| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Difficulty | Standard +0.8 This is a two-part mechanics problem requiring impulse-momentum theorem (straightforward) and then a collision problem with two simultaneous equations from conservation of momentum and the given energy loss condition. Part (b) requires solving a quadratic or system of equations with careful attention to direction signs, which is moderately challenging but follows standard Further Maths mechanics techniques. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Impulse = \(m_B v_B - m_B u_B\) (change in momentum) | M1 | Used |
| Answer | Marks | Guidance |
|---|---|---|
| \( | Impulse | = 32\) (Ns) |
| Opposite direction to original motion | A1 | |
| (b) Conservation of momentum: | M1 | Attempted |
| \((4)(3) + (-6)(2) = 3v_A + 2v_B\) | A1 | All correct |
| \(3v_A + 2v_B = 0\) (\(v_A = -\frac{2}{3}v_B\)) | ||
| Loss in Kinetic energy: | M1 | Attempted |
| \[\frac{1}{2}(3)(3)^2 + \frac{1}{2}(2)(-6)^2 = \frac{1}{2}(3)v_A^2 + \frac{1}{2}(2)v_B^2 = 45\] | A1 | Before = 60, After = \(\frac{1}{2}v_A^2 + v_B^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \[3\left(\frac{-2v_B}{3}\right)^2 + 2v_B^2 = 30\] | m1 | One variable eliminated |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_A = -\frac{2}{3}(\pm 3) = \mp 2\) | m1 | |
| \((v_A, v_B) = (2, -3)\) or \((-2, 3)\) | ||
| speed \(v_A = 2\) (ms\(^{-1}\)), speed \(v_B = 3\) (ms\(^{-1}\)) | A1 | cao |
| Both objects are moving in the opposite direction to their original motion | A1 |
(a) Impulse = $m_B v_B - m_B u_B$ (change in momentum) | M1 | Used
$= 2(-6 - 10)$
$|Impulse| = 32$ (Ns) | A1 |
Opposite direction to original motion | A1 |
(b) Conservation of momentum: | M1 | Attempted
$(4)(3) + (-6)(2) = 3v_A + 2v_B$ | A1 | All correct
$3v_A + 2v_B = 0$ ($v_A = -\frac{2}{3}v_B$) |
Loss in Kinetic energy: | M1 | Attempted
$$\frac{1}{2}(3)(3)^2 + \frac{1}{2}(2)(-6)^2 = \frac{1}{2}(3)v_A^2 + \frac{1}{2}(2)v_B^2 = 45$$ | A1 | Before = 60, After = $\frac{1}{2}v_A^2 + v_B^2$
$60 - \frac{3}{2}v_A^2 - v_B^2 = 45$
$3v_A^2 + 2v_B^2 = 30$
$$3\left(\frac{-2v_B}{3}\right)^2 + 2v_B^2 = 30$$ | m1 | One variable eliminated
$v_B = \pm 3$ or $v_A = \pm 2$
$v_A = -\frac{2}{3}(\pm 3) = \mp 2$ | m1 |
$(v_A, v_B) = (2, -3)$ or $(-2, 3)$ |
speed $v_A = 2$ (ms$^{-1}$), speed $v_B = 3$ (ms$^{-1}$) | A1 | cao
Both objects are moving in the opposite direction to their original motion | A1 |
---The diagram below shows two spheres $A$ and $B$, of equal radii, moving in the same direction on a smooth horizontal surface. Sphere $A$, of mass $3$ kg, is moving with speed $4$ ms$^{-1}$ and sphere $B$, of mass $2$ kg, is moving with speed $10$ ms$^{-1}$.
\includegraphics{figure_5}
Sphere $B$ is then given an impulse after which it moves in the opposite direction with speed $6$ ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the magnitude and direction of the impulse exerted on $B$. [3]
\end{enumerate}
Sphere $B$ continues to move with speed $6$ ms$^{-1}$ so that it collides directly with sphere $A$. The kinetic energy lost due to the collision is $45$ J.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate the speed of $A$ and the speed of $B$ immediately after the two spheres collide. State the direction in which each sphere is moving relative to its motion immediately before the collision. [8]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2023 Q5 [11]}}