| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Collision with fixed wall |
| Difficulty | Standard +0.8 This M4 question requires understanding of oblique collisions with walls using perpendicular unit vectors (n and p directions), applying coefficient of restitution in the normal direction while preserving tangential velocity, then calculating energy loss. The vector decomposition framework and multi-step reasoning (verify decomposition, reverse the collision using e=3/5, compute kinetic energies) elevates this above routine mechanics problems, though the individual steps follow standard procedures once the setup is understood. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation6.02d Mechanical energy: KE and PE concepts6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a): \(\frac{2}{5}\mathbf{n} + \frac{13}{5}\mathbf{p} = \frac{2}{5}\left(-\frac{3}{4}\mathbf{i} + \frac{4}{3}\mathbf{j}\right) + \frac{13}{5}\left(\frac{3}{4}\mathbf{i} + \frac{4}{3}\mathbf{j}\right) = \frac{-23}{20}\mathbf{i} + \frac{232}{15}\mathbf{j} = \mathbf{i} + 3\mathbf{j}\) | M1 A1 | Correct calculation of components; final answer |
| Part (b): Velocity parallel to wall unchanged: \(v_1 = \frac{13}{5}\mathbf{p}\) | B1 | Recognition that parallel component is conserved |
| Newton's law of restitution perpendicular to wall: \(ev_2 = -\frac{2}{5}\mathbf{n}\) | M1 | Application of coefficient of restitution |
| Put in values: \(v_2 = -\frac{16}{5}\left(-\frac{3}{4}\mathbf{i} + \frac{3}{3}\mathbf{j}\right) = \frac{39}{20}\mathbf{i} - \frac{64}{15}\mathbf{j}\) | M1 A1 | Correct perpendicular velocity component |
| \(\mathbf{v}_1 + \mathbf{v}_2 = \frac{13}{5}\left(\frac{3}{4}\mathbf{i} + \frac{4}{3}\mathbf{j}\right) - \frac{16}{5}\left(-\frac{3}{4}\mathbf{i} + \frac{3}{3}\mathbf{j}\right) = 4\mathbf{i} - \mathbf{j}\) | A1 | Final velocity vector |
| Part (c): Change in KE: \(\Delta KE = \frac{1}{2} \times 1 \times (4^2 + 1^2) - \frac{1}{2} \times 1 \times (3^2 + 1^2) = 1.75\text{ J}\) | M1 A1 | Correct calculation of kinetic energy change |
**Part (a):** $\frac{2}{5}\mathbf{n} + \frac{13}{5}\mathbf{p} = \frac{2}{5}\left(-\frac{3}{4}\mathbf{i} + \frac{4}{3}\mathbf{j}\right) + \frac{13}{5}\left(\frac{3}{4}\mathbf{i} + \frac{4}{3}\mathbf{j}\right) = \frac{-23}{20}\mathbf{i} + \frac{232}{15}\mathbf{j} = \mathbf{i} + 3\mathbf{j}$ | M1 A1 | Correct calculation of components; final answer |
**Part (b):** Velocity parallel to wall unchanged: $v_1 = \frac{13}{5}\mathbf{p}$ | B1 | Recognition that parallel component is conserved |
Newton's law of restitution perpendicular to wall: $ev_2 = -\frac{2}{5}\mathbf{n}$ | M1 | Application of coefficient of restitution |
Put in values: $v_2 = -\frac{16}{5}\left(-\frac{3}{4}\mathbf{i} + \frac{3}{3}\mathbf{j}\right) = \frac{39}{20}\mathbf{i} - \frac{64}{15}\mathbf{j}$ | M1 A1 | Correct perpendicular velocity component |
$\mathbf{v}_1 + \mathbf{v}_2 = \frac{13}{5}\left(\frac{3}{4}\mathbf{i} + \frac{4}{3}\mathbf{j}\right) - \frac{16}{5}\left(-\frac{3}{4}\mathbf{i} + \frac{3}{3}\mathbf{j}\right) = 4\mathbf{i} - \mathbf{j}$ | A1 | Final velocity vector |
**Part (c):** Change in KE: $\Delta KE = \frac{1}{2} \times 1 \times (4^2 + 1^2) - \frac{1}{2} \times 1 \times (3^2 + 1^2) = 1.75\text{ J}$ | M1 A1 | Correct calculation of kinetic energy change |
---[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal perpendicular unit vectors.]
The vector $\mathbf{n} = (-\frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j})$ and the vector $\mathbf{p} = (-\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j})$ are perpendicular unit vectors.
\begin{enumerate}[label=(\alph*)]
\item Verify that $\frac{3}{5}\mathbf{n} + \frac{4}{5}\mathbf{p} = (\mathbf{i} + 3\mathbf{j})$. [2]
\end{enumerate}
A smooth uniform sphere $S$ of mass 0.5 kg is moving on a smooth horizontal plane when it collides with a fixed vertical wall which is parallel to $\mathbf{p}$. Immediately after the collision the velocity of $S$ is $(\mathbf{i} + 3\mathbf{j})$ m s$^{-1}$. The coefficient of restitution between $S$ and the wall is $\frac{3}{5}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in terms of $\mathbf{i}$ and $\mathbf{j}$, the velocity of $S$ immediately before the collision. [5]
\item Find the energy lost in the collision. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q5 [10]}}