# Physics 9702 Doubts | Help Page 3

__Question 16: [Force > Momentum]__
(a) State relation between force and
momentum.

(b) Rigid bar of mass 450 g is held horizontally
by two supports A and B, as shown.

Support A is 45 cm from centre of
gravity C of bar and support B is 25 cm from C.

Ball of mass 140 g falls vertically
onto bar such that it hits bar at distance of 50 cm from C. Variation with time
t of velocity v of ball before, during and after hitting bar is shown.

For time that ball is in contact
with bar, use Fig

(i) to determine change in momentum
of ball

(ii) to show that force exerted by
ball on bar is 33N

(c) For time that ball is in contact
with bar, use data from (1

^{st}) Fig and (b)(ii) to calculate force exerted on bar by
(i) Support A

(ii) Support B

**Reference:**

*Past Exam Paper – November 2010 Paper 21 Q3*

__Solution 16:__**(a)**

Force is the rate of change of momentum

(F = Δp / t)

**(b)**

(i) Change in momentum of ball:

{Let’s try interpreting
the graph. The ball falls from rest (velocity = 0) and hits the bar at time t =
0.56s with a velocity of 5.5ms

^{-1}. Collision starts as soon as the speed drops from 5.5ms^{-1}and ends at time t = 0.6s where the ball reaches a velocity of 4.4ms^{-1}in the opposite direction. So, the time of contact of the ball with the bar is 0.6 – 0.56 = 0.04s. Then, the ball moves upwards until it reaches a maximum height (velocity = 0) and then falls again. But the last part is not necessary here.}
{For collision, initial
velocity = 5.5ms

^{-1}and final velocity = - 4.4ms^{-1}. So, change in velocity, Δv = 5.5 + 4.4 }
Δp (= mΔv) = (140x10

^{-3}) x (5.5 + 4.4) = 1.33kgms^{-1}
(ii) Show that force exerted by ball
on bar is 33N:

{As stated in part (a),
force is the rate of change of momentum. This change in momentum occurs in a
time (time of contact) of 0.04s}

Force exerted by ball (= Δp/t) =
1.33 / 0.04 = 33.3N

**(c)**

(i)

Support A:

Taking moments about support B,

33(75) + (0.45 x g)(25) = F

_{A}(20)
Force exerted by support A, F

_{A}= 129N
(ii)

Support B:

Force exerted by support B, F

_{B}= 33 + 129 + 0.45g = 166N

__Question 17: [Dynamics > Laws of motion]__
Show that Newton's 2

^{nd}law contains the 3^{rd}law.

__Solution 17:__
First, Newton’s laws of motion are
as follows:

**Newton’s first law**of motion states that a body continues at rest or constant velocity unless acted on by a resultant (external) force.**Newton’s second law**of motion states that the (resultant) force on a body is equal to the rate of change of momentum. {From this law (F = Δp / t), it can be shown that, since Δp = mΔv, the law may also be written as F = m(Δv/ t) = ma.}**Newton’s third law**of motion states that the force on a body A is equal in magnitude to the force on a body B (from body A). The forces are in opposite directions and are of the same kind.

{Note that these are the
correct definitions accepted by Cambridge}

Consider 2 bodies, A and B,
interacting with each other. Body A exerts a force F

_{A}on body B and body B would exert a force F_{B}on body B.
From Newton’s 2

^{nd}law, F_{A}= dp_{A}/dt and F_{B}= dp_{B}/dt where p_{A}and p_{B}are the momenta of body A and B respectively and d / dt is the rate of change [differentiation with respect to time – this is actually a more appropriate way to write F = Δp / t even if both are accepted at A-Level]
So, F

_{A}+ F_{B}= [dp_{A}/dt] + [dp_{B}/dt] = d(p_{A}+ p_{B}) / dt
Assuming no external forces are
acting, the momentum must be conserved
[this is the law of conservation of momentum]

So, the sum of p

_{A}+ p_{B}is always constant. [p_{A}+ p_{B}= constant]
The differentiation of a
constant is zero {dp / dt is the rate of change of momentum with time. Since
momentum is constant, “it does not change”. So, the rate of change of momentum
[which does not change] is zero}

F

_{A}+ F_{B}= d(p_{A}+ p_{B}) / dt = d(constant) / dt = 0
Thus, F

_{A}= - F_{B}
This is Newton’s 3

^{rd}law. F_{A}and F_{B}are equal in magnitude and opposite in direction.
Similarly, Newton’s 1

^{st}law may also be obtained from the 2^{nd}law. The equation F = ma can be obtained from the 2^{nd}law (as shown above). F is the force acting on a body.
If there is no force, F = 0.
Therefore, acceleration a is also equal to zero. When there is no acceleration,
the velocity is unchanged. So, the body continues at constant velocity, and if
its velocity was zero, it would remain zero in the absence of a force.

__Question 18: [Radioactivity]__
Isotope Manganese-56 decays and
undergoes β-particle emission to form stable isotope Iron-56. Half-life for
decay is 2.6 hours. Initially, at time t = 0, sample of Manganese-56 has mass
of 1.4 μg and there is no Iron-56.

**(c)**

Determine time at which

**ratio**of mass of iron-56**to**mass of Manganese-56 = 9.0:**Reference:**

*Past Exam Paper – June 2005 Paper 4 Q7(c)*

__Solution 18:__
{Since the ratio = 9 = 9 /
1, the mass of iron-56 is reduced to 9 while that of Manganese-56 is reduced to
1. So, the total relative mass = 9 + 1 = 10}

1/10 of the original mass of
Manganese remains {at that time}

{The ratio N / N

_{o}is 1 / 10 = 0.10. The decay constant λ = ln2 / T_{½}= ln 2 / 2.6. The ratio N / N_{o}= exp(-λt) = 1 / 10 where t is the time at which the ratio is 1 /10}
0.10 = exp(-[ln2 / 2.6] t)

Time, t = 8.63 hours

__Question 19: [SI units]__**(a)**Two of SI base quantities are mass and time. State 3 other SI base quantities.

**(b)**Sphere of radius r is moving at speed v through air of density ρ. Resistive force F acting on sphere is given by expression F = Br

^{2}ρv

^{k}where B and k are constants without units.

(i) State SI base units of F, ρ and v:

(ii) Use base units to determine
value of k:

**Reference:**

*Past Exam Paper – November 2010 Paper 21 Q1*

__Solution 19:__**(a)**

Choose
any 3 from:

Length,
Current, Temperature, Amount of substance, (luminous intensity)

**(b)**

(i)

SI base units of force F {= ma} = kg ms

^{-2}
SI base units of density ρ {= mass / volume} = kg m

^{-3}
SI base units of velocity v {= distance / time} = m s

^{-1}
(ii)

__Some working__e.g. kgms

^{-2}= m

^{2}kgm

^{-3}(ms

^{-1})

^{k}

OR

{Make v

^{k}the subject of formula}
v

^{k}= F / Br^{2}ρ
{Now wire in terms of SI
base units and simplify such that the units of velocity is in brackets on both
sides and any other value is outside the bracket as a power}

[ms

^{-1}]^{k}= [kg ms^{-2}] / [m^{2}] [kg m^{-3}] = [kg ms^{-2}] [m^{-2}] [kg^{-1}m^{3}] = m^{2}s^{-2}= [ms^{-1}]^{2}
Hence, the value of k =2

__Question 20: [Force > Air resistance]__**(a)**Derive SI base unit of force.

**(b)**Spherical ball of radius r experiences resistive force F due to air as it moves through air at speed v. Resistive force F is given by expression F = crv, where c is a constant. Derive SI base unit of constant c.

**(c)**Ball is dropped from rest through height of 4.5 m.

(i) Assuming air resistance to be
negligible, calculate final speed of ball.

(ii) Ball has mass 15 g and radius
1.2 cm. Numerical value of constant c in equation in (b) is equal to 3.2 × 10

^{–4}when measured using SI system of units. Show quantitatively whether assumption made in (i) is justified.**Reference:**

*Past Exam Paper – June 2006 Paper 2 Q1*

__Solution 20:__**(a)**

{Force = ma}

SI base unit of Force = kg ms

^{-2}**(b)**

{c = F / rv}

SI base unit of c = [kg ms

^{-2}] / [m][ms^{-1}] = [kg ms^{-2}] [m^{-1}] [m^{-1}s^{1}] = kg m^{-1}s^{-1}
SI base unit of c = kg m

^{-1}s^{-1 }**(c)**

(i)

{Initial velocity of ball
= 0 (dropped from rest)}

v

^{2}= u^{2}+ 2as
v

^{2}= 0 + 2(9.8)(4.5)
Final velocity, v = 9.4ms

^{-1}
(ii)

{The assumption is (i) is
that air resistance is negligible}

EITHER

{In this method, we try to
calculate the resistive force F, given by the expression F = crv, and compare
it with the force on the ball (the weight). If the resistive force F is much
less than the weight, then it can be neglected and the assumption is correct.}

Resistive force, F {= (3.2x10

^{-4}) (1.2x10^{-2}) (9.4)} = 3.6x10^{-5}N
Weight of sphere ball {= mg = (1.5x10

^{-3}) (9.8)} = 0.15N
Since 3.6x10

^{-5}N << 0.15N, the assumption is justified
OR

{In this method, we try to
calculate the terminal velocity of the ball and compare it with the final
velocity reached by the ball. Terminal velocity is reached when air resistance
equals the weight.

(As the ball starts
falling, air resistance keeps on increasing until terminal velocity is reached.
So, at the time before terminal velocity is reached, air resistance is less
than the weight).

Therefore, a comparison of
the weight (which is constant at all time) and the air resistance at a
particular time t can be obtained by considering the terminal velocity and the
velocity reached by the ball at that time t. So, at any particular time that
the instantaneous velocity is much less than the terminal velocity, then at the
instant, air resistance is much less than the weight.}

(Weight) mg = (Resistive force) crv

_{T}
Terminal velocity, v

_{T}= 3.8x10^{4}ms^{-1}
Since the final velocity reached by
the ball, 9.4 << 3.8x104ms-1, the assumption (that air resistance is
negligible) is justified.

__Question 21: [Measurements]__
What does ‘answers to the nearest mm’
mean on a metre rule?

__Solution 21:__
When taking measurements using any
measuring instrument, the reading should be recorded to the smallest number of
decimal place that can be measured by the instrument (all instruments have a
level of uncertainty associated with it – they cannot measure smaller than these
uncertainties). Obviously this will vary from the instruments used.

For example, the readings from a
metre rule can (and should) be measured to the nearest mm. Consider the metre
rule below:

The labels 1, 2, 3, … represent the
1cm, 2cm, 3cm, … It can also be seen that the distance between any 2 successive
labels contains 10 (smallest) divisions. This means that 10 divisions
correspond to 1cm. So, 1 division represents 1 / 10 = 0.1cm = 1mm.

That is, the smallest reading that
can be read from the metre rule is 1mm. This is the nearest mm. {Note that many would say that readings can be taken to half
the division (even on this metre rule) by inferring whether it is midway
between the smallest divisions, but this is a mistake.}

Now consider the rule below.

The labels also represent readings
to cm but here, each cm is division by 5 divisions. So, 1 division would
represent 1 / 5 = 0.2mm. ‘The nearest mm’ on this ruler is 0.2mm.

This method I just explained is very
basic and important as it allows you to determine the smallest division on any measuring
instrument. There is no need to memorize the smallest division of all
instruments.

(Note that these pictures are found on the internet and were not drawn by me. All credits go to their owner)

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