Question 18M.2.SL.TZ2.1

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

Do not accept towards the centre of the bowl

[1 mark]

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: 

[2 marks]

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

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

[3 marks]

$\frac{mg}{\tan \theta }$ = 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]

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]

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]