| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on piecewise motion with non-constant acceleration. It requires systematic integration (a→v→s) with careful attention to initial conditions and piecewise definitions, but follows predictable patterns. Part (v) adds mild problem-solving through comparison of scenarios, but the calculations remain routine. Slightly easier than average due to straightforward integration and clear structure. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate \(a\) to obtain \(v\) | M1 | Attempt to integrate |
| \(v = 10t - \frac{1}{2}t^2\) \((+c)\) | A1 | |
| \(t = 10 \Rightarrow v = 100 - 50 = 50\) | M1 | Substitution of \(t = 10\) to find \(v\) |
| Since \(a = 0\) for \(t > 10\), \(v = 50\) for \(t > 10\) | A1 | Sound argument required for given answer. It must in some way refer to \(a = 0\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Continuous two part \(v\)-\(t\) graph | B1 | The graph must cover \(t = 0\) to \(t = 20\) |
| Curve for \(0 \leq t \leq 10\) | B1 | |
| Horizontal straight line for \(10 \leq t \leq 20\) | B1 | B0 if no vertical scale is given |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance fallen \(= \int\left(10t - \frac{1}{2}t^2\right)dt\) | M1 | Attempt to integrate |
| \(d = 5t^2 - \frac{1}{6}t^3 + c \quad (c = 0)\) | A1 | |
| Height \(= 1000 - 5t^2 + \frac{1}{6}t^3\) | A1 | This mark should only be given if the signs are correctly obtained |
| When \(t = 10\), \(h = 667\) | B1 | oe |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Time at constant vel \(= 667 \div 50 = 13.3\) | B1 | FT for \(h\) from part (iii) |
| Total time \(t = 10 + 13.3 = 23.3\) | B1 | FT |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Since \(500 > 333\) | M1 | For finding the height at which the crate reaches terminal velocity, e.g. \(h = 167\), or equivalent relevant calculation. FT for \(h\) from part (iii) if used |
| The box will have reached terminal speed. So there is no improvement | A1 | Allow either one (or both) of these two statements |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 10t - t^2\) (for \(t \leq 5\)) | M1 | Integration to find \(v\) |
| Terminal velocity is \(25\ \text{ms}^{-1}\) | A1 | |
| So better | A1 | |
| [3] |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate $a$ to obtain $v$ | M1 | Attempt to integrate |
| $v = 10t - \frac{1}{2}t^2$ $(+c)$ | A1 | |
| $t = 10 \Rightarrow v = 100 - 50 = 50$ | M1 | Substitution of $t = 10$ to find $v$ |
| Since $a = 0$ for $t > 10$, $v = 50$ for $t > 10$ | A1 | Sound argument required for given answer. It must in some way refer to $a = 0$ |
| **[4]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Continuous two part $v$-$t$ graph | B1 | The graph must cover $t = 0$ to $t = 20$ |
| Curve for $0 \leq t \leq 10$ | B1 | |
| Horizontal straight line for $10 \leq t \leq 20$ | B1 | B0 if no vertical scale is given |
| **[3]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance fallen $= \int\left(10t - \frac{1}{2}t^2\right)dt$ | M1 | Attempt to integrate |
| $d = 5t^2 - \frac{1}{6}t^3 + c \quad (c = 0)$ | A1 | |
| Height $= 1000 - 5t^2 + \frac{1}{6}t^3$ | A1 | This mark should only be given if the signs are correctly obtained |
| When $t = 10$, $h = 667$ | B1 | oe |
| **[4]** | | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Time at constant vel $= 667 \div 50 = 13.3$ | B1 | FT for $h$ from part (iii) |
| Total time $t = 10 + 13.3 = 23.3$ | B1 | FT |
| **[2]** | | |
### Part (v)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Since $500 > 333$ | M1 | For finding the height at which the crate reaches terminal velocity, e.g. $h = 167$, or equivalent relevant calculation. FT for $h$ from part (iii) if used |
| The box will have reached terminal speed. So there is no improvement | A1 | Allow either one (or both) of these two statements |
| **[2]** | | |
### Part (v)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 10t - t^2$ (for $t \leq 5$) | M1 | Integration to find $v$ |
| Terminal velocity is $25\ \text{ms}^{-1}$ | A1 | |
| So better | A1 | |
| **[3]** | | |
---2 A box of emergency supplies is dropped to victims of a natural disaster from a stationary helicopter at a height of 1000 metres. The initial velocity of the box is zero.
At time $t \mathrm {~s}$ after being dropped, the acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, of the box in the vertically downwards direction is modelled by
$$\begin{aligned}
& a = 10 - t \text { for } 0 \leqslant t \leqslant 10 \\
& a = 0 \quad \text { for } \quad t > 10
\end{aligned}$$
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of the box in the vertically downwards direction in terms of $t$ for $0 \leqslant t \leqslant 10$.
Show that for $t > 10 , v = 50$.
\item Draw a sketch graph of $v$ against $t$ for $0 \leqslant t \leqslant 20$.
\item Show that the height, $h \mathrm {~m}$, of the box above the ground at time $t$ s is given, for $0 \leqslant t \leqslant 10$, by
$$h = 1000 - 5 t ^ { 2 } + \frac { 1 } { 6 } t ^ { 3 }$$
Find the height of the box when $t = 10$.
\item Find the value of $t$ when the box hits the ground.
\item Some of the supplies in the box are damaged when the box hits the ground. So measures are considered to reduce the speed with which the box hits the ground the next time one is dropped. Two different proposals are made. Carry out suitable calculations and then comment on each of them.\\
(A) The box should be dropped from a height of 500 m instead of 1000 m .\\
(B) The box should be fitted with a parachute so that its acceleration is given by
$$\begin{gathered}
\quad a = 10 - 2 t \text { for } 0 \leqslant t \leqslant 5 , \\
a = 0 \quad \text { for } \quad t > 5 .
\end{gathered}$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q2 [18]}}