| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision with mass ratio parameter |
| Difficulty | Standard +0.3 This is a standard two-part momentum and collisions question requiring conservation of momentum and Newton's experimental law. Part (i) involves algebraic manipulation to derive an inequality, while part (ii) applies the coefficient of restitution formula and impulse calculation. The techniques are routine for Further Maths mechanics, though the mass ratio parameter adds mild algebraic complexity. Slightly easier than average A-level due to straightforward application of standard formulas. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(0.1v_A + mv_B = 0.1 \times 5 - m \times 2\) | M1 | Use conservation of momentum |
| \(m = (0.5 - 0.1v_A)/(2 + v_B)\) | A1 | Find \(m\) |
| \(v_B > 0\), \(m < 0.5/2 = 0.25\) A.G. | M1 A1 | Use \(v_A > 0\) to find lower bound on \(m\); Part total: 4 |
| (ii) \(v_B - v_A = \frac{1}{2}(2 + 5) = 7/2\) | M1 A1 | Use Newton's law of restitution |
| \(2v_B + v_A = 1\), \(v_A = -2\) or \(v_B = 1.5\) | M1 A1 | Put \(m = 0.2\) and find one of \(v_A\), \(v_B\) |
| \(0.1(5 - v_A)\) or \(0.2(1.5 + 2) = 0.7\) | M1 A1 | Find magnitude of impulse [N s]; Part total: 6 |
| Total: 10 |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $0.1v_A + mv_B = 0.1 \times 5 - m \times 2$ | M1 | Use conservation of momentum |
| $m = (0.5 - 0.1v_A)/(2 + v_B)$ | A1 | Find $m$ |
| $v_B > 0$, $m < 0.5/2 = 0.25$ **A.G.** | M1 A1 | Use $v_A > 0$ to find lower bound on $m$; **Part total: 4** |
| **(ii)** $v_B - v_A = \frac{1}{2}(2 + 5) = 7/2$ | M1 A1 | Use Newton's law of restitution |
| $2v_B + v_A = 1$, $v_A = -2$ or $v_B = 1.5$ | M1 A1 | Put $m = 0.2$ and find one of $v_A$, $v_B$ |
| $0.1(5 - v_A)$ or $0.2(1.5 + 2) = 0.7$ | M1 A1 | Find magnitude of impulse [N s]; **Part total: 6** |
| | | **Total: 10** |
---4 Two smooth spheres $A$ and $B$, of equal radii, have masses 0.1 kg and $m \mathrm {~kg}$ respectively. They are moving towards each other in a straight line on a smooth horizontal table and collide directly. Immediately before collision the speed of $A$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the speed of $B$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Assume that in the collision $A$ does not change direction. The speeds of $A$ and $B$ after the collision are $v _ { A } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $v _ { B } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. Express $m$ in terms of $v _ { A }$ and $v _ { B }$, and hence show that $m < 0.25$.\\
(ii) Assume instead that $m = 0.2$ and that the coefficient of restitution between the spheres is $\frac { 1 } { 2 }$. Find the magnitude of the impulse acting on $A$ in the collision.
\hfill \mbox{\textit{CAIE FP2 2008 Q4 [10]}}