DP Physics (last assessment 2024)

Question 18M.2.HL.TZ2.1

Date	May 2018	Marks available	[Maximum mark: 21]	Reference code	18M.2.HL.TZ2.1
Level	HL	Paper	2	Time zone	TZ2
Command term	Construct, Determine, Draw, Outline, Show that, State	Question number	1	Adapted from	N/A

[Maximum mark: 21]

18M.2.HL.TZ2.1

A small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless hemispherical bowl.

M18/4/PHYSI/SP2/ENG/TZ2/01.a

The normal reaction force N makes an angle θ to the horizontal.

(a.i)

State the direction of the resultant force on the ball.

[1]

Markscheme

towards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

(a.ii)

On the diagram, construct an arrow of the correct length to represent the weight of the ball.

[2]

Markscheme

downward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required

eg: M18/4/PHYSI/SP2/ENG/TZ2/01.a.ii

[2 marks]

(a.iii)

Show that the magnitude of the net force F on the ball is given by the following equation.

$F = \frac{{mg}}{{\tan \theta }}$

[3]

Markscheme

ALTERNATIVE 1

F = N cos θ

mg = N sin θ

dividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

M18/4/PHYSI/SP2/ENG/TZ2/01.a.ii

tan θ = $\frac{{\text{O}}}{A} = \frac{{mg}}{F}$

[3 marks]

(b)

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

[4]

Markscheme

$\frac{{mg}}{{\tan \theta }}$ = m $\frac{{{v^2}}}{r}$

r = R cos θ

v = $\sqrt {\frac{{gR{{\cos }^2}\theta }}{{\sin \theta }}} /\sqrt {\frac{{gR\cos \theta }}{{\tan \theta }}} /\sqrt {\frac{{9.81 \times 8.0\cos 22}}{{\tan 22}}}$

v = 13.4/13 «ms ^–¹»

Award [4] for a bald correct answer

Award [3] for an answer of 13.9/14 «ms ^–¹». MP2 omitted

[4 marks]

(c)

Outline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.

[2]

Markscheme

there is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

The ball is now displaced through a small distance x from the bottom of the bowl and is then released from rest.

M18/4/PHYSI/HP2/ENG/TZ2/01.d

The magnitude of the force on the ball towards the equilibrium position is given by

$\frac{{mgx}}{R}$

where R is the radius of the bowl.

(d.i)

Outline why the ball will perform simple harmonic oscillations about the equilibrium position.

[1]

Markscheme

the «restoring» force/acceleration is proportional to displacement

Direction is not required

[1 mark]

(d.ii)

Show that the period of oscillation of the ball is about 6 s.

[2]

Markscheme

ω = « $\sqrt {\frac{g}{R}}$ » = $\sqrt {\frac{{9.81}}{{8.0}}}$ «= 1.107 s^–1»

T = « $\frac{{2\pi }}{\omega }$ = $\frac{{2\pi }}{{1.107}}$ =» 5.7 «s»

Allow use of or g = 9.8 or 10

Award [0] for a substitution into T = 2π $\sqrt {\frac{I}{g}}$

[2 marks]

(d.iii)

The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time t of the velocity v of the ball during one period.

[3]

Markscheme

sine graph

correct amplitude «0.13 m s^–1»

correct period and only 1 period shown

Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.

Amplitude should be correct to ± $\frac{1}{2}$ square for MP2

eg: v /m s^–1 M18/4/PHYSI/HP2/ENG/TZ2/01.d.iii

[3 marks]

(e)

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.

M18/4/PHYSI/SP2/ENG/TZ2/01.d

The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.

[3]

Markscheme

speed before collision v = « $\sqrt {2gR}$ =» 12.5 «ms^–1»

«from conservation of momentum» common speed after collision is $\frac{1}{2}$ initial speed «v_c = $\frac{{12.5}}{2}$ = 6.25 ms^–1»

h = « $\frac{{{v_c}^2}}{{2g}} = \frac{{{{6.25}^2}}}{{2 \times 9.81}}$ » 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

Question 18M.2.HL.TZ2.1

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