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Date May 2018 Marks available 2 Reference code 18M.2.SL.TZ2.1
Level Standard level Paper Paper 2 Time zone 2
Command term Construct Question number 1 Adapted from N/A

Question

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.

State the direction of the resultant force on the ball.

[1]
a.i.

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

[2]
a.ii.

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

                                          F = m g tan θ

[3]
a.iii.

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

[4]
b.

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

[2]
c.

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]
d.

Markscheme

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

 

Do not accept towards the centre of the bowl

[1 mark]

a.i.

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

egM18/4/PHYSI/SP2/ENG/TZ2/01.a.ii

[2 marks]

a.ii.

ALTERNATIVE 1

F = N cos θ

mgN 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 O A = m g F

[3 marks]

a.iii.

m g tan θ m v 2 r

r = R cos θ

v g R cos 2 θ sin θ / g R cos θ tan θ / 9.81 × 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]

b.

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]

c.

speed before collision v = « 2 g R =» 12.5 «ms–1»

«from conservation of momentum» common speed after collision is 1 2  initial speed «vc 12.5 2 = 6.25 ms–1»

h = « v c 2 2 g = 6.25 2 2 × 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]

d.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Core » Topic 1: Measurements and uncertainties » 1.3 – Vectors and scalars
Core » Topic 1: Measurements and uncertainties
Core

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