CAIE M1 2021 June — Question 7 11 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2021
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeParticle on slope then horizontal
DifficultyStandard +0.3 This is a standard work-energy problem with straightforward application of conservation of energy and friction calculations. Part (a) uses basic energy conservation on a smooth slope, then work against resistance. Part (b) requires working backwards with friction, but follows a predictable method. The multi-step nature and friction coefficient calculation add slight complexity, but this remains a routine M1 mechanics question requiring no novel insight.
Spec3.03f Weight: W=mg3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle
\includegraphics{figure_7} A slide in a playground descends at a constant angle of 30° for 2.5 m. It then has a horizontal section in the same vertical plane as the sloping section. A child of mass 35 kg, modelled as a particle \(P\), starts from rest at the top of the slide and slides straight down the sloping section. She then continues along the horizontal section until she comes to rest (see diagram). There is no instantaneous change in speed when the child goes from the sloping section to the horizontal section. The child experiences a resistance force on the horizontal section of the slide, and the work done against the resistance force on the horizontal section of the slide is 250 J per metre.
  1. It is given that the sloping section of the slide is smooth.
    1. Find the speed of the child when she reaches the bottom of the sloping section. [3]
    2. Find the distance that the child travels along the horizontal section of the slide before she comes to rest. [2]
  2. It is given instead that the sloping section of the slide is rough and that the child comes to rest on the slide 1.05 m after she reaches the horizontal section. Find the coefficient of friction between the child and the sloping section of the slide. [6]
Question 7:
AnswerMarks Guidance
7(a)(i)PE =35g×2.5sin30 M1
1
×35v2 =35g×2.5sin30
AnswerMarks Guidance
2M1 Use of conservation of energy, 2 terms, correct dimensions
v=5 m s–1A1
Alternative method for Question 7(a)(i)
AnswerMarks Guidance
mgsin30=ma leading to a=5M1 For applying Newton’s 2nd law down the plane, 2 terms, correct
dimensions
AnswerMarks Guidance
v2 =0+2×5×2.5M1 For using v2 = u2 + 2as, using their a≠±g
v=5 m s–1A1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(a)(ii)1
×35×52 =250d
AnswerMarks Guidance
2M1 Use of work-energy from the bottom of the slide until motion stops,
2 terms, correct dimensions, using their v
AnswerMarks
d =1.75 mA1
Alternative method for Question 7(a)(ii)
AnswerMarks Guidance
35g×2.5sin30=250dM1 Use of work-energy from the start until motion stops, 2 terms,
correct dimensions.
AnswerMarks
d =1.75 mA1
Alternative method for Question 7(a)(ii)
50
−250=35a leading to a=− =−7.14
7
AnswerMarks Guidance
0=52 +2 ( a ) dM1 Newton’s 2nd law on the horizontal section with resistance = 250 N
to find a and use v2 =u2 +2as with v=0, u=5 and s=d.
AnswerMarks
d =1.75 mA1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)1
×35v2 =250×1.05 v2 =15
 
2
or
50
−250=35a leading to a=−
7
50
0=v2 +2×− ×1.05 v2 =15
 
AnswerMarks Guidance
7B1 Either use the correct work energy equation for motion on the
horizontal section or use the fact that the frictional force on the
horizontal section is 250 N in order to set up an equation that would
lead to finding the speed at the bottom of the slide.
[ ]
AnswerMarks
R=35gcos30 =303.11B1
v2 =0+2×a×2.5=15 leading to a = 3
or
[ ]
AnswerMarks Guidance
PE change =35g×2.5sin30 =437.5M1 For using v2 =u2 +2as, with their v2 to set up an equation that
would lead to finding a.
[ ]
35gsin30−F =35a or 175−F =35a
or
1 [ ]
35g×2.5sin30=F×2.5+ ×35×15 437.5=F×2.5+262.5
AnswerMarks Guidance
2M1 For using Newton’s 2nd law down the slope with correct
dimensions.
or
For using energy equation, 3 relevant terms with correct
dimensions.
AnswerMarks Guidance
F =μ×RM1 For using F = µR, where R is a component of 35g.
μ=0.231A1 2 3
Allow μ= OE
15
AnswerMarks Guidance
QuestionAnswer Marks
7(b)Alternative method for Question 7(b)
R=35gcos30B1
[ ]
AnswerMarks Guidance
PE change =35g×2.5sin30 =437.5B1
WD against friction on the flat =250×1.05B1 WD = 262.5
[ ]
AnswerMarks Guidance
35g×2.5sin30=F×2.5+250×1.05 437.5=F×2.5+262.5M1 For using energy equation, 3 relevant terms with correct
dimensions.
AnswerMarks Guidance
F =μ×RM1 For using F = µR at any stage, where R is a component of 35g.
μ=0.231A1 2 3
Allow μ= OE
15
6
Question 7:
--- 7(a)(i) ---
7(a)(i) | PE =35g×2.5sin30 | M1
1
×35v2 =35g×2.5sin30
2 | M1 | Use of conservation of energy, 2 terms, correct dimensions
v=5 m s–1 | A1
Alternative method for Question 7(a)(i)
mgsin30=ma leading to a=5 | M1 | For applying Newton’s 2nd law down the plane, 2 terms, correct
dimensions
v2 =0+2×5×2.5 | M1 | For using v2 = u2 + 2as, using their a≠±g
v=5 m s–1 | A1
3
Question | Answer | Marks | Guidance
--- 7(a)(ii) ---
7(a)(ii) | 1
×35×52 =250d
2 | M1 | Use of work-energy from the bottom of the slide until motion stops,
2 terms, correct dimensions, using their v
d =1.75 m | A1
Alternative method for Question 7(a)(ii)
35g×2.5sin30=250d | M1 | Use of work-energy from the start until motion stops, 2 terms,
correct dimensions.
d =1.75 m | A1
Alternative method for Question 7(a)(ii)
50
−250=35a leading to a=− =−7.14
7
0=52 +2 ( a ) d | M1 | Newton’s 2nd law on the horizontal section with resistance = 250 N
to find a and use v2 =u2 +2as with v=0, u=5 and s=d.
d =1.75 m | A1
2
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | 1
×35v2 =250×1.05 v2 =15
 
2
or
50
−250=35a leading to a=−
7
50
0=v2 +2×− ×1.05 v2 =15
 
7 | B1 | Either use the correct work energy equation for motion on the
horizontal section or use the fact that the frictional force on the
horizontal section is 250 N in order to set up an equation that would
lead to finding the speed at the bottom of the slide.
[ ]
R=35gcos30 =303.11 | B1
v2 =0+2×a×2.5=15 leading to a = 3
or
[ ]
PE change =35g×2.5sin30 =437.5 | M1 | For using v2 =u2 +2as, with their v2 to set up an equation that
would lead to finding a.
[ ]
35gsin30−F =35a or 175−F =35a
or
1 [ ]
35g×2.5sin30=F×2.5+ ×35×15 437.5=F×2.5+262.5
2 | M1 | For using Newton’s 2nd law down the slope with correct
dimensions.
or
For using energy equation, 3 relevant terms with correct
dimensions.
F =μ×R | M1 | For using F = µR, where R is a component of 35g.
μ=0.231 | A1 | 2 3
Allow μ= OE
15
Question | Answer | Marks | Guidance
7(b) | Alternative method for Question 7(b)
R=35gcos30 | B1
[ ]
PE change =35g×2.5sin30 =437.5 | B1
WD against friction on the flat =250×1.05 | B1 | WD = 262.5
[ ]
35g×2.5sin30=F×2.5+250×1.05 437.5=F×2.5+262.5 | M1 | For using energy equation, 3 relevant terms with correct
dimensions.
F =μ×R | M1 | For using F = µR at any stage, where R is a component of 35g.
μ=0.231 | A1 | 2 3
Allow μ= OE
15
6
\includegraphics{figure_7}

A slide in a playground descends at a constant angle of 30° for 2.5 m. It then has a horizontal section in the same vertical plane as the sloping section. A child of mass 35 kg, modelled as a particle $P$, starts from rest at the top of the slide and slides straight down the sloping section. She then continues along the horizontal section until she comes to rest (see diagram). There is no instantaneous change in speed when the child goes from the sloping section to the horizontal section.

The child experiences a resistance force on the horizontal section of the slide, and the work done against the resistance force on the horizontal section of the slide is 250 J per metre.

\begin{enumerate}[label=(\alph*)]
\item It is given that the sloping section of the slide is smooth.
\begin{enumerate}[label=(\roman*)]
\item Find the speed of the child when she reaches the bottom of the sloping section. [3]
\item Find the distance that the child travels along the horizontal section of the slide before she comes to rest. [2]
\end{enumerate}
\item It is given instead that the sloping section of the slide is rough and that the child comes to rest on the slide 1.05 m after she reaches the horizontal section.

Find the coefficient of friction between the child and the sloping section of the slide. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2021 Q7 [11]}}
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