Tuesday, March 31, 2015

Physics 9702 Doubts | Help Page 100

  • Physics 9702 Doubts | Help Page 100

Question 513: [Forces > Momentum]
Which row correctly states whether momentum and kinetic energy are conserved in inelastic collision in which there are no external forces?
momentum                  kinetic energy
A         conserved                    conserved
B         conserved                    not conserved
C         not conserved              conserved
D         not conserved              not conserved

Reference: Past Exam Paper – June 2012 Paper 12 Q11

Solution 513:
Answer: B.
For any system in which there are no external forces, the momentum is always conserved. [C and D are incorrect]

If the system is elastic, kinetic energy is conserved.
If the system is inelastic, kinetic energy is not conserved. The energy may be lost as heat, sound and other forms of energy. [A is incorrect]

Question 514: [Kinematics]
Graph shows how velocity v varies with time t for a bungee jumper.

At which point is bungee jumper momentarily at rest and at which point does she have zero acceleration?
jumper at rest              jumper with zero acceleration
A                     Q                                             P
B                     Q                                             R
C                     R                                             Q
D                     R                                             R

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q7

Solution 514:
Answer: C.
The graph shown is a velocity-time graph.
When the jumper is at rest, her velocity is zero. Zero velocity is indicated when the curve touches the time axis. This is at point R. [A and B are incorrect]

The acceleration of the jumper is given by the gradient of the velocity-time graph. The gradient at a point in the curve is given by the gradient of the tangent at that point. For the gradient to be zero, the tangent should be a horizontal line. This occurs at point Q. [D is incorrect]

Question 515: [Current of Electricity]
Battery is marked 9.0 V.
What does this mean?
A Each coulomb of charge from the battery supplies 9.0 J of electrical energy to the whole circuit.
B The battery supplies 9.0 J to an external circuit for each coulomb of charge.
C The potential difference across any component connected to the battery will be 9.0 V.
D There will always be 9.0 V across the battery terminals.

Reference: Past Exam Paper – June 2011 Paper 11 Q32 & November 2014 Paper 11 & 12 Q31

Solution 515:
Answer: A.
The e.m.f. of a battery does not tell anything about its internal resistance. If the battery has some internal resistance, the p.d. across the battery terminals will not be exactly 9.0V, but less. [D is incorrect] This would also means that the amount of energy that the battery supplies to an external circuit for each coulomb of charge is NOT 9.0J. [B is incorrect]

Choice C is obviously incorrect. It should have been as follows. IF there is no internal resistance in the battery (and the wires have negligible resistance), the SUM of p.d. across the componentS connected to the battery will be 9.0V. [C is incorrect]

A battery being marked 9.0V means that each coulomb of charge from the battery supplies 9.0 J of electrical energy to the whole circuit (this includes the internal resistance of the battery).

Question 516: [Measurements > Uncertainties]
Micrometer is used to measure diameters of two cylinders.
diameter of first cylinder = 12.78 ± 0.02 mm
diameter of second cylinder = 16.24 ± 0.03 mm
Difference in the diameters is calculated.
What is uncertainty in this difference?
A ± 0.01 mm               B ± 0.02 mm               C ± 0.03 mm               D ± 0.05 mm

Reference: Past Exam Paper – November 2011 Paper 12 Q4

Solution 516:
Answer: D.
Let diameter of first cylinder = d1 = 12.78 ± 0.02 mm
Let diameter of second cylinder = d2 = 16.24 ± 0.03 mm

Difference in diameters = d2 – d1
Uncertainty in (difference in diameters) = Sum of uncertainties in d1 and d2
Uncertainty in (difference in diameters) = 0.02 + 0.03 = ±0.05mm

Question 517: [Kinematics + Momentum]
Small ball is thrown horizontally with speed of 4.0ms-1. It falls through vertical height of 1.96m before bouncing off a horizontal plate, as illustrated in Fig. 

Air resistance is negligible.
(a) For ball, as it hits the horizontal plate,
(i) state magnitude of horizontal component of its velocity
(ii) show that vertical component of velocity is 6.2ms-1

(b) Components of velocity in (a) are both vectors.
Complete Fig to draw a vector diagram, to scale, to determine velocity of ball as it hits horizontal plate

(c) After bouncing on plate, the ball rises to vertical height of 0.98m.
(i) Calculate vertical component of velocity of ball as it leaves the plate
(ii) Ball of mass 34g is in contact with plate for time 0.12s.
Use answer in (c)(i) and data in (a)(ii) to calculate, for the ball as it bounces on the plate,
1. change in momentum
2. magnitude of average force exerted by plate on ball due to this momentum change

Reference: Past Exam Paper – November 2009 Paper 22 Q3

Solution 517:
(i) Horizontal component of velocity = 4.0ms-1

(v2 = u2 + 2as. Here u = 0, a = g = 9.8ms-2 and s = h = 1.96m)
v2 = 2gh = 2(9.8)(1.96)
Vertical component of velocity, v = 6.2ms-1

(b) Diagram must have a correct basic shape with the correct directions for the vectors.
Speed = (7.4 ± 0.2) ms-1 at (33 ± 2)o to the vertical

(v2 = u2 + 2as. At (max) vertical height, v = 0. So, 0 = u2 + 2(-9.8)(0.98))
EITHER v2 = 2(9.8)(0.98)      OR v = 6.2 / √2
Vertical component of velocity = 4.4ms-1

Momentum, p = mv
Change in momentum, Δp = 0.034 (6.2 – (-4.4)) = 0.36kgms-1

2. Force = Δp / Δt (however expressed) = 0.36 / 012 = 3.0N


  1. how would I know what number of significant figures is correct to write?

    1. Normally, it's 3 significant figures, unless stated otherwise in the question

    2. But in Question 517c(i) you have given the answer in 2 significant figures.

    3. Normally, it would be 3sf. But in this question, you may have noticed that all values are given to 2 significant figures. So, here you can give the answer to 2sf.

      Even the 4m/s is given as 4.0 m/s. If it was given as 4.00 m/s, then you would have needed to provide 3 sf

    4. Ok, but what if in the question the data has a value which has two sig figs and another value which has 3 sig figs? Then how many sig figs should my answer have?

    5. I personally don't think this would happen. If it did, then it's a mistake from the question paper and I think they would accept either of them. I THINK, but I'm not one of the examiners, so I can't say for sure.

      If this happened, give your answer to 3s.f.

      Sometimes, you may also look at the number of decimal place from the question and give your answer to the same number of decimal places.

      Note that since the beginning, I have been talking about final answers, not intermediate calculations. For intermediate calculations, keep your working to 1 more sf or 1 more decimal place than the final answer.

    6. Ok thanks but can you check November 2010 paper 21 question 2a(iii), why is the answer in 3 sig figs and not two since in the data, that is, 8.2, has 2 sig figs? I hope you don't mind :/

    7. First, Nov 2010 Paper 21 Q2 is explained as solution 54 at

      Well, you should look at the question carefully. Some values are given to 2sf and some to 3sf. But all the values are given to 1 decimal place.

      So, the final answer is also given to 1decimal place. At the same time it is to 3sf. But, the important thing is that the answer is given to 1 decimal place.

  2. I don't understand. For the last question, why is the change in momentum p(v'-v), the final velocity is 6.2? Shouldn't the final be 4.4 and initial velocity is 6.2 m/s?

    1. Here, we are considering the magnitude of the change in momentum. Taking m (final velocity – initial velocity) = m (-4.4 – 6.6) would still give the same answer, but with a negative sign. We need to be able to apply the formulae in the different cases, not just use them directly.

  3. For question no.517 (c).ii, Don't we have to consider the velocity "v" in the change in momentum formulae, i.e. p(v'-v) as the resultant landing velocity before and after the collision ? And why have you taken only the vertical component of the velocity "v" for the before and after collision with the plate ?

    1. There is a change in momentum, and thus a 'change' in speed. We cannot consider a single value of speed. We need 2 values to show how the speed and hence momentum has changed.

      Since the ball is thrown horizontally and air resistance is negligible, the horizontal component of the speed will remain constant throughout the motion. Thus, only the vertical component needs to be considered.


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