total momentum before
collision = Total momentum after collision}

0.53 = (0.4×0.41) + (0.6×v)

Velocity v = 0.366 / 0.6 = 0.61 m s

^{–1}
(iii)

KE = ½ mv

^{2}
Total initial KE = ½×0.4×(0.65)

^{2}+ ½×0.6×(0.45)^{2}= 0.0845 + 0.06075 = 0.15 (0.145) J**(c)**

EITHER Check if the relative speed
of approach equals the relative speed of separation

OR Total final kinetic energy equals
the total initial kinetic energy

**(d)**

{Newton’s 3

^{rd}law:} The forces on the two bodies (or on X and Y) are equal and opposite.
{Force is defined as the
rate of change of momentum. F = Δp / Δt where Δp is the change in momentum and Δt
is the time of collision, which is the same for both spheres.

Δp = F Δt. Newton’s 3

^{rd}law states that F is equal and opposite for the spheres. We know that Δt is the same for both spheres. So, Δp is also equal and opposite for the spheres.}
The time (of
collision) is the same for both forces

__and__force is defined as change in momentum / time.

__Question 954: [Dynamics > Momentum]__
A resultant force of 10 N acts on a
body for a time of 2.0 s.

Which graph could show variation
with time t of the momentum p of the body?

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q11*

__Solution 954:__**Answer: B.**

A graph of momentum against time is
given. So, we need a relationship that relates these 2 quantities.

Newton’s 2

^{nd}law of motion: Force, F = Δp / Δt
Change in momentum, Δp = FΔt

Gradient of graph = Δp / Δt. Thus, the
gradient of the graph represents the force.

The gradient of the graph should have
a magnitude equal to 10N.

For choice B, gradient = (20 – 0) /
(0 – 2) = (-) 10N

__Question 955: [Current of Electricity > Resistance]__
Diagram shows part of a circuit.

What is the resistance between the
points P and Q due to the resistance network?

A 1.3 Ω B 4.0 Ω C
10 Ω D 37 Ω

**Reference:**

*Past Exam Paper – November 2010 Paper 12 Q35*

__Solution 955:__**Answer: B.**

Consider the parallel combination of
the right section of the network.

Combined resistance = [¼ + ¼ + ¼ + ¼]

^{-1}= 1 Ω
Now, consider the two 6.0 Ω resistors
in parallel at the lower part of the left section. The combined resistance of
these 2 is in series with the 3.0 Ω resistor.

Combined resistance = 3 + [1/6 + 1/6]

^{-1}= 3 + 3 = 6 Ω
The above calculated resistance is
in parallel with the 6.0 Ω resistor at the top.

Total combined resistance on the
left section = [1/6 + 1/6]

^{-1}= 3 Ω
Overall resistance between P and Q =
3 + 1 = 4 Ω

__Question 956: [Current of Electricity > Thermistor]__
A thermistor and another component
are connected to a constant voltage supply. A voltmeter is connected across one
of the components. Temperature of the thermistor is then reduced but no other
changes are made.

In which circuit will the voltmeter
reading increase?

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q36*

__Solution 956:__**Answer: D.**

The resistance of a thermistor
decreases as the temperature increases.

The resistance of an LDR decreases
as the light intensity increases.

For circuit A, the voltmeter reading
remains constant (and equal to the e.m.f. of the supply) since all the 3
components are connected in parallel to each other.

The temperature of the thermistor is
reduced and no other changes are made. So, its resistance increases. From Ohm’s
law, V = IR, an increase in the resistance of the thermistor would cause the
p.d. across it to increase.

Thus, for the voltmeter reading to
increase, the voltmeter should be connected in parallel to the thermistor (so
that it reads the p.d. across the thermistor).

__Question 957: [Pressure > Liquid]__
A submarine carries a pressure meter
so that the crew can work out how far they are below the surface of the sea. At
surface, the meter indicates a pressure of 100 kPa. The density of seawater is
1030 kg m

^{–3}.
What is the depth below the surface
when the meter reads 450 kPa?

A 34.6 m B 44.5 m C
340 m D 437 m

**Reference:**

*Past Exam Paper – November 2007 Paper 1 Q18*

__Solution 957:__**Answer: A.**

Pressure P in a liquid = hρg

Since the pressure at the surface is
not zero, we should account for the change in pressure.

ΔP = Δhρg

Change in depth Δh = ΔP / ρg = (450000
– 100000) / (1030 × 9.81) = 34.6m

(The depth is the surface is 0.)

Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

04/M/J/09 Q.11(b)(iii)

41/O/N/09 Q.6(a),(b)(i),Q.10

42/O/N/09 Q.7(b)(ii)

41/M/J/10 Q.6(a),Q.7(a)

51/M/J/10 Q.2(d)

41/M/J/11 Q.8(a)

For 42/O/N/09 Q.7(b)(ii), go to

Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2009-paper-42-worked.html