CAIE M1 2024 June — Question 6 11 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeMultiple sequential collisions
DifficultyStandard +0.3 This is a standard multi-stage collision problem requiring systematic application of conservation of momentum and energy equations. While it involves two successive collisions and some algebraic manipulation, the steps are routine for M1: apply momentum conservation twice, calculate KE loss, solve for v, then use kinematics. The 'show that' parts guide students through the solution, and no novel insight is required beyond standard mechanics techniques.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form
Three particles \(A\), \(B\) and \(C\) of masses 5 kg, 1 kg and 2 kg respectively lie at rest in that order on a straight smooth horizontal track \(XYZ\). Initially \(A\) is at \(X\), \(B\) is at \(Y\) and \(C\) is at \(Z\). Particle \(A\) is projected towards \(B\) with a speed of \(6\text{ ms}^{-1}\) and at the same instant \(C\) is projected towards \(B\) with a speed of \(v\text{ ms}^{-1}\). In the subsequent motion, \(A\) collides and coalesces with \(B\) to form particle \(D\). Particle \(D\) then collides and coalesces with \(C\) to form particle \(E\) and \(E\) moves towards \(Z\).
  1. Show that after the second collision the speed of \(E\) is \(\frac{15-v}{4}\text{ ms}^{-1}\). [3]
  2. The total loss of kinetic energy of the system due to the two collisions is 63 J. Use the result from (a) to show that \(v = 3\). [3]
  3. It is given that the distance \(XY\) is 36 m and the distance \(YZ\) is 98 m.
    1. Find the time between the two collisions. [4]
    2. Find the time between the instant that \(A\) is projected from \(X\) and the instant that \(E\) reaches \(Z\). [1]
Question 6:
AnswerMarks
6(a)Attempt at conservation of momentum for the 1st collision
5651v 
AnswerMarks Guidance
 D*M1 3 non-zero terms; allow sign errors; using correct masses.
For reference v 5.
D
If mgv used, allow M1 M1 A0 max.
Attempt at conservation of momentum for the 2nd collision
51their v 2v512v 
AnswerMarks Guidance
 D EDM1 6 non-zero terms; allow sign errors; using correct masses;
allow their numerical v .
D
Allow vv for this mark.
E
Note: 562v512v is M2.
E
If mgv used, allow M1 M1 A0 max.
15v
v 
AnswerMarks Guidance
E 4A1 AG
Must in terms of v, as v is given in the question or
explicitly defined their letter used as v.
Do not allow vv for this mark.
E
Any error seen is A0 but condone saying ‘divide by 2’ or
equivalent.
If mgv used, allow M1 M1 A0 max.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)1 1
KE  562  2v2 90v2
initial 2 2  
2
1 15v 
AnswerMarks Guidance
KE final  2 512   4  B1 For either KE or KE correct.
initial final
Attempt difference in KE is 63 to get an equation
1 1 1 15v  2 
 562  2v2  512   63
AnswerMarks Guidance
 2 2 2  4  M1 Using sum of two initial KEsKE 63.
final
Correct number of relevant terms – correct masses, must
be adding 2 KE terms for KE .
initial
sum of two initial KEs and KE coming from use of
final
correct formula and of the correct form.
AnswerMarks Guidance
Solve algebraically 3v2 30v1170 OE to get v3 ONLYA1 AG
Any error seen is A0.
Allow solving correct quadratic expression, rather than
correct quadratic equation, for full marks.
If v13 seen it must be discarded.
Must see solving for this mark. A quadratic equation
followed by the answer is insufficient.
AnswerMarks Guidance
QuestionAnswer Marks
6(b)Alternative Method for Question 6(b): Using loss of KE in second collision
1 1
KE  652  2v2 75v2
after 1st collision 2 2  
2
1 15v 
KE  512
AnswerMarks Guidance
final 2   4  (B1) For either KE or KE correct.
after 1stcollision final
1 1 
Attempt difference in KE is 63  562  652  631548 to
2 2 
get an equation
1 1 1 15v  2 
 652  2v2  512   6315
AnswerMarks Guidance
 2 2 2  4  (M1) Using KE KE 63their1 5.
after 1st collision final
Correct number of relevant terms.
KE , KE and their 15 coming from use of
after 1st collision final
correct formula and of the correct form.
AnswerMarks Guidance
Solve algebraically 3v2 30v1170 OE to get v3 ONLY(A1) AG
Any error seen is A0.
If v13 seen it must be discarded.
Must see solving for this mark. A quadratic equation
followed by the answer is insufficient.
Alternative Method 2 for Question 6(b): Verifying that v3
1 1
KE  562  232 99
AnswerMarks
initial 2 2(B1)
2
1 153 
AnswerMarks
KE final  2 512   4   36(B1)
KE KE 63, hence loss in KE is 63J
AnswerMarks Guidance
initial final(B1) Must have a conclusion for this mark.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(c)(i)Time A to B = 6s B1
Distance BC 983their 6 80*B1FT FT their 6 which MUST come from 6t36.
Use sum of distance moved by D and distance moved by C is 80m
their 5t3t their 80
 
OR
use distance moved by C divided by relative velocity
 80 
 
their 53
AnswerMarks Guidance
DM1 Using their v from part (a).
D
v 6or 3 andtheir 8098.
D
AnswerMarks Guidance
Time = 10 sA1 Do not ISW.
4
AnswerMarks
6(c)(ii) 31036 
610  32s
 
AnswerMarks
 3 B1
1
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | Attempt at conservation of momentum for the 1st collision
5651v 
 D | *M1 | 3 non-zero terms; allow sign errors; using correct masses.
For reference v 5.
D
If mgv used, allow M1 M1 A0 max.
Attempt at conservation of momentum for the 2nd collision
51their v 2v512v 
 D E | DM1 | 6 non-zero terms; allow sign errors; using correct masses;
allow their numerical v .
D
Allow vv for this mark.
E
Note: 562v512v is M2.
E
If mgv used, allow M1 M1 A0 max.
15v
v 
E 4 | A1 | AG
Must in terms of v, as v is given in the question or
explicitly defined their letter used as v.
Do not allow vv for this mark.
E
Any error seen is A0 but condone saying ‘divide by 2’ or
equivalent.
If mgv used, allow M1 M1 A0 max.
3
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 1 1
KE  562  2v2 90v2
initial 2 2  
2
1 15v 
KE final  2 512   4   | B1 | For either KE or KE correct.
initial final
Attempt difference in KE is 63 to get an equation
1 1 1 15v  2 
 562  2v2  512   63
 2 2 2  4   | M1 | Using sum of two initial KEsKE 63.
final
Correct number of relevant terms – correct masses, must
be adding 2 KE terms for KE .
initial
sum of two initial KEs and KE coming from use of
final
correct formula and of the correct form.
Solve algebraically 3v2 30v1170 OE to get v3 ONLY | A1 | AG
Any error seen is A0.
Allow solving correct quadratic expression, rather than
correct quadratic equation, for full marks.
If v13 seen it must be discarded.
Must see solving for this mark. A quadratic equation
followed by the answer is insufficient.
Question | Answer | Marks | Guidance
6(b) | Alternative Method for Question 6(b): Using loss of KE in second collision
1 1
KE  652  2v2 75v2
after 1st collision 2 2  
2
1 15v 
KE  512
final 2   4   | (B1) | For either KE or KE correct.
after 1stcollision final
1 1 
Attempt difference in KE is 63  562  652  631548 to
2 2 
get an equation
1 1 1 15v  2 
 652  2v2  512   6315
 2 2 2  4   | (M1) | Using KE KE 63their1 5.
after 1st collision final
Correct number of relevant terms.
KE , KE and their 15 coming from use of
after 1st collision final
correct formula and of the correct form.
Solve algebraically 3v2 30v1170 OE to get v3 ONLY | (A1) | AG
Any error seen is A0.
If v13 seen it must be discarded.
Must see solving for this mark. A quadratic equation
followed by the answer is insufficient.
Alternative Method 2 for Question 6(b): Verifying that v3
1 1
KE  562  232 99
initial 2 2 | (B1)
2
1 153 
KE final  2 512   4   36 | (B1)
KE KE 63, hence loss in KE is 63J
initial final | (B1) | Must have a conclusion for this mark.
3
Question | Answer | Marks | Guidance
--- 6(c)(i) ---
6(c)(i) | Time A to B = 6s | B1
Distance BC 983their 6 80 | *B1FT | FT their 6 which MUST come from 6t36.
Use sum of distance moved by D and distance moved by C is 80m
their 5t3t their 80
 
OR
use distance moved by C divided by relative velocity
 80 
 
their 53
 | DM1 | Using their v from part (a).
D
v 6or 3 andtheir 8098.
D
Time = 10 s | A1 | Do not ISW.
4
--- 6(c)(ii) ---
6(c)(ii) |  31036 
610  32s
 
 3  | B1
1
Question | Answer | Marks | Guidance
Three particles $A$, $B$ and $C$ of masses 5 kg, 1 kg and 2 kg respectively lie at rest in that order on a straight smooth horizontal track $XYZ$. Initially $A$ is at $X$, $B$ is at $Y$ and $C$ is at $Z$. Particle $A$ is projected towards $B$ with a speed of $6\text{ ms}^{-1}$ and at the same instant $C$ is projected towards $B$ with a speed of $v\text{ ms}^{-1}$. In the subsequent motion, $A$ collides and coalesces with $B$ to form particle $D$. Particle $D$ then collides and coalesces with $C$ to form particle $E$ and $E$ moves towards $Z$.

\begin{enumerate}[label=(\alph*)]
\item Show that after the second collision the speed of $E$ is $\frac{15-v}{4}\text{ ms}^{-1}$. [3]

\item The total loss of kinetic energy of the system due to the two collisions is 63 J.

Use the result from (a) to show that $v = 3$. [3]

\item It is given that the distance $XY$ is 36 m and the distance $YZ$ is 98 m.

\begin{enumerate}[label=(\roman*)]
\item Find the time between the two collisions. [4]

\item Find the time between the instant that $A$ is projected from $X$ and the instant that $E$ reaches $Z$. [1]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q6 [11]}}
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