| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Coalescence collision |
| Difficulty | Standard +0.3 This is a standard mechanics question requiring conservation of momentum for coalescence, impulse calculation, and kinetic energy loss. While it involves vectors and multiple parts (7 marks total), each step follows routine procedures: proving collision by equating position vectors, applying momentum conservation m₁v₁ + m₂v₂ = (m₁+m₂)v, finding impulse from momentum change, and calculating KE before/after. No novel insight required, just systematic application of standard formulae. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(\mathbf{r}_A = 11\mathbf{i} + 6\mathbf{j} + (2\mathbf{i} + 7\mathbf{j})t\) | M1 | AO3 |
| \(\mathbf{r}_B = 7\mathbf{i} + 10\mathbf{j} + (5\mathbf{i} + 4\mathbf{j})t\) | A1 | AO1 |
| If particles collide, \(\mathbf{r}_A = \mathbf{r}_B\) for some value of \(t\) | ||
| For \(\mathbf{i}\) component: \(11 + 2t = 7 + 5t\), \(t = \frac{4}{3}\) | M1 | AO2 |
| For \(\mathbf{j}\) component: \(6 + 7t = 10 + 4t\), \(t = \frac{4}{3}\) | A1 | AO2 |
| Since the value for \(t\) for both components are equal, the particles collide. | A1 | AO2 |
| Conservation of momentum: \(m(2\mathbf{i} + 7\mathbf{j}) + 2m(5\mathbf{i} + 4\mathbf{j}) = 3m(x\mathbf{i} + y\mathbf{j})\) | M1 | AO3 |
| \(12\mathbf{i} + 15\mathbf{j} = 3x\mathbf{i} + 3y\mathbf{j}\) | A1 | AO2 |
| \(x = 4,\ y = 5\) | m1 | AO2 |
| \(x\mathbf{i} + y\mathbf{j} = 4\mathbf{i} + 5\mathbf{j}\) (Ns) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(\mathbf{I}\) = change in momentum | M1 | AO3, used |
| \(\mathbf{I} = 2m(4\mathbf{i} + 5\mathbf{j}) - 2m(5\mathbf{i} + 4\mathbf{j})\) | ||
| \(\mathbf{I} = m(-2\mathbf{i} + 2\mathbf{j})\) | ||
| \(\mathbf{I} = 2m(-\mathbf{i} + \mathbf{j})\) (Ns) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Loss in KE \(= \frac{1}{2}m(4+49) + \frac{1}{2}2m(25+16) - \frac{1}{2}\times 3m(16+25)\) | M1 | AO3 |
| Loss in KE \(= 6m\) (J) | A1 | AO1 |
| Total | [13] |
# Question 5:
## Part (a)
| Answer | Mark | Notes |
|--------|------|-------|
| $\mathbf{r}_A = 11\mathbf{i} + 6\mathbf{j} + (2\mathbf{i} + 7\mathbf{j})t$ | M1 | AO3 |
| $\mathbf{r}_B = 7\mathbf{i} + 10\mathbf{j} + (5\mathbf{i} + 4\mathbf{j})t$ | A1 | AO1 |
| If particles collide, $\mathbf{r}_A = \mathbf{r}_B$ for some value of $t$ | | |
| For $\mathbf{i}$ component: $11 + 2t = 7 + 5t$, $t = \frac{4}{3}$ | M1 | AO2 |
| For $\mathbf{j}$ component: $6 + 7t = 10 + 4t$, $t = \frac{4}{3}$ | A1 | AO2 |
| Since the value for $t$ for both components are equal, the particles collide. | A1 | AO2 |
| Conservation of momentum: $m(2\mathbf{i} + 7\mathbf{j}) + 2m(5\mathbf{i} + 4\mathbf{j}) = 3m(x\mathbf{i} + y\mathbf{j})$ | M1 | AO3 |
| $12\mathbf{i} + 15\mathbf{j} = 3x\mathbf{i} + 3y\mathbf{j}$ | A1 | AO2 |
| $x = 4,\ y = 5$ | m1 | AO2 |
| $x\mathbf{i} + y\mathbf{j} = 4\mathbf{i} + 5\mathbf{j}$ (Ns) | A1 | AO1 |
## Part (b)
| Answer | Mark | Notes |
|--------|------|-------|
| $\mathbf{I}$ = change in momentum | M1 | AO3, used |
| $\mathbf{I} = 2m(4\mathbf{i} + 5\mathbf{j}) - 2m(5\mathbf{i} + 4\mathbf{j})$ | | |
| $\mathbf{I} = m(-2\mathbf{i} + 2\mathbf{j})$ | | |
| $\mathbf{I} = 2m(-\mathbf{i} + \mathbf{j})$ (Ns) | A1 | AO1 |
## Part (c)
| Answer | Mark | Notes |
|--------|------|-------|
| Loss in KE $= \frac{1}{2}m(4+49) + \frac{1}{2}2m(25+16) - \frac{1}{2}\times 3m(16+25)$ | M1 | AO3 |
| Loss in KE $= 6m$ (J) | A1 | AO1 |
| **Total** | **[13]** | |
---5. A particle $A$, of mass $m \mathrm {~kg}$, has position vector $11 \mathbf { i } + 6 \mathbf { j }$ and a velocity $2 \mathbf { i } + 7 \mathbf { j }$. At the same moment, second particle $B$, of mass $2 m \mathrm {~kg}$, has position vector $7 \mathbf { i } + 10 \mathbf { j }$ and a velocity $5 \mathbf { i } + 4 \mathbf { j }$.
\begin{enumerate}[label=(\alph*)]
\item If the particles continue to move with these velocities, prove that the particles will collide. Given that the particles coalesce after collision, find the common velocity of the particles after collision.
\item Determine the impulse exerted by $A$ on $B$.
\item Calculate the loss of kinetic energy caused by the collision.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 Q5 [13]}}