| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring integration of F=ma to find velocity, then solving equations and integrating again for displacement. While it has multiple parts and requires careful bookkeeping, the techniques are routine for this module with no novel problem-solving insight needed—slightly easier than average overall. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(F = ma\) | M1 | \(\frac{3}{2}t - 1 = 0.2\frac{\text{d}v}{\text{d}t}\); allow sign errors or \(m\) omitted |
| Integrate correctly | A1 | \(v = \frac{15}{4}t^2 - 5t(+c)\); allow if \(c\) missing or wrong |
| \(v = \frac{15}{4}t^2 - 5t + 0.8\) | A1 | oe |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use vel \(= 0.8\) | M1 | \(\frac{15}{4}t^2 - 5t + 0.8 = 0.8\); ft their (i) |
| \(t = 1.33\) (s) or \(1\frac{1}{3}\) (s) | A1 | Must come from correct equation for \(v\); Accept 4/3 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate to find \(x\) | M1* | At least 2 terms with powers increased by 1 |
| \(x = \frac{15}{12}t^3 - \frac{5}{2}t^2 + 0.8t\) | A1 | Need to state \(c = 0\), or use limits |
| Solve for \(x = 0\) | *M1 | |
| \(t = 1.6\) (s) or \(0.4\) (s) | A1 | Both answers needed; must be from correct work to find equation; Ignore \(t = 0\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x(3) - x(2)\) | M1 | Allow for \(x(2)\) or \(x(3)\) worked out from (iii); 13.65 or 1.6 |
| Distance is \(12.05\) (m) | A1 | Accept 12 or 12.1 |
| [2] |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $F = ma$ | M1 | $\frac{3}{2}t - 1 = 0.2\frac{\text{d}v}{\text{d}t}$; allow sign errors or $m$ omitted |
| Integrate correctly | A1 | $v = \frac{15}{4}t^2 - 5t(+c)$; allow if $c$ missing or wrong |
| $v = \frac{15}{4}t^2 - 5t + 0.8$ | A1 | oe |
| **[3]** | | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use vel $= 0.8$ | M1 | $\frac{15}{4}t^2 - 5t + 0.8 = 0.8$; ft their (i) |
| $t = 1.33$ (s) or $1\frac{1}{3}$ (s) | A1 | Must come from correct equation for $v$; Accept 4/3 |
| **[2]** | | |
## Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to find $x$ | M1* | At least 2 terms with powers increased by 1 |
| $x = \frac{15}{12}t^3 - \frac{5}{2}t^2 + 0.8t$ | A1 | Need to state $c = 0$, or use limits |
| Solve for $x = 0$ | *M1 | |
| $t = 1.6$ (s) or $0.4$ (s) | A1 | Both answers needed; must be from correct work to find equation; Ignore $t = 0$ |
| **[4]** | | |
## Question 3(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x(3) - x(2)$ | M1 | Allow for $x(2)$ or $x(3)$ worked out from (iii); 13.65 or 1.6 |
| Distance is $12.05$ (m) | A1 | Accept 12 or 12.1 |
| **[2]** | | |3 A particle $P$ of mass 0.2 kg moves on a smooth horizontal plane. Initially it is projected with velocity $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a fixed point $O$ towards another fixed point $A$. At time $t$ s after projection, $P$ is $x \mathrm {~m}$ from $O$ and is moving with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, with the direction $O A$ being positive. A force of $( 1.5 t - 1 ) \mathrm { N }$ acts on $P$ in the direction parallel to $O A$.\\
(i) Find an expression for $v$ in terms of $t$.\\
(ii) Find the time when the velocity of $P$ is next $0.8 \mathrm {~ms} ^ { - 1 }$.\\
(iii) Find the times when $P$ subsequently passes through $O$.\\
(iv) Find the distance $P$ travels in the third second of its motion.
\hfill \mbox{\textit{OCR M3 2013 Q3 [11]}}