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Thursday, September 10, 2015

Physics 9702 Doubts | Help Page 197

  • 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.)





4 comments:

  1. Please consider answering ALL of the following questions:
    4/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)

    ReplyDelete
    Replies
    1. For 42/O/N/09 Q.7(b)(ii), go to
      http://physics-ref.blogspot.com/2014/09/9702-november-2009-paper-42-worked.html

      Delete
  2. Solution 956 keeps contradicting itself.

    "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.

    ReplyDelete
    Replies
    1. i have looked at the explanation again and have NOT seen anything contracting.

      we 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.

      Delete

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