A car is travelling at constant velocity. Its brakes are then applied, causing uniform deceleration.

Question 31

A car is travelling at constant velocity. Its brakes are then applied, causing uniform deceleration.

Which graph shows the variation with distance s of the velocity v of the car?

Reference: Past Exam Paper – March 2016 Paper 12 Q7

Solution:

Answer: A.

The motion of the car can be broken down into two: one with constant speed and another with constant deceleration.

We need to identify the correct velocity-distance graph. So we need to find equations that relate the velocity and distance travelled.

Consider the car moving at constant velocity.

Speed / Velocity = distance / time

v = s / t

Since the velocity is constant, the line is obviously horizontal on a velocity-distance graph.

Now consider the car moving with constant deceleration.

Equation that relates the velocity to the distance:

v² = u² + 2as

Deceleration is negative acceleration, so the value of ‘a’ would be negative.

v² = u² – 2as

v = √(u² – 2as)

The line should satisfy the above relationship.

Plotting a graph of v against s gives a curve similar to that of option A.

For example, put u = 100 and a = 10 and plot points for x = 0 to x = 10. Joining the points would results in a curve as in option A. It has a negative gradient (due to the deceleration) with an increasing magnitude.

This can also be understood as such.

The gradient of a velocity-distance graph is

Gradient = Δv / Δs

We are told that the deceleration is uniform, i.e. the velocity changes by equal amount. So Δv is the same at any distance.

But what happens to the change in distance travelled (Δs) as the velocity decreases?

Consider the equation

s = ut + ½ at² 

Put the deceleration to be 10 m s^-2. That is, a = - 10 m s^-2

When initial speed u = 100 m s^-1, the distance travelled in 1 s is

s = 100(1) + (½ × -10 × 1²) = 95 m

When initial speed u = 50 m s^-1, the distance travelled in 1 s is

s = 50(1) + (½ × -10 × 1²) = 45 m

So, as the velocity decreases, the change in distance travelled each second becomes smaller.

Gradient = Δv / Δs

Since Δv is constant and Δs is decreasing with velocity, the value of gradient increases as the velocity decreases by equal amount.

Graph B shows the velocity-time graph of the car.

Graph C tells us that the distance travelled increases as velocity decreases.

Graph D indicates the car comes immediately to rest after travelling at constant speed.

The diagrams show four pairs of waves. In each case the displacement y measured at a fixed point is plotted against time t.

Question 26

The diagrams show four pairs of waves. In each case the displacement y measured at a fixed point is plotted against time t.

Which pair of waves is not coherent?

Reference: Past Exam Paper – November 2017 Paper 13 Q30

Solution:

Answer: C.

Waves are said to be coherent if they have a constant phase difference. That is, the phase difference does not change with time.

Note that a constant phase difference does not mean that the phase difference is zero. There may be a phase difference but this value needs to be constant – it should not be changing.

However 2 waves have a constant phase differences implies that the periods / frequencies need to be the same. If the 2 waves have different periods, the phase difference would keep on changing with time.

So for a pair of waves to be coherent they should have a constant phase difference. This implies that they should have the same period.

Option C is incorrect as the waves have different periods.

A graph of nucleon number A against proton number Z is shown in Fig. 7.1.

Question 15

A graph of nucleon number A against proton number Z is shown in Fig. 7.1.

Fig. 7.1

The graph shows a cross (labelled P) that represents a nucleus P.

Nucleus P decays by emitting an α particle to form a nucleus Q.

Nucleus Q then decays by emitting a β^– particle to form a nucleus R.

(a) On Fig. 7.1, use a cross to represent

(i) nucleus Q (label this cross Q), [1]

(ii) nucleus R (label this cross R). [1]

(b) State the name of the class (group) of particles that includes the β^– particle. [1]

(c) The quark composition of one nucleon in Q is changed during the emission of the β^– particle.

Describe this change to the quark composition. [1]

[Total: 4]

Reference: Past Exam Paper – November 2018 Paper 23 Q7

Solution:

(a)

(i) Q plotted at (82, 210)

{²¹⁴₈₄P             - - - >   ^A_ZQ      +          ⁴₂α

Proton number and nucleon number should be conserved.

214 = A + 4

A = 214 – 4 = 210

84 = Z + 2

Z = 84 – 2 = 82

The emission of an α-particle from a nucleus will cause the proton number to decrease by two and the nucleon number to decrease by four.}

(ii) R plotted at (83, 210)

{²¹⁰₈₂Q - - ->                ^A_ZR      +          ⁰_-1β

210 = A + 0

A = 210

82 = Z – 1

Z = 82 + 1 = 83}

(b) lepton(s)

{The β^– particle is a lepton. It is not made up of quarks, so it is not a hadron.}

(c)

up down down changes to up up down

or udd → uud

or down changes to up

or d → u         

{The nucleus Q is ²¹⁰₈₂Q.

Number of proton in Q = 82

Number of neutron in Q = 210 – 82 = 128

Nucleus R is ²¹⁰₈₃R.

Number of proton in R = 83

Number of neutron in Q = 210 – 83 = 127

So during the decay of Q into R, a neutron is changed into a proton.

Quark composition of neutron = udd

Quark composition of proton = uud

A d-quark has changed into a u-quark.}