Pre-U Pre-U 9794/3 2016 June — Question 7 5 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2016
SessionJune
Marks5
TopicFriction
TypeProjectile on rough surface
DifficultyModerate -0.8 This is a straightforward mechanics problem requiring only Newton's second law, friction force calculation (F = μR), and a single SUVAT equation. The setup is direct with no conceptual subtlety—students simply find the deceleration from friction and use kinematics to find distance. Easier than average due to its routine single-method approach.
Spec3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition
A stone that weighs 15 kg is propelled across the ice in an ice rink with an initial speed of \(4 \text{ m s}^{-1}\). The coefficient of friction between the stone and the ice is \(0.017\). How far does the stone slide before it comes to rest? [5]
AnswerMarks Guidance
\(15a = -F\)M1 N2 applied.
\(F = \mu N = 0.017 \times 15 \times 10 (= 2.55)\)B1 Limiting friction and \(N = mg\). c.a.o.
\(\therefore a = -0.017 \times 10 = -0.17\) (ms\(^{-2}\))M1 Use of appropriate 'suvat' equation.
\(0^2 = 4^2 - 2 \times 0.17 \times s\)M1 Use of appropriate 'suvat' equation.
\(\therefore s = \frac{16}{2 \times 0.17} = 47.058\ldots \approx 47.1\) (m)A1 Ft c's \(a\). 
$15a = -F$ | M1 | N2 applied.

$F = \mu N = 0.017 \times 15 \times 10 (= 2.55)$ | B1 | Limiting friction and $N = mg$. c.a.o.

$\therefore a = -0.017 \times 10 = -0.17$ (ms$^{-2}$) | M1 | Use of appropriate 'suvat' equation.

$0^2 = 4^2 - 2 \times 0.17 \times s$ | M1 | Use of appropriate 'suvat' equation.

$\therefore s = \frac{16}{2 \times 0.17} = 47.058\ldots \approx 47.1$ (m) | A1 | Ft c's $a$. | [5]

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A stone that weighs 15 kg is propelled across the ice in an ice rink with an initial speed of $4 \text{ m s}^{-1}$. The coefficient of friction between the stone and the ice is $0.017$. How far does the stone slide before it comes to rest? [5]

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q7 [5]}}
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