CAIE M1 2022 June — Question 7 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeMultiple sequential collisions
DifficultyStandard +0.3 This is a multi-part momentum conservation problem with constant acceleration on an inclined plane. Part (a) is straightforward application of conservation of momentum. Part (b) requires using kinematics (SUVAT) to find B's speed at the barrier, then applying the 90% reduction, followed by another momentum calculation for the coalescing collision. While it involves multiple steps and careful tracking of directions, all techniques are standard M1 content with no novel problem-solving required—slightly easier than average A-level.
Spec3.02h Motion under gravity: vector form3.03f Weight: W=mg6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact
Two particles \(A\) and \(B\), of masses 0.4 kg and 0.2 kg respectively, are moving down the same line of greatest slope of a smooth plane. The plane is inclined at 30° to the horizontal, and \(A\) is higher up the plane than \(B\). When the particles collide, the speeds of \(A\) and \(B\) are 3 m s\(^{-1}\) and 2 m s\(^{-1}\) respectively. In the collision between the particles, the speed of \(A\) is reduced to 2.5 m s\(^{-1}\).
  1. Find the speed of \(B\) immediately after the collision. [2]
After the collision, when \(B\) has moved 1.6 m down the plane from the point of collision, it hits a barrier and returns back up the same line of greatest slope. \(B\) hits the barrier 0.4 s after the collision, and when it hits the barrier, its speed is reduced by 90%. The two particles collide again 0.44 s after their previous collision, and they then coalesce on impact.
  1. Show that the speed of \(B\) immediately after it hits the barrier is 0.5 m s\(^{-1}\). Hence find the speed of the combined particle immediately after the second collision between \(A\) and \(B\). [7]
Question 7:
AnswerMarks Guidance
7(a)0.430.220.42.50.2v M1
v = 3 ms−1A1 Allow M1A0 if g included with the masses.
2
AnswerMarks Guidance
7(b)For A 0.4gsin300.4a or for B 0.2gsin300.2a
or mgsin30maM1 For either. Allow sin/cos mix.
a =  5 or gsin30A1 Allow gsin30 without working for M1A1
For B when hits barrier v2 32 251.6 [⇒ v = 5]
AnswerMarks Guidance
OR vuat v350.4  v  5M1 Using their ag and their v from part (a)
uv 3v
OR: use of s t 1.6 0.4v5
2 2
1 1
OR 0.2v2  0.232 0.21.6gsin30
2 2
AnswerMarks Guidance
Speed after hitting barrier = 0.1  5 = 0.5A1 AG
v = 2.5 + 5  0.44 [= 4.7] v = −0.5 + 5  0.04 [= −0.3]
A B
or v = 0.5 + (–5)  0.04 [= 0.3]
AnswerMarks Guidance
B*M1 Use of vuat for either with correct t-value, with initial speeds
2.5 or0.5 their ag
0.4  4.7 + 0.2  (−0.3) = 0.6 v
AnswerMarks Guidance
combDM1 Use of vuat for BOTH with correct t-values, initial speeds
2.5, 0.5 and  their acceleration (same for both) and use of
conservation of momentum with correct number of terms. Allow
sign errors.
v = 3.03 ms−1
AnswerMarks Guidance
combA1 91 1
Allow v 3
30 30
Allow DM1A0 if g included with the masses.
7
Question 7:
--- 7(a) ---
7(a) | 0.430.220.42.50.2v | M1 | Use of conservation of momentum with 4 terms. Allow sign errors.
v = 3 ms−1 | A1 | Allow M1A0 if g included with the masses.
2
--- 7(b) ---
7(b) | For A 0.4gsin300.4a or for B 0.2gsin300.2a
or mgsin30ma | M1 | For either. Allow sin/cos mix.
a =  5 or gsin30 | A1 | Allow gsin30 without working for M1A1
For B when hits barrier v2 32 251.6 [⇒ v = 5]
OR vuat v350.4  v  5 | M1 | Using their ag and their v from part (a)
uv 3v
OR: use of s t 1.6 0.4v5
2 2
1 1
OR 0.2v2  0.232 0.21.6gsin30
2 2
Speed after hitting barrier = 0.1  5 = 0.5 | A1 | AG
v = 2.5 + 5  0.44 [= 4.7] v = −0.5 + 5  0.04 [= −0.3]
A B
or v = 0.5 + (–5)  0.04 [= 0.3]
B | *M1 | Use of vuat for either with correct t-value, with initial speeds
2.5 or0.5 their ag
0.4  4.7 + 0.2  (−0.3) = 0.6 v
comb | DM1 | Use of vuat for BOTH with correct t-values, initial speeds
2.5, 0.5 and  their acceleration (same for both) and use of
conservation of momentum with correct number of terms. Allow
sign errors.
v = 3.03 ms−1
comb | A1 | 91 1
Allow v 3
30 30
Allow DM1A0 if g included with the masses.
7
Two particles $A$ and $B$, of masses 0.4 kg and 0.2 kg respectively, are moving down the same line of greatest slope of a smooth plane. The plane is inclined at 30° to the horizontal, and $A$ is higher up the plane than $B$. When the particles collide, the speeds of $A$ and $B$ are 3 m s$^{-1}$ and 2 m s$^{-1}$ respectively. In the collision between the particles, the speed of $A$ is reduced to 2.5 m s$^{-1}$.

\begin{enumerate}[label=(\alph*)]
\item Find the speed of $B$ immediately after the collision. [2]
\end{enumerate}

After the collision, when $B$ has moved 1.6 m down the plane from the point of collision, it hits a barrier and returns back up the same line of greatest slope. $B$ hits the barrier 0.4 s after the collision, and when it hits the barrier, its speed is reduced by 90%. The two particles collide again 0.44 s after their previous collision, and they then coalesce on impact.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the speed of $B$ immediately after it hits the barrier is 0.5 m s$^{-1}$. Hence find the speed of the combined particle immediately after the second collision between $A$ and $B$. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q7 [9]}}
This paper (7 questions)
View full paper
Q1 6 Q2 5 Q3 5 Q4 6 Q5 9 Q6 10 Q7 9