Edexcel M1 — Question 7 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks12
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Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeMultiple sequential collisions
DifficultyStandard +0.3 This is a standard M1 collision question requiring conservation of momentum and impulse calculations. Part (a) involves straightforward momentum conservation with one unknown, part (b) is direct impulse calculation, and part (c) repeats the process with sign considerations. While it requires careful attention to direction and multiple parts, the techniques are routine textbook applications with no novel problem-solving required, making it slightly easier than average.
Spec6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation

Two smooth spheres \(A\) and \(B\), of masses 60 grams and 90 grams respectively, are at rest on a smooth horizontal table. \(A\) is projected towards \(B\) with speed 4 ms\(^{-1}\) and the particles collide. After the collision, \(A\) and \(B\) move in the same direction as each other, with speeds \(u\) ms\(^{-1}\) and \(6u\) ms\(^{-1}\) respectively. Calculate
  1. the value of \(u\), [4 marks]
  2. the magnitude of the impulse exerted by \(A\) on \(B\), stating the units of your answer. [3 marks]
\(A\) and \(B\) are now replaced in their original positions and projected towards each other with speeds 2 ms\(^{-1}\) and 8 ms\(^{-1}\) respectively. They collide again, after which \(A\) moves with speed 7 ms\(^{-1}\), its direction of motion being reversed.
  1. Find the speed of \(B\) after this collision and state whether its direction of motion has been reversed. [5 marks]

AnswerMarks Guidance
(a) \(60 \times 4 = 60u + 90 \times 6u\)\(600u = 240\) \(u = 0.4\)
(b) Change in momentum of \(B = 0.09 \times 2.4 = 0.216 \text{ Ns}\) M1 A1 B1
(c) \(60(2) + 90(-8) = 60(-7) + 90v_B\)\(-180 = 90v_B\) \(v_B = -2\), so speed \(= 2 \text{ ms}^{-1}\), direction unchanged
(a) $60 \times 4 = 60u + 90 \times 6u$ | $600u = 240$ | $u = 0.4$ | M1 A1 M1 A1

(b) Change in momentum of $B = 0.09 \times 2.4 = 0.216 \text{ Ns}$ | | M1 A1 B1

(c) $60(2) + 90(-8) = 60(-7) + 90v_B$ | $-180 = 90v_B$ | $v_B = -2$, so speed $= 2 \text{ ms}^{-1}$, direction unchanged | M1 A1 A1 | **12 marks**
Two smooth spheres $A$ and $B$, of masses 60 grams and 90 grams respectively, are at rest on a smooth horizontal table. $A$ is projected towards $B$ with speed 4 ms$^{-1}$ and the particles collide.
After the collision, $A$ and $B$ move in the same direction as each other, with speeds $u$ ms$^{-1}$ and $6u$ ms$^{-1}$ respectively. Calculate
\begin{enumerate}[label=(\alph*)]
\item the value of $u$, [4 marks]
\item the magnitude of the impulse exerted by $A$ on $B$, stating the units of your answer. [3 marks]
\end{enumerate}
$A$ and $B$ are now replaced in their original positions and projected towards each other with speeds 2 ms$^{-1}$ and 8 ms$^{-1}$ respectively. They collide again, after which $A$ moves with speed 7 ms$^{-1}$, its direction of motion being reversed.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the speed of $B$ after this collision and state whether its direction of motion has been reversed. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q7 [12]}}