| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix conformability and dimensions |
| Difficulty | Easy -1.8 The text is severely corrupted and unreadable, making it impossible to assess the actual mathematical content. However, based on the topic 'Matrix conformability and dimensions' for CAIE FP2, this would typically involve basic matrix multiplication rules and determining valid dimensions - a routine procedural skill requiring only recall of conformability conditions (columns of first = rows of second). Such questions are generally straightforward and below average difficulty for Further Maths students. |
| Spec | 3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mv_A + kmv_B = mu + \frac{2}{3}kmu\) | B1 | Conservation of momentum |
| \(v_A + kv_B = u(1 + \frac{2}{3}k)\) | B1 | |
| \(v_A - v_B = -\frac{4}{5}(u - \frac{2}{3}u)\) | M1 | Restitution (4/5 on wrong side is M0; signs inconsistent with prev. eqn is A0) |
| \(v_A - v_B = -\frac{4u}{15}\) | A1 | |
| \((1+k)v_A = u(1 + \frac{2}{3}k - \frac{4k}{15})\) | M1 | Solve for \(v_A\) (allow verification) |
| \(v_A = \frac{u(2k+5)}{5(k+1)}\) A.G. | A1 | |
| \([v_B = \frac{u(10k+19)}{15(k+1)}]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mu - \frac{2}{5}mu = mv_A\) | M1 | Equate impulse to momentum change for \(A\) |
| \(\frac{3}{5} = \frac{(2k+5)}{5(k+1)}\), \(k=2\) | A1 | |
| *OR B:* \(\frac{2}{3}kmu + \frac{2}{5}mu = kmv_B\) | M1 | |
| \(\frac{2}{3}k + \frac{2}{5} = \frac{k(10k+19)}{15(k+1)}\) | ||
| \(10k^2 + 16k + 6 = 10k^2 + 19k\), \(k=2\) | A1 |
# Question 1:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mv_A + kmv_B = mu + \frac{2}{3}kmu$ | B1 | Conservation of momentum |
| $v_A + kv_B = u(1 + \frac{2}{3}k)$ | B1 | |
| $v_A - v_B = -\frac{4}{5}(u - \frac{2}{3}u)$ | M1 | Restitution (4/5 on wrong side is M0; signs inconsistent with prev. eqn is A0) |
| $v_A - v_B = -\frac{4u}{15}$ | A1 | |
| $(1+k)v_A = u(1 + \frac{2}{3}k - \frac{4k}{15})$ | M1 | Solve for $v_A$ (allow verification) |
| $v_A = \frac{u(2k+5)}{5(k+1)}$ **A.G.** | A1 | |
| $[v_B = \frac{u(10k+19)}{15(k+1)}]$ | | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mu - \frac{2}{5}mu = mv_A$ | M1 | Equate impulse to momentum change for $A$ |
| $\frac{3}{5} = \frac{(2k+5)}{5(k+1)}$, $k=2$ | A1 | |
| *OR B:* $\frac{2}{3}kmu + \frac{2}{5}mu = kmv_B$ | M1 | |
| $\frac{2}{3}k + \frac{2}{5} = \frac{k(10k+19)}{15(k+1)}$ | | |
| $10k^2 + 16k + 6 = 10k^2 + 19k$, $k=2$ | A1 | |
---1 Two small smooth spheres $A$ and $B$ have equal radii and have masses $m$ and $k m$ respectively. They are moving in a straight line in the same direction on a smooth horizontal table. The speed of $A$ is $u$ and the speed of $B$ is $\frac { 2 } { 3 } u$. Sphere $A$ collides directly with sphere $B$. The coefficient of restitution between the spheres is $\frac { 4 } { 5 }$.\\
(i) Show that the speed of $A$ after the collision is $\frac { u ( 2 k + 5 ) } { 5 ( k + 1 ) }$.\\
(ii) Given that the magnitude of the impulse experienced by $B$ during the collision is $\frac { 2 } { 5 } m u$, find the value of $k$.
\hfill \mbox{\textit{CAIE FP2 2014 Q1}}