CAIE M1 2022 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCollision with two possible outcomes
DifficultyStandard +0.3 This is a standard two-body collision problem requiring conservation of momentum and the given speed constraint to find final velocities, then calculating kinetic energy loss. It involves straightforward algebraic manipulation with no conceptual surprises, making it slightly easier than average for A-level mechanics.
Spec6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles

Small smooth spheres \(A\) and \(B\), of equal radii and of masses 6 kg and 2 kg respectively, lie on a smooth horizontal plane. Initially \(A\) is moving towards \(B\) with speed 5 m s\(^{-1}\) and \(B\) is moving towards \(A\) with speed 3 m s\(^{-1}\). After the spheres collide, both \(A\) and \(B\) move in the same direction and the difference in the speeds of the spheres is 2 m s\(^{-1}\). Find the loss of kinetic energy of the system due to the collision. [5]

Question 2:
AnswerMarks
2Use conservation of momentum
65+2(−3 )=6v +2v
AnswerMarks Guidance
A B*M1 4 dimensionally correct terms.
Allow sign errors, v and v must be
A B
different.
Use v =v +2 or v =v −2 with their momentum equation and solve for v or v
AnswerMarks Guidance
B A A B A BDM1 Allow v =v 2 or v =v 2.
B A A B
v = 2.5 or v = 4.5
AnswerMarks
A BA1
Attempt at initial KE, or final KE, or change in KE for A, or change in KE for B
1 1
Initial KE = 652 + 2(−3 )2=84 
2 2
1 1
Final KE = 6(their2.5 )2 + 2(their4.5 )2
2 2
1 1 
Change in KE for A=  652 − 6(their2.5 )2 
2 2 
1 1 
Change in KE for B=  2(−3 )2− 2(their4.5 )2 
AnswerMarks Guidance
2 2 M1 Allow use or their v and/or v .
A B
Allow if 2 KE equations seen.
AnswerMarks Guidance
Loss of KE = 45JA1 Allow –45J.
Allow if mgv used in momentum equation.
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | Use conservation of momentum
65+2(−3 )=6v +2v
A B | *M1 | 4 dimensionally correct terms.
Allow sign errors, v and v must be
A B
different.
Use v =v +2 or v =v −2 with their momentum equation and solve for v or v
B A A B A B | DM1 | Allow v =v 2 or v =v 2.
B A A B
v = 2.5 or v = 4.5
A B | A1
Attempt at initial KE, or final KE, or change in KE for A, or change in KE for B
1 1
Initial KE = 652 + 2(−3 )2=84 
2 2
1 1
Final KE = 6(their2.5 )2 + 2(their4.5 )2
2 2
1 1 
Change in KE for A=  652 − 6(their2.5 )2 
2 2 
1 1 
Change in KE for B=  2(−3 )2− 2(their4.5 )2 
2 2  | M1 | Allow use or their v and/or v .
A B
Allow if 2 KE equations seen.
Loss of KE = 45J | A1 | Allow –45J.
Allow if mgv used in momentum equation.
5
Question | Answer | Marks | Guidance
Small smooth spheres $A$ and $B$, of equal radii and of masses 6 kg and 2 kg respectively, lie on a smooth horizontal plane. Initially $A$ is moving towards $B$ with speed 5 m s$^{-1}$ and $B$ is moving towards $A$ with speed 3 m s$^{-1}$. After the spheres collide, both $A$ and $B$ move in the same direction and the difference in the speeds of the spheres is 2 m s$^{-1}$.

Find the loss of kinetic energy of the system due to the collision. [5]

\hfill \mbox{\textit{CAIE M1 2022 Q2 [5]}}