AQA M1 2010 June — Question 3 6 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, vector velocity form
DifficultyModerate -0.8 This is a straightforward application of conservation of momentum in two dimensions. Students apply the principle separately to each component (horizontal and vertical), leading to two simple linear equations that directly yield m and b. No problem-solving insight is required—just routine substitution into a standard formula.
Spec6.03b Conservation of momentum: 1D two particles6.03d Conservation in 2D: vector momentum

3 Two particles, \(A\) and \(B\), are moving on a smooth horizontal plane when they collide. The mass of \(A\) is 6 kg and the mass of \(B\) is \(m \mathrm {~kg}\). Before the collision, the velocity of \(A\) is \(\left[ \begin{array} { l } 2 \\ 4 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }\) and the velocity of \(B\) is \(\left[ \begin{array} { r } 3 \\ - 2 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }\). After the collision, the velocity of \(A\) is \(\left[ \begin{array} { l } 1 \\ 3 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }\) and the velocity of \(B\) is \(\left[ \begin{array} { l } 7 \\ b \end{array} \right] \mathrm { ms } ^ { - 1 }\).
  1. Find \(m\).
  2. \(\quad\) Find \(b\).
    (2 marks)
    .......... \(\_\_\_\_\) \includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-07_40_118_529_159} \includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-07_39_117_623_159}

Question 3:
(a)
Conservation of momentum (component form):
AnswerMarks Guidance
\(6\begin{pmatrix}2\\4\end{pmatrix} + m\begin{pmatrix}3\\-2\end{pmatrix} = 6\begin{pmatrix}1\\3\end{pmatrix} + m\begin{pmatrix}7\\b\end{pmatrix}\)M1 Setting up momentum equation
Using \(x\)-components: \(12 + 3m = 6 + 7m\)A1
\(6 = 4m\)DM1
\(m = 1.5\) kgA1
(b)
AnswerMarks
Using \(y\)-components: \(24 + 1.5(-2) = 6(3) + 1.5b\)M1
\(24 - 3 = 18 + 1.5b\)
\(1.5b = 3\)
AnswerMarks
\(b = 2\)A1
# Question 3:

**(a)**
Conservation of momentum (component form):
$6\begin{pmatrix}2\\4\end{pmatrix} + m\begin{pmatrix}3\\-2\end{pmatrix} = 6\begin{pmatrix}1\\3\end{pmatrix} + m\begin{pmatrix}7\\b\end{pmatrix}$ | M1 | Setting up momentum equation

Using $x$-components: $12 + 3m = 6 + 7m$ | A1 |
$6 = 4m$ | DM1 |
$m = 1.5$ kg | A1 |

**(b)**
Using $y$-components: $24 + 1.5(-2) = 6(3) + 1.5b$ | M1 |
$24 - 3 = 18 + 1.5b$
$1.5b = 3$
$b = 2$ | A1 |
3 Two particles, $A$ and $B$, are moving on a smooth horizontal plane when they collide. The mass of $A$ is 6 kg and the mass of $B$ is $m \mathrm {~kg}$. Before the collision, the velocity of $A$ is $\left[ \begin{array} { l } 2 \\ 4 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $\left[ \begin{array} { r } 3 \\ - 2 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }$. After the collision, the velocity of $A$ is $\left[ \begin{array} { l } 1 \\ 3 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $\left[ \begin{array} { l } 7 \\ b \end{array} \right] \mathrm { ms } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $m$.
\item $\quad$ Find $b$.\\
(2 marks)\\

.......... $\_\_\_\_$\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-07_40_118_529_159}\\

\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-07_39_117_623_159}
\end{enumerate}

\hfill \mbox{\textit{AQA M1 2010 Q3 [6]}}