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 = m1v1
+ m2v2 = (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 mv2
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 3rd
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 3rd
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 2nd 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
Solution 956 keeps contradicting itself.
ReplyDelete"The resistance of a thermistor decreases as the temperature increases."
"The temperature of the thermistor is reduced and no other changes are made. So, its resistance increases."
I know it wasn't specified what type of thermistor was being used, so I'm assuming we could've gone with EITHER explanation rather than both?
Also it feels like the question wasn't answered. Could you please try explaining again, ignoring the other components and sticking to one type of thermistor?
It would be very appreciated, thank you.
i have looked at the explanation again and have NOT seen anything contracting.
Deletewe are not discussing on the 2 types of thermistor. Anyway, the only thermistor in the syllabus is the one reffered to here.
If temperature increases, the resistance of the thermistor decrease. That is what we know. But this is not happening here.
In this question, we are informed that the temperature is reduced. SO, the opposite occur. The resistance of the thermistor increases.
From Ohm's law (V = IR), the p.d. across the thermistor also increases. So, we only need to connect the voltmeter across the thermistor and its reading would increase.