Edexcel M2 — Question 1 4 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCollision with vector velocities
DifficultyModerate -0.3 This is a straightforward application of conservation of momentum in 2D. Students must equate total momentum before and after collision in vector form, then solve a simple linear equation for the unknown mass. The calculation is routine with no conceptual subtleties, making it slightly easier than average for M2.
Spec6.03b Conservation of momentum: 1D two particles

A small ball \(A\) is moving with velocity \((7\mathbf{i} + 12\mathbf{j})\) ms\(^{-1}\). It collides in mid-air with another ball \(B\), of mass \(0.4\) kg, moving with velocity \((-\mathbf{i} + 7\mathbf{j})\) ms\(^{-1}\). Immediately after the collision, \(A\) has velocity \((-3\mathbf{i} + 4\mathbf{j})\) ms\(^{-1}\) and \(B\) has velocity \((6.5\mathbf{i} + 13\mathbf{j})\) ms\(^{-1}\). Calculate the mass of \(A\). [4 marks]

AnswerMarks Guidance
\(m(7i + 12j) + 0 \cdot 4(-i + 7j) = m(-3i + 4j) + 0 \cdot 4(6 \cdot 5i + 13j)\)M1 A1
\(7m - 0 \cdot 4 = -3m + 2 \cdot 6\)M1 A1
\(10m = 3\)
\(m = 0.3\) Total: 4 marks
$m(7i + 12j) + 0 \cdot 4(-i + 7j) = m(-3i + 4j) + 0 \cdot 4(6 \cdot 5i + 13j)$ | M1 A1 |
$7m - 0 \cdot 4 = -3m + 2 \cdot 6$ | M1 A1 |
$10m = 3$ | |
$m = 0.3$ | | **Total: 4 marks**
A small ball $A$ is moving with velocity $(7\mathbf{i} + 12\mathbf{j})$ ms$^{-1}$. It collides in mid-air with another ball $B$, of mass $0.4$ kg, moving with velocity $(-\mathbf{i} + 7\mathbf{j})$ ms$^{-1}$. Immediately after the collision, $A$ has velocity $(-3\mathbf{i} + 4\mathbf{j})$ ms$^{-1}$ and $B$ has velocity $(6.5\mathbf{i} + 13\mathbf{j})$ ms$^{-1}$.

Calculate the mass of $A$. [4 marks]

\hfill \mbox{\textit{Edexcel M2  Q1 [4]}}