| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Moderate -0.5 This is a straightforward mechanics problem requiring direct application of impulse-momentum theorem and constant acceleration equations. All three parts follow standard procedures: (i) impulse = change in momentum, (ii) work-energy or kinematics with constant force, (iii) impulse = force × time. No novel insight or complex problem-solving required, making it slightly easier than average A-level standard. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\pm 0.01) \times (320 - 20)\) | M1 A1 | if \(320 + 20\), can allow M1 but not MR |
| \(= (\pm) 3\) [N s\(^{-1}\)] | Total: [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1000d = \frac{1}{2} \times 0.01 \times (320^2 - 20^2)\) | M1 A1 | EITHER: Find thickness \(d\) by change in energy; M0 if \(\Delta v^2 = 320^2 + 20^2\) |
| \(d = 0.51\) [m] | ||
| \(d = \frac{1}{2}(320 + 20)t\) | (M1 A1) | OR: Find \(t\) from impulse and use \(t\) to find \(d\); M0 if \(320 - 20\) |
| \(= 170 \times 3/1000 = 0.51\) [m] | (M1 A1) | |
| \(a = 10^5\), \(d = 0.51\) [m] | OR: Find \(d\) using \(F = ma\), \(v^2 = u^2 + 2as\); Total: [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(d = \frac{1}{2}(320 + 20)t\), \(t = 0.003\) [s] | B1 | EITHER: Find time \(t\) from constant deceleration formula |
| \(1000t = 3\), \(t = 0.003\) [s] | (B1) | OR: Find time \(t\) from impulse; Total: [1] |
## Question 1:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\pm 0.01) \times (320 - 20)$ | M1 A1 | if $320 + 20$, can allow M1 but not MR |
| $= (\pm) 3$ [N s$^{-1}$] | | Total: [2] |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1000d = \frac{1}{2} \times 0.01 \times (320^2 - 20^2)$ | M1 A1 | EITHER: Find thickness $d$ by change in energy; M0 if $\Delta v^2 = 320^2 + 20^2$ |
| $d = 0.51$ [m] | | |
| $d = \frac{1}{2}(320 + 20)t$ | (M1 A1) | OR: Find $t$ from impulse and use $t$ to find $d$; M0 if $320 - 20$ |
| $= 170 \times 3/1000 = 0.51$ [m] | (M1 A1) | |
| $a = 10^5$, $d = 0.51$ [m] | | OR: Find $d$ using $F = ma$, $v^2 = u^2 + 2as$; Total: [2] |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $d = \frac{1}{2}(320 + 20)t$, $t = 0.003$ [s] | B1 | EITHER: Find time $t$ from constant deceleration formula |
| $1000t = 3$, $t = 0.003$ [s] | (B1) | OR: Find time $t$ from impulse; Total: [1] |
---1 A bullet of mass 0.01 kg is fired horizontally into a fixed vertical barrier which exerts a constant resisting force of magnitude 1000 N . The bullet enters the barrier with speed $320 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and emerges with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. You may assume that the motion takes place in a horizontal straight line. Find\\
(i) the magnitude of the impulse that acts on the bullet,\\
(ii) the thickness of the barrier,\\
(iii) the time taken for the bullet to pass through the barrier.
\hfill \mbox{\textit{CAIE FP2 2016 Q1 [5]}}