CAIE M1 2020 November — Question 7 12 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2020
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeEnergy methods on slope
DifficultyStandard +0.3 This is a standard two-part mechanics problem involving motion on an inclined plane with smooth and rough sections. Part (a) requires applying energy methods or equations of motion across two sections (smooth then rough), calculating work done against friction. Part (b) reverses the problem to find μ. While it requires careful bookkeeping of energy/forces across sections, the techniques are routine M1 content with no novel insight needed. The multi-step nature and 12 total marks place it slightly above average difficulty.
Spec3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings
\includegraphics{figure_7} Three points \(A\), \(B\) and \(C\) lie on a line of greatest slope of a plane inclined at an angle of \(30°\) to the horizontal, with \(AB = 1\) m and \(BC = 1\) m, as shown in the diagram. A particle of mass 0.2 kg is released from rest at \(A\) and slides down the plane. The part of the plane from \(A\) to \(B\) is smooth. The part of the plane from \(B\) to \(C\) is rough, with coefficient of friction \(\mu\) between the plane and the particle.
  1. Given that \(\mu = \frac{1}{2}\sqrt{3}\), find the speed of the particle at \(C\). [8]
  2. Given instead that the particle comes to rest at \(C\), find the exact value of \(\mu\). [4]
Question 7:
AnswerMarks
7(a)1
0.2×10×0.5= ×0.2×v2
AnswerMarks Guidance
2 BM1 Attempt PE or KE for motion from A to B
M1Attempt PE loss = KE gain from A to B
v2 =10
AnswerMarks
BA1
Alternative method for the first 3 marks
AnswerMarks Guidance
0.2 × 10 × sin 30 = 0.2a, a = 5(M1) Attempt to find acceleration a for motion from A to B
v2 =02 +2×5×1
AnswerMarks Guidance
B(M1) Use v2 = u2 + 2as in attempt to find speed at B
v2 =10
AnswerMarks Guidance
B(A1)
QuestionAnswer Marks
7(a)THEN, either this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3B1
3 3
F = ×0.2× ×10=1.5
AnswerMarks Guidance
2 2M1 For using F = µR where R must be a component of 0.2g
PE loss = 0.2 × 10 × 0.5 = 1
AnswerMarks Guidance
WD against F = 1.5 × 1M1 Attempt to find either PE loss or WD against F from B to C
1 1
0.2 ×10+0.2×10×0.5=1.5×1+ 0.2v2
AnswerMarks Guidance
2 2 CM1 Apply work-energy equation for motion from B to C as
KE at B + PE at B = WD against F + KE at C with v ≠ 0
B
v = 5 = 2.24 ms–1
AnswerMarks
cA1
OR, this method for the next 5 marks
AnswerMarks
R = 0.2 × 10 × cos 30 = √3(B1)
3 3
F = ×0.2× ×10=1.5
AnswerMarks Guidance
2 2(M1) For using F = µR where R must be a component of 0.2g
0.2 × 10 sin 30 – 1.5 = 0.2a a = –2.5(M1) Attempt to find acceleration a for motion from B to C
v2 =10+2×−2.5×1
AnswerMarks Guidance
c(M1) Use v2 = u2 + 2as in attempt to find v using v ≠ 0
c B
v = 5 = 2.24 ms–1
AnswerMarks
c(A1)
8
AnswerMarks Guidance
QuestionAnswer Marks
7(a)Alternative method for question 7(a)
PE loss = 0.2 × 10 × 2 sin 30 = 2M1 Attempt PE loss for motion from A to C
1
= ×0.2×v2
KE gain
AnswerMarks Guidance
2 CM1 Attempt KE gain for motion from A to C
Both PE loss and KE gain correctA1
R = 0.2 × 10 × cos 30 = √3B1
3 3
F = ×0.2× ×10=1.5
AnswerMarks Guidance
2 2M1 For using F = µR where R must be a component of 0.2g
WD against F = 1.5 × 1M1 Attempt WD against F
1
0.2×10×1=1.5×1+ ×0.2×v2
AnswerMarks Guidance
2 CM1 Attempt work-energy equation for motion from A to C
v = 5 = 2.24 ms–1
AnswerMarks
cA1
8
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)0 = 10 + 2a [a = –5] M1
B C
AnswerMarks Guidance
0.2 × 10 × sin 30 – F = 0.2 × -5M1 Attempt Newton’s 2nd law for motion from B to C
2=μ 3M1 Use F = µR where R is a component of 0.2g but R = 0.2g is M0
2
μ=
AnswerMarks Guidance
3A1 Any correct exact form such as 2/ √3
3
Alternative method for question 7(b)
AnswerMarks Guidance
PE loss = 0.2 × 10 × 1 sin 30 = 1M1 Attempt PE loss for motion from B to C
1 + ½ × 0.2 × 10 = F × 1M1 Work-Energy equation for motion from B to C in the form
PE at B + KE at B = WD against F using v2 =10 , v =0
B C
AnswerMarks Guidance
F =μ 3M1 Use F = µR leading to an equation in µ where R is a component of
0.2g
2
μ=
AnswerMarks Guidance
3A1 Any correct exact form such as 2/ √3
3
AnswerMarks Guidance
QuestionAnswer Marks
7(b)Alternative method for question 7(b)
PE loss = 0.2 × 10 × 2 sin 30 = 2M1 Attempt PE loss for motion from A to C
2 = F × 1M1 Work-Energy equation for motion from B to C
F =μ 3M1 Use F = µR leading to an equation in µ where R is a component of
0.2g
2
μ=
AnswerMarks Guidance
3A1 Any correct exact form such as 2/ √3
3
4
Question 7:
--- 7(a) ---
7(a) | 1
0.2×10×0.5= ×0.2×v2
2 B | M1 | Attempt PE or KE for motion from A to B
M1 | Attempt PE loss = KE gain from A to B
v2 =10
B | A1
Alternative method for the first 3 marks
0.2 × 10 × sin 30 = 0.2a, a = 5 | (M1) | Attempt to find acceleration a for motion from A to B
v2 =02 +2×5×1
B | (M1) | Use v2 = u2 + 2as in attempt to find speed at B
v2 =10
B | (A1)
Question | Answer | Marks | Guidance
7(a) | THEN, either this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3 | B1
3 3
F = ×0.2× ×10=1.5
2 2 | M1 | For using F = µR where R must be a component of 0.2g
PE loss = 0.2 × 10 × 0.5 = 1
WD against F = 1.5 × 1 | M1 | Attempt to find either PE loss or WD against F from B to C
1 1
0.2 ×10+0.2×10×0.5=1.5×1+ 0.2v2
2 2 C | M1 | Apply work-energy equation for motion from B to C as
KE at B + PE at B = WD against F + KE at C with v ≠ 0
B
v = 5 = 2.24 ms–1
c | A1
OR, this method for the next 5 marks
R = 0.2 × 10 × cos 30 = √3 | (B1)
3 3
F = ×0.2× ×10=1.5
2 2 | (M1) | For using F = µR where R must be a component of 0.2g
0.2 × 10 sin 30 – 1.5 = 0.2a a = –2.5 | (M1) | Attempt to find acceleration a for motion from B to C
v2 =10+2×−2.5×1
c | (M1) | Use v2 = u2 + 2as in attempt to find v using v ≠ 0
c B
v = 5 = 2.24 ms–1
c | (A1)
8
Question | Answer | Marks | Guidance
7(a) | Alternative method for question 7(a)
PE loss = 0.2 × 10 × 2 sin 30 = 2 | M1 | Attempt PE loss for motion from A to C
1
= ×0.2×v2
KE gain
2 C | M1 | Attempt KE gain for motion from A to C
Both PE loss and KE gain correct | A1
R = 0.2 × 10 × cos 30 = √3 | B1
3 3
F = ×0.2× ×10=1.5
2 2 | M1 | For using F = µR where R must be a component of 0.2g
WD against F = 1.5 × 1 | M1 | Attempt WD against F
1
0.2×10×1=1.5×1+ ×0.2×v2
2 C | M1 | Attempt work-energy equation for motion from A to C
v = 5 = 2.24 ms–1
c | A1
8
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | 0 = 10 + 2a [a = –5] | M1 | Attempt to find a for motion from B to C, using v2 =10, v =0
B C
0.2 × 10 × sin 30 – F = 0.2 × -5 | M1 | Attempt Newton’s 2nd law for motion from B to C
2=μ 3 | M1 | Use F = µR where R is a component of 0.2g but R = 0.2g is M0
2
μ=
3 | A1 | Any correct exact form such as 2/ √3
3
Alternative method for question 7(b)
PE loss = 0.2 × 10 × 1 sin 30 = 1 | M1 | Attempt PE loss for motion from B to C
1 + ½ × 0.2 × 10 = F × 1 | M1 | Work-Energy equation for motion from B to C in the form
PE at B + KE at B = WD against F using v2 =10 , v =0
B C
F =μ 3 | M1 | Use F = µR leading to an equation in µ where R is a component of
0.2g
2
μ=
3 | A1 | Any correct exact form such as 2/ √3
3
Question | Answer | Marks | Guidance
7(b) | Alternative method for question 7(b)
PE loss = 0.2 × 10 × 2 sin 30 = 2 | M1 | Attempt PE loss for motion from A to C
2 = F × 1 | M1 | Work-Energy equation for motion from B to C
F =μ 3 | M1 | Use F = µR leading to an equation in µ where R is a component of
0.2g
2
μ=
3 | A1 | Any correct exact form such as 2/ √3
3
4
\includegraphics{figure_7}

Three points $A$, $B$ and $C$ lie on a line of greatest slope of a plane inclined at an angle of $30°$ to the horizontal, with $AB = 1$ m and $BC = 1$ m, as shown in the diagram. A particle of mass 0.2 kg is released from rest at $A$ and slides down the plane. The part of the plane from $A$ to $B$ is smooth. The part of the plane from $B$ to $C$ is rough, with coefficient of friction $\mu$ between the plane and the particle.

\begin{enumerate}[label=(\alph*)]
\item Given that $\mu = \frac{1}{2}\sqrt{3}$, find the speed of the particle at $C$. [8]

\item Given instead that the particle comes to rest at $C$, find the exact value of $\mu$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2020 Q7 [12]}}
This paper (7 questions)
View full paper
Q1 5 Q2 5 Q3 6 Q4 6 Q5 7 Q6 9 Q7 12