| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics question on sequential collisions requiring conservation of momentum, Newton's restitution law, and kinetic energy calculations. Part (a) is routine application of collision formulas with equal masses, part (b) is straightforward energy calculation, and part (c) requires checking a condition for no further collisions but follows a predictable method. While it has multiple parts and requires careful algebra, it involves well-practiced techniques without requiring novel insight, making it slightly easier than average for Further Maths. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i 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 |
|---|---|---|
| (a) Conservation of momentum: | M1 | Allow 1 sign error |
| \(mu + 0 = mv_A + mv_B\) | A1 | All correct |
| Restitution: | M1 | Allow one sign error |
| \(v_B - v_A = -e(-u)\) | A1 | All correct, any form |
| \(v_A + v_B = u\) | ||
| \(-v_A + v_B = eu\) | ||
| \(2v_A = (1 - e)u\) | m1 | One variable eliminated |
| \(v_A = \frac{1}{2}(1 - e)u\) | A1 | cao, oe |
| \(v_B = \frac{1}{2}(1 + e)u\) | A1 | cao, oe |
| Answer | Marks | Guidance |
|---|---|---|
| (b) Loss in KE \(= \frac{1}{2}mu^2 - \frac{1}{2}m\left[\left(\frac{1}{2}u\right)^2 + \left(\frac{1}{2}u\right)^2\right]\) | M1 | |
| \(= \frac{1}{2}mu^2\left(1 - \frac{5}{8}\right) = \frac{3}{16}mu^2\) (1) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| (c) Velocity of \(B\) after 2\(^{\text{nd}}\) collision \(= \frac{1}{2}(1 - e_1) \times \frac{3}{4}u\) | M1 | FT (a) |
| For no further collisions to occur, | ||
| Vel. of \(B\) after 2\(^{\text{nd}}\) collision | ||
| \(\geq\) Vel. of \(A\) after 1\(^{\text{st}}\) collision | ||
| \(\frac{1}{2}(1 - e_1) \times \frac{3}{4}u \geq \frac{1}{4}u\) | M1 | FT (a) |
| \(3 - 3e_1 \geq 2\) | ||
| \(e_1 \leq \frac{1}{3}\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| Velocity of \(B\) after 2\(^{\text{nd}}\) collision \(= \frac{1}{2}(1 - e_1) \times \frac{3}{4}u\) | (M1) | FT (a) |
| If \(e_1 \leq \frac{1}{3}\) then \(1 - e_1 \geq \frac{2}{3}\) | (M1) | FT (a) |
| Vel. of \(B\) after 2\(^{\text{nd}}\) collis \(\geq \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}u = \frac{1}{4}u = v_A\) | (M1) | Convincing |
| Vel. of \(B\) after 2\(^{\text{nd}}\) collision | ||
| \(\geq\) Vel. of \(A\) after 1\(^{\text{st}}\) collision |
**(a)** Conservation of momentum: | M1 | Allow 1 sign error
$mu + 0 = mv_A + mv_B$ | A1 | All correct
Restitution: | M1 | Allow one sign error
$v_B - v_A = -e(-u)$ | A1 | All correct, any form
$v_A + v_B = u$ | |
$-v_A + v_B = eu$ | |
$2v_A = (1 - e)u$ | m1 | One variable eliminated
$v_A = \frac{1}{2}(1 - e)u$ | A1 | cao, oe
$v_B = \frac{1}{2}(1 + e)u$ | A1 | cao, oe
**Subtotal: [7]**
**(b)** Loss in KE $= \frac{1}{2}mu^2 - \frac{1}{2}m\left[\left(\frac{1}{2}u\right)^2 + \left(\frac{1}{2}u\right)^2\right]$ | M1 |
$= \frac{1}{2}mu^2\left(1 - \frac{5}{8}\right) = \frac{3}{16}mu^2$ (1) | A1 | cao
**Subtotal: [2]**
**(c)** Velocity of $B$ after 2$^{\text{nd}}$ collision $= \frac{1}{2}(1 - e_1) \times \frac{3}{4}u$ | M1 | FT (a)
For no further collisions to occur, | |
Vel. of $B$ after 2$^{\text{nd}}$ collision | |
$\geq$ Vel. of $A$ after 1$^{\text{st}}$ collision | |
$\frac{1}{2}(1 - e_1) \times \frac{3}{4}u \geq \frac{1}{4}u$ | M1 | FT (a)
$3 - 3e_1 \geq 2$ | |
$e_1 \leq \frac{1}{3}$ | A1 | Convincing
**Subtotal: [3]**
**Alternative solution**
Velocity of $B$ after 2$^{\text{nd}}$ collision $= \frac{1}{2}(1 - e_1) \times \frac{3}{4}u$ | (M1) | FT (a)
If $e_1 \leq \frac{1}{3}$ then $1 - e_1 \geq \frac{2}{3}$ | (M1) | FT (a)
Vel. of $B$ after 2$^{\text{nd}}$ collis $\geq \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}u = \frac{1}{4}u = v_A$ | (M1) | Convincing
Vel. of $B$ after 2$^{\text{nd}}$ collision | |
$\geq$ Vel. of $A$ after 1$^{\text{st}}$ collision | |
**Subtotal: [3]**
**Total for Question 7: 12**7. Three spheres $A , B , C$, of equal radii and each of mass $m \mathrm {~kg}$, lie at rest on a smooth horizontal surface such that their centres are in a straight line with $B$ between $A$ and $C$. The coefficient of restitution between $A$ and $B$ is $e$. Sphere $A$ is projected towards $B$ with speed $u \mathrm {~ms} ^ { - 1 }$ so that it collides with $B$.
\begin{enumerate}[label=(\alph*)]
\item Find expressions, in terms of $e$ and $u$, for the speed of $A$ and the speed of $B$ after they collide.
You are now given that $e = \frac { 1 } { 2 }$.
\item Find, in terms of $m$ and $u$, the loss in kinetic energy due to the collision between $A$ and $B$.
\item After the collision between $A$ and $B$, sphere $B$ then collides with $C$. The coefficient of restitution between $B$ and $C$ is $e _ { 1 }$. Show that there will be no further collisions if $e _ { 1 } \leqslant \frac { 1 } { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2019 Q7 [12]}}