# Physics 9702 Doubts | Help Page 156

__Question 772: [Dynamics > Collisions]__
Diagram shows a particle P,
travelling at speed v, about to collide with a stationary particle Q of the
same mass. The collision is perfectly elastic.

Which statement describes motion of
P and of Q immediately after the collision?

A P rebounds with speed ½ v and Q
acquires speed ½ v.

B P rebounds with speed v and Q
remains stationary.

C P and Q both travel in the same
direction with speed ½ v.

D P comes to a standstill and Q
acquires speed v.

**Reference:**

*Past Exam Paper – June 2011 Paper 12 Q13*

__Solution 772:__**Answer: D.**

For a perfectly elastic collision,
both momentum and energy are conserved.

From the law of conservation of
momentum,

Sum of momentum before collision =
Sum of collision after collision

Before collision,

Sum of momentum, p = mv + 0 = mv

Total kinetic energy = ½ mv

^{2}
After collision,

Consider choice A,

Sum of momentum, p = m(0.5v) +
m(-0.5v) = 0. [A is incorrect]

Consider choice B,

Sum of momentum, p = - mv [B is incorrect]

Consider choice C,

Sum of momentum, p = m(0.5v) +
m(0.5v) = mv

Total kinetic energy = ½ m (0.5v)

^{2}+ ½ m (0.5v)^{2}= 0.25mv^{2}[C is incorrect]
Consider choice D,

Sum of momentum, p = m(0) + mv = mv

Total kinetic energy = ½ mv

^{2}
Both momentum and energy are
conserved.

__Question 773: [Waves]__
Diagram shows a transverse wave on a
rope. Wave is travelling from left to right.

At the instant shown, points P and Q
on the rope have zero displacement and maximum displacement respectively.

Which of the following describes the
direction of motion, if any, of points P and Q at this instant?

point
P point Q

A downwards
stationary

B stationary
downwards

C stationary
upwards

D upwards
stationary

**Reference:**

*Past Exam Paper – June 2004 Paper 1Q25*

__Solution 773:__**Answer: A.**

The direction of motion of the wave
is shown to be to the right in the diagram.

The easiest way to answer this
question is to draw, on the diagram itself, the new wave a little bit later in
time. This is the same wave as shown, but a little displaced to the right.

This shows that P must go down and
that Q barely moves – even if some stated that Q would go down as well, they
would not have been able to give that as an answer because it is not there.

Otherwise, it is known that a point
at the maximum displacement on a wave is momentarily stationary. At the
equilibrium position, the point moves with maximum velocity.

__Question 774: [Dynamics]__
A resultant force causes body to
accelerate.

What is equal to the resultant
force?

A the acceleration of the body per
unit mass

B the change in kinetic energy of
the body per unit time

C the change in momentum of the body
per unit time

D the change in velocity of the body
per unit time

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q8*

__Solution 774:__**Answer: C.**

A resultant force causes body to
accelerate.

The resultant force is equal to the
change in momentum of the body per unit time (rate of change of momentum) – as stated
by Newton’s 2

^{nd}law.
Newton’s 2

^{nd}law: F = Î”p / t = m(Î”v) / t = ma
In symbol, ‘the acceleration of the
body per unit mass’ means (= m / a), which is wrong.

The rate of change of energy (change
in energy per unit time) gives the power while the change in velocity of the body
per unit time gives the acceleration.

__Question 775: [Dynamics > Equilibrium]__
Diagram shows an experiment to
measure force exerted on a ball by a horizontal air flow.

Ball is suspended by a light string
and weighs 0.15 N.

Deflection of the string from
vertical is 30°.

What is the force on the ball from
the air flow?

A 0.075 N B 0.087 N C
0.26 N D 0.30 N

**Reference:**

*Past Exam Paper – November 2012 Paper 11 Q15*

__Solution 775:__**Answer: B.**

Consider the ball suspended by the
string when there is no air flow. The weight acts vertically downwards. From
Newton’s 3

^{rd}law, there is a vertical upward force in the string which is equal in magnitude to the weight. We call this force the ‘**tension**’.
Now, consider the air flow. This
causes a force F, horizontally to the left on the ball. The weight W is always
downwards and the tension T is always in the string, which is now 30° to the vertical.

At the maximum displacement, the
ball is momentarily at rest. At this position, the force acting on the ball are
balanced.

W = T cos30°

F = T sin30°

From the 1

^{st}equation, T = W / cos30°
So, Force F = (W / cos30°) sin30° = W (sin30°/ cos30°) = W tan30°

Force F = 0.15 × tan30° = 0.087 N

__Question 776: [Current of Electricity]__
In circuit below, a voltmeter of
resistance R

_{V}and an ammeter of resistance R_{A}are used to measure the resistance R of the fixed resistor.
Which condition is necessary for an
accurate value to be obtained for R?

A R is much smaller than R

_{V}.
B R is much smaller than R

_{A}.
C R is much greater than R

_{V}.
D R is much greater than R

_{A}.**Reference:**

*Past Exam Paper – June 2014 Paper 11 Q33*

__Solution 776:__**Answer: A.**

To obtain the resistance of resistor
R from the set-up shown, the potential difference and current flowing through
the resistor are measured.

Ohm’s law: V = IR

Resistance = V / I

(Voltmeter and ammeter cannot give
resistance on their own. Resistance should be calculated.)

Ideally, a voltmeter should have an
infinite resistance and is connected in parallel to the component being
investigated. This is to avoid a large current to flow through it. Current is
highest when the resistance is smallest.

Ideally, an ammeter should have negligible
resistance and is connected in series to the component being investigated so
that it does not affect the value of the current flowing.

This, R

_{V}should be very large (larger than R) and R_{A}should be very small (smaller than R). Hence, the necessary condition for an accurate value to be obtained for R is than R is much smaller than R_{V}.
The voltmeter must have much greater
resistance than R, otherwise there will be significant current in voltmeter and
the value obtained from V/I will not be an accurate value for R.

However: If the statement "R

_{V}should be very large (larger than R) and R_{A}should be very small (smaller than R)" is true, then surely both A and D are correct options?
The idea is
that If Rv is too small then the current flows through V and the ammeter
reading is not only measuring the current through R. Therefore the equation
R=V/I cannot do this as although it is the correct V it is not the I through
resistor R - it is instead the current through V and R.

However, if R

_{A}is large this will have no effect as both I and V will change. Whatever the value of R_{A}, the voltmeter is only across R and will only measure the V across R. And as the ammeter is in series with R, the ammeter will always measure the current through R. Using R = V/I will give the resistance.
If R

_{A}is large both V and I in R=V/I decrease proportionally and R wouldn't change.
For 776:

ReplyDeleteIf the statement "RV should be very large (larger than R) and RA should be very small (smaller than R)" is true, then surely both A and D are correct options?

For 776:

ReplyDeleteThe idea you have not explained is that If Rv is too small then the current flows through V and the ammeter reading is not only measuring the current through R. Therefore the equation R=V/I cannot do this as although it is the correct V it is not the I through resistor R - it is instead the current through V and R.

However, if Ra is large this will have no effect as both I and V will change. Whatever the value of Ra , the voltmeter is only across R and will only measure the V across R. And as the ammeter is in series with R, the ammeter will always measure the current through R. Using R = V/I will give the resistance. If Ra is large both V and I in R=V/I decrease proportionally and R wouldn't change.

Thanks a lot. I did not distinguish why A is correct here.

DeleteI have added you explanation above, hope you don't mind.

Suggestions, corrections, ideas, .... from readers are very much appreciated.

Hmmmm nice nice

ReplyDelete