CAIE M1 2020 Specimen — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2020
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeMultiple sequential collisions
DifficultyStandard +0.3 This is a straightforward two-stage collision problem using conservation of momentum twice. The first collision uses standard momentum conservation with given masses and velocities, then the second applies the coalescence formula. Both are routine M1 applications with no conceptual challenges beyond careful arithmetic with the given masses.
Spec6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation

3 Three small smooth spheres \(A , B\) and \(C\) of equal radii and of masses \(4 \mathrm {~kg} , 2 \mathrm {~kg}\) and 3 kg respectively, lie in that order in a straight line on a smooth horizontal plane. Initially, \(B\) and \(C\) are at rest and \(A\) is moving towards \(B\) with speed \(6 \mathrm {~ms} ^ { - 1 }\). After the collison with \(B\), sphere \(A\) continues to move in the same direction but with speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Find the speed of \(B\) after this collison.
    Sphere \(B\) collides with \(C\). In this collison these two spheres coalesce to form an object \(D\).
  2. Find the speed of \(D\) after this collision.
  3. Show that the total loss of kinetic energy in the system due to the two collisions is 38.4 J .

AnswerMarks Guidance
3(a)Conservation of momentum \(4 \times 6 + 0 = 8[\text{ms}^{-1}]\) \(v = 8[\text{ms}^{-1}]\) M1 A1
3(b)\(2 \times \text{chert}(8) [= 0] = 2[\text{ms}^{-1}]\) \(v = 3.2[\text{ms}^{-1}]\) M1 A1
3(c)Kinetic energy (KE) initial = \(\frac{1}{2} \times 4 \times 6^2\) KE final = \(\frac{1}{2} \times 4 \times 2^2 + \frac{1}{2} \times 2 \times 5 \times 3.2^2\) Loss of KE = 72 – 33.6 [J] M1 M1 A1
**3(a)** | Conservation of momentum $4 \times 6 + 0 = 8[\text{ms}^{-1}]$ $v = 8[\text{ms}^{-1}]$ | M1 A1 | 

**3(b)** | $2 \times \text{chert}(8) [= 0] = 2[\text{ms}^{-1}]$ $v = 3.2[\text{ms}^{-1}]$ | M1 A1 | 

**3(c)** | Kinetic energy (KE) initial = $\frac{1}{2} \times 4 \times 6^2$ KE final = $\frac{1}{2} \times 4 \times 2^2 + \frac{1}{2} \times 2 \times 5 \times 3.2^2$ Loss of KE = 72 – 33.6 [J] | M1 M1 A1 | 

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3 Three small smooth spheres $A , B$ and $C$ of equal radii and of masses $4 \mathrm {~kg} , 2 \mathrm {~kg}$ and 3 kg respectively, lie in that order in a straight line on a smooth horizontal plane. Initially, $B$ and $C$ are at rest and $A$ is moving towards $B$ with speed $6 \mathrm {~ms} ^ { - 1 }$. After the collison with $B$, sphere $A$ continues to move in the same direction but with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Find the speed of $B$ after this collison.\\

Sphere $B$ collides with $C$. In this collison these two spheres coalesce to form an object $D$.\\
(b) Find the speed of $D$ after this collision.\\
(c) Show that the total loss of kinetic energy in the system due to the two collisions is 38.4 J .\\

\hfill \mbox{\textit{CAIE M1 2020 Q3 [6]}}