CAIE M1 2023 March — Question 7 13 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionMarch
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeMultiple sequential collisions
DifficultyStandard +0.3 This is a multi-part mechanics question involving energy conservation, collision principles, and kinematics. Part (a) uses straightforward conservation of momentum with given values. Part (b) applies energy conservation to find an angle. Part (c) requires tracking two particles meeting, using constant acceleration equations. All techniques are standard M1 content with clear setup and routine calculations, making it slightly easier than average.
Spec6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form
\includegraphics{figure_7} The diagram shows a smooth track which lies in a vertical plane. The section \(AB\) is a quarter circle of radius 1.8 m with centre \(O\). The section \(BC\) is a horizontal straight line of length 7.0 m and \(OB\) is perpendicular to \(BC\). The section \(CFE\) is a straight line inclined at an angle of \(\theta°\) above the horizontal. A particle \(P\) of mass 0.5 kg is released from rest at \(A\). Particle \(P\) collides with a particle \(Q\) of mass 0.1 kg which is at rest at \(B\). Immediately after the collision, the speed of \(P\) is \(4\,\text{m}\,\text{s}^{-1}\) in the direction \(BC\). You should assume that \(P\) is moving horizontally when it collides with \(Q\).
  1. Show that the speed of \(Q\) immediately after the collision is \(10\,\text{m}\,\text{s}^{-1}\). [4] When \(Q\) reaches \(C\), it collides with a particle \(R\) of mass 0.4 kg which is at rest at \(C\). The two particles coalesce. The combined particle comes instantaneously to rest at \(F\). You should assume that there is no instantaneous change in speed as the combined particle leaves \(C\), nor when it passes through \(C\) again as it returns down the slope.
  2. Given that the distance \(CF\) is 0.4 m, find the value of \(\theta\). [4]
  3. Find the distance from \(B\) at which \(P\) collides with the combined particle. [5]
Question 7:
AnswerMarks
7(a)Attempt to use conservation of energy
1  1 
0.5v2 =0.5g1.8 or mv2 =mg1.8
   
AnswerMarks Guidance
2  2 M1 2 terms, dimensionally correct.
Do not allow from use of constant acceleration.
AnswerMarks Guidance
v=6A1 Do not allow from use of constant acceleration.
Attempt at conservation of momentum
 0.56(+0)=0.54+0.1w
AnswerMarks Guidance
M1 3 terms; allow sign errors; allow their v=6 or just v; allow if
using mgv (consistently in all terms).
AnswerMarks Guidance
Speed of Q (= w) =1 0m s-1A1 AG
Do not allow from use of constant acceleration.
Do not allow if using mgv.
Use of constant acceleration gets M0 A0 M1 A0 maximum.
AnswerMarks
4SC Assuming elastic collision
1 1
M1A1 0.5g1.8= 0.1w2 + 0.542
2 2
M1 For attempt at conservation of energy, 3 terms; allow sign
errors.
B1 Speed of Q (= w) =1 0m s-1
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)Attempt at conservation of momentum
 0.110=(0.1+0.4)z (z=2)
AnswerMarks Guidance
*M1 3 terms, allow sign errors, allow if using mgv.
Attempt to use conservation of energy
1 
(0.1+0.4)(their 2)2 =(0.1+0.4)gh (h=0.2)
 
AnswerMarks Guidance
2 *DM1 Dependent on previous M mark.
4 terms, dimensionally correct.
Do not allow from use of constant acceleration.
their 210.
Use trigonometry to get an equation in  and solve for 
 their 0.2
= sin−1
  
AnswerMarks Guidance
  0.4 DM1 Dependent on previous 2 M marks.
Using their h and 0.4 .
Allow sin/cos mix.
AnswerMarks Guidance
θ = 30A1 Do not allow if using mgv.
Alternative method for Question 7(b): Using constant acceleration
Attempt at conservation of momentum
0.110=0.5z (z=2)
AnswerMarks Guidance
 *M1 2 terms, allow sign errors, allow if using mgv.
Attempt at use of constant acceleration
 02 =(their 2)2 2a0.4 (a= 5)
AnswerMarks Guidance
 *DM1 Dependent on previous M mark.
Uses constant acceleration with u=their 2 and s=0.4 to get an
equation in a; their 210.
Use N2L to get an equation in  leading to a positive value of 
and solve for 
AnswerMarks Guidance
  (0.5)their a =(0.5)gsin DM1 Dependent on previous 2 M marks.
Using their a; May have m for 0.5 .
Allow sin/cos mix.
AnswerMarks Guidance
θ = 30A1 Do not allow if using mgv.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(c)Q takes 0.7 s to travel from B to C B1
(their 2)+0
0.4= t t=0.4
AnswerMarks Guidance
2B1FT SOI
0.8
FT their 2 from (b), t= .
their 2
u+v
For use of s= t to get a time up the slope.
 2 
Allow for total time on slope from
1
0=(their 2)t− (their a)t2 t=0.8 .
2
AnswerMarks Guidance
Distance between P moved is (0.7+0.8)4(=6)B1 Allow 1m from point C.
Set up equation in t using 4t, (their 2)t and their 6 and solve for
t
AnswerMarks Guidance
 4t+(their 2)t=(their1 ) OR (their 6)+4t+(their 2)t=7 M1 Must have considered all parts of motion to find times from
relevant equations.
2
Distance from B = 6 m
AnswerMarks Guidance
3A1
QuestionAnswer Marks
7(c)Alternative method for last 3 marks of Question 7(c)
b 7−b
[Time for P =] and [Time for QR =]
4 2
7−c c
OR [Time for P =] and [Time for QR =]
AnswerMarks Guidance
4 2B1 Where b is distance from B
OR Where c is distance from C.
Attempt to form an equation from use of total time and solve for b
(or c)
 7−b b  2 
+0.7+0.4+0.4= b=6
   
2 4  3
 
 c 7−c  1
 OR +0.7+0.4+0.4=  c= 
AnswerMarks Guidance
 2 4  3M1 Where b is distance from B
OR Where c is distance from C.
Must have considered all parts of motion to find times from
relevant equations.
2
Distance from B = 6 m
AnswerMarks
3A1
5
Question 7:
--- 7(a) ---
7(a) | Attempt to use conservation of energy
1  1 
0.5v2 =0.5g1.8 or mv2 =mg1.8
   
2  2  | M1 | 2 terms, dimensionally correct.
Do not allow from use of constant acceleration.
v=6 | A1 | Do not allow from use of constant acceleration.
Attempt at conservation of momentum
 0.56(+0)=0.54+0.1w
 | M1 | 3 terms; allow sign errors; allow their v=6 or just v; allow if
using mgv (consistently in all terms).
Speed of Q (= w) =1 0m s-1 | A1 | AG
Do not allow from use of constant acceleration.
Do not allow if using mgv.
Use of constant acceleration gets M0 A0 M1 A0 maximum.
4 | SC Assuming elastic collision
1 1
M1A1 0.5g1.8= 0.1w2 + 0.542
2 2
M1 For attempt at conservation of energy, 3 terms; allow sign
errors.
B1 Speed of Q (= w) =1 0m s-1
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Attempt at conservation of momentum
 0.110=(0.1+0.4)z (z=2)
 | *M1 | 3 terms, allow sign errors, allow if using mgv.
Attempt to use conservation of energy
1 
(0.1+0.4)(their 2)2 =(0.1+0.4)gh (h=0.2)
 
2  | *DM1 | Dependent on previous M mark.
4 terms, dimensionally correct.
Do not allow from use of constant acceleration.
their 210.
Use trigonometry to get an equation in  and solve for 
 their 0.2
= sin−1
  
  0.4  | DM1 | Dependent on previous 2 M marks.
Using their h and 0.4 .
Allow sin/cos mix.
θ = 30 | A1 | Do not allow if using mgv.
Alternative method for Question 7(b): Using constant acceleration
Attempt at conservation of momentum
0.110=0.5z (z=2)
  | *M1 | 2 terms, allow sign errors, allow if using mgv.
Attempt at use of constant acceleration
 02 =(their 2)2 2a0.4 (a= 5)
  | *DM1 | Dependent on previous M mark.
Uses constant acceleration with u=their 2 and s=0.4 to get an
equation in a; their 210.
Use N2L to get an equation in  leading to a positive value of 
and solve for 
  (0.5)their a =(0.5)gsin  | DM1 | Dependent on previous 2 M marks.
Using their a; May have m for 0.5 .
Allow sin/cos mix.
θ = 30 | A1 | Do not allow if using mgv.
4
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | Q takes 0.7 s to travel from B to C | B1
(their 2)+0
0.4= t t=0.4
2 | B1FT | SOI
0.8
FT their 2 from (b), t= .
their 2
u+v
For use of s= t to get a time up the slope.
 2 
Allow for total time on slope from
1
0=(their 2)t− (their a)t2 t=0.8 .
2
Distance between P moved is (0.7+0.8)4(=6) | B1 | Allow 1m from point C.
Set up equation in t using 4t, (their 2)t and their 6 and solve for
t
 4t+(their 2)t=(their1 ) OR (their 6)+4t+(their 2)t=7  | M1 | Must have considered all parts of motion to find times from
relevant equations.
2
Distance from B = 6 m
3 | A1
Question | Answer | Marks | Guidance
7(c) | Alternative method for last 3 marks of Question 7(c)
b 7−b
[Time for P =] and [Time for QR =]
4 2
7−c c
OR [Time for P =] and [Time for QR =]
4 2 | B1 | Where b is distance from B
OR Where c is distance from C.
Attempt to form an equation from use of total time and solve for b
(or c)
 7−b b  2 
+0.7+0.4+0.4= b=6
   
2 4  3
 
 c 7−c  1
 OR +0.7+0.4+0.4=  c= 
 2 4  3 | M1 | Where b is distance from B
OR Where c is distance from C.
Must have considered all parts of motion to find times from
relevant equations.
2
Distance from B = 6 m
3 | A1
5
\includegraphics{figure_7}

The diagram shows a smooth track which lies in a vertical plane. The section $AB$ is a quarter circle of radius 1.8 m with centre $O$. The section $BC$ is a horizontal straight line of length 7.0 m and $OB$ is perpendicular to $BC$. The section $CFE$ is a straight line inclined at an angle of $\theta°$ above the horizontal.

A particle $P$ of mass 0.5 kg is released from rest at $A$. Particle $P$ collides with a particle $Q$ of mass 0.1 kg which is at rest at $B$. Immediately after the collision, the speed of $P$ is $4\,\text{m}\,\text{s}^{-1}$ in the direction $BC$. You should assume that $P$ is moving horizontally when it collides with $Q$.

\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ immediately after the collision is $10\,\text{m}\,\text{s}^{-1}$. [4]

When $Q$ reaches $C$, it collides with a particle $R$ of mass 0.4 kg which is at rest at $C$. The two particles coalesce. The combined particle comes instantaneously to rest at $F$. You should assume that there is no instantaneous change in speed as the combined particle leaves $C$, nor when it passes through $C$ again as it returns down the slope.

\item Given that the distance $CF$ is 0.4 m, find the value of $\theta$. [4]

\item Find the distance from $B$ at which $P$ collides with the combined particle. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q7 [13]}}
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