| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (graph-based) |
| Difficulty | Standard +0.3 This is a standard two-particle kinematics problem using velocity-time graphs. Students must find when A decelerates (simple calculation: 8 = 10 - 5t gives t = 0.4s after deceleration starts) and use areas under graphs to find initial separation. Requires understanding that area = distance and careful bookkeeping of the 1m offset, but follows routine M1 procedures without novel insight. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((10-8)/5 = T_\text{dec}\) OR \(8 = 10 - 5T_\text{dec}\) | M1 | Attempt to find \(T_\text{dec} = \pm0.4 = \pm2/5\) |
| \(t\) \((= 2 - 0.4) = 1.6\) | A1 | Exact. Accept 1 3/5, not 8/5, www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_B = \frac{1}{2} \times 8 \times 2\) | B1 | \(S_B = 8\) |
| \(S_A = 10 \times 1.6 + \frac{1}{2} \times (10+8) \times 0.4\) OR \(S_A = 10 \times 2 - \frac{1}{2} \times (2-1.6) \times (10-8)\) | M1 | Using area under graph is distance (at least two parts). Complete method for \(S_A\) run in first 2s, using \(\text{cv}(t)\) |
| \(S_A = 19.6\) | A1 | Accept as 16+3.6 or 20-0.40, from \(t=1.6\) |
| \(AB = 19.6 - 8 + 1\) | M1 | \(AB = +/-(S_A - S_B +/- 1)\) |
| \(AB = 12.6 \text{ m}\) | A1 | Exact. Or \(AB = -12.6\) m |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(10-8)/5 = T_\text{dec}$ OR $8 = 10 - 5T_\text{dec}$ | M1 | Attempt to find $T_\text{dec} = \pm0.4 = \pm2/5$ |
| $t$ $(= 2 - 0.4) = 1.6$ | A1 | Exact. Accept 1 3/5, not 8/5, www |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_B = \frac{1}{2} \times 8 \times 2$ | B1 | $S_B = 8$ |
| $S_A = 10 \times 1.6 + \frac{1}{2} \times (10+8) \times 0.4$ OR $S_A = 10 \times 2 - \frac{1}{2} \times (2-1.6) \times (10-8)$ | M1 | Using area under graph is distance (at least two parts). Complete method for $S_A$ run in first 2s, using $\text{cv}(t)$ |
| $S_A = 19.6$ | A1 | Accept as 16+3.6 or 20-0.40, from $t=1.6$ |
| $AB = 19.6 - 8 + 1$ | M1 | $AB = +/-(S_A - S_B +/- 1)$ |
| $AB = 12.6 \text{ m}$ | A1 | Exact. Or $AB = -12.6$ m |
---3\\
$\mathrm { v } \left( \mathrm { ms } ^ { - 1 } \right)$\\
\includegraphics[max width=\textwidth, alt={}, center]{f0813713-d677-4ed7-87e1-971a64bdb6ff-2_449_1121_1500_440}\\
not to scale
The diagram shows the $( t , v )$ graphs for two athletes, $A$ and $B$, who run in the same direction in the same straight line while they exchange the baton in a relay race. $A$ runs with constant velocity $10 \mathrm {~ms} ^ { - 1 }$ until he decelerates at $5 \mathrm {~ms} ^ { - 2 }$ and subsequently comes to rest. $B$ has constant acceleration from rest until reaching his constant speed of $10 \mathrm {~ms} ^ { - 1 }$. The baton is exchanged 2 s after $B$ starts running, when both athletes have speed $8 \mathrm {~ms} ^ { - 1 }$ and $B$ is 1 m ahead of $A$.\\
(i) Find the value of $t$ at which $A$ starts to decelerate.\\
(ii) Calculate the distance between $A$ and $B$ at the instant when $B$ starts to run.
\hfill \mbox{\textit{OCR M1 2012 Q3 [7]}}