| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Reaction time and stopping distance |
| Difficulty | Standard +0.3 This is a straightforward M1 mechanics question requiring standard SUVAT equations and Newton's second law. Part (i) is a 'show that' using v²=u²+2as then F=ma. Parts (ii)-(vi) are routine applications of the same principles with no conceptual challenges—slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.02z Models in context: use functions in modelling3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 - u^2 = 2as\) | ||
| \(0^2 - 40^2 = 2 \times a \times 125\) | M1 | Substitution required. For \(u\), \(v\) interchange award up to M1 A0 |
| \(\Rightarrow a = -6.4\) | A1 | Condone no – sign |
| \(F = ma\) | M1 | |
| \(F = 800 \times (-)6.4 = (-)5120\) | E1 | Allow +5120 or −5120 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = u + at\) | ||
| \(0 = 40 - 6.4 \times t\) | M1 | FT for \(a\) |
| \(t = 6.25\) — It takes 6.25 seconds to stop | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = \frac{1}{2}(u+v)t\) | ||
| \(125 = \frac{1}{2}(40+0) \times t\) | (M1) | |
| \(t = 6.25\) — it takes 6.25 seconds to stop | (A1) | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = ut + \frac{1}{2}at^2\) | ||
| \(125 = 40t + \frac{1}{2} \times (-6.4)t^2\) | (M1) | |
| \(3.2t^2 - 40t + 125 = 0\) | ||
| \(t = 6.25\) | (A1) | |
| ([2]) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reaction distance \(< 155 - 125 = 30\) m | M1 | 30 must be seen and used |
| Time taken to travel 30 m at 40 m s\(^{-1}\) is 0.75 s | E1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance travelled before braking \(= 20 \times 0.675 = 13.5\) m | B1 | |
| Distance travelled while braking \(= \frac{20^2}{2 \times 6.4} = 31.25\) m | B1 | |
| Stopping distance \(= 13.5 + 31.25 = 44.75\) m | B1 | cao |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance during reaction time is not affected by slope. It is \(20 \times 0.675 = 13.5\) m | 13.5 is rewarded later | |
| Component of car's weight down the slope \(= mg\sin\alpha = 800 \times 9.8 \times \sin 5°\ (= 683.3\text{ N})\) | M1 | Allow cos for sin for M1. Allow omission of \(g\) for this mark only |
| \(= 683.3\) N | A1 | cao |
| Force opposing motion when brakes applied \(= 5120 - 683.3 = 4436.9\) | M1 | The resistance (5120) and weight component (683.3) must have opposite signs |
| Acceleration \(= (-)\frac{4436.7}{800} = (-)5.546\) ms\(^{-2}\) | A1 | |
| Distance travelled while braking \(= -\frac{u^2}{2a} = -\frac{400}{2\times(-5.546)} = 36.06\) m | A1 | |
| Stopping distance \(= 13.5 + 36.06 = 49.56\) m | F1 | Allow FT for 36.06 from previous answer. Allow FT of 13.5 from part (iv) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Increase in stopping distance on account of slope \(= 49.56 - 44.75 = 4.81\) m | ||
| Percentage increase \(= \frac{4.81}{44.75} \times 100 = 11\%\) | B1 | cao. This mark is dependent on a correct final answer to part (v) |
| [1] |
## Question 2:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 - u^2 = 2as$ | | |
| $0^2 - 40^2 = 2 \times a \times 125$ | M1 | Substitution required. For $u$, $v$ interchange award up to M1 A0 |
| $\Rightarrow a = -6.4$ | A1 | Condone no – sign |
| $F = ma$ | M1 | |
| $F = 800 \times (-)6.4 = (-)5120$ | E1 | Allow +5120 or −5120 |
| **[4]** | | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = u + at$ | | |
| $0 = 40 - 6.4 \times t$ | M1 | FT for $a$ |
| $t = 6.25$ — It takes 6.25 seconds to stop | A1 | |
| **[2]** | | |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = \frac{1}{2}(u+v)t$ | | |
| $125 = \frac{1}{2}(40+0) \times t$ | (M1) | |
| $t = 6.25$ — it takes 6.25 seconds to stop | (A1) | |
| **[2]** | | |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = ut + \frac{1}{2}at^2$ | | |
| $125 = 40t + \frac{1}{2} \times (-6.4)t^2$ | (M1) | |
| $3.2t^2 - 40t + 125 = 0$ | | |
| $t = 6.25$ | (A1) | |
| **([2])** | | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reaction distance $< 155 - 125 = 30$ m | M1 | 30 must be seen and used |
| Time taken to travel 30 m at 40 m s$^{-1}$ is 0.75 s | E1 | |
| **[2]** | | |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance travelled before braking $= 20 \times 0.675 = 13.5$ m | B1 | |
| Distance travelled while braking $= \frac{20^2}{2 \times 6.4} = 31.25$ m | B1 | |
| Stopping distance $= 13.5 + 31.25 = 44.75$ m | B1 | cao |
| **[3]** | | |
### Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance during reaction time is not affected by slope. It is $20 \times 0.675 = 13.5$ m | | 13.5 is rewarded later |
| Component of car's weight down the slope $= mg\sin\alpha = 800 \times 9.8 \times \sin 5°\ (= 683.3\text{ N})$ | M1 | Allow cos for sin for M1. Allow omission of $g$ for this mark only |
| $= 683.3$ N | A1 | cao |
| Force opposing motion when brakes applied $= 5120 - 683.3 = 4436.9$ | M1 | The resistance (5120) and weight component (683.3) must have opposite signs |
| Acceleration $= (-)\frac{4436.7}{800} = (-)5.546$ ms$^{-2}$ | A1 | |
| Distance travelled while braking $= -\frac{u^2}{2a} = -\frac{400}{2\times(-5.546)} = 36.06$ m | A1 | |
| Stopping distance $= 13.5 + 36.06 = 49.56$ m | F1 | Allow FT for 36.06 from previous answer. Allow FT of 13.5 from part (iv) |
| **[6]** | | |
### Part (vi)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Increase in stopping distance on account of slope $= 49.56 - 44.75 = 4.81$ m | | |
| Percentage increase $= \frac{4.81}{44.75} \times 100 = 11\%$ | B1 | cao. This mark is dependent on a correct final answer to part (v) |
| **[1]** | | |
---2 Robin is driving a car of mass 800 kg along a straight horizontal road at a speed of $40 \mathrm {~ms} ^ { - 1 }$.\\
Robin applies the brakes and the car decelerates uniformly; it comes to rest after travelling a distance of 125 m .\\
(i) Show that the resistance force on the car when the brakes are applied is 5120 N .\\
(ii) Find the time the car takes to come to rest.
For the rest of this question, assume that when Robin applies the brakes there is a constant resistance force of 5120 N on the car.
The car returns to its speed of $40 \mathrm {~ms} ^ { - 1 }$ and the road remains straight and horizontal.\\
Robin sees a red light 155 m ahead, takes a short time to react and then applies the brakes.\\
The car comes to rest before it reaches the red light.\\
(iii) Show that Robin's reaction time is less than 0.75 s .
The 'stopping distance' is the total distance travelled while a driver reacts and then applies the brakes to bring the car to rest. For the rest of this question, assume that Robin is still driving the car described above and has a reaction time of 0.675 s . (This is the figure used in calculating the stopping distances given in the Highway Code.)\\
(iv) Calculate the stopping distance when Robin is driving at $20 \mathrm {~ms} ^ { - 1 }$ on a horizontal road.
The car then travels down a hill which has a slope of $5 ^ { \circ }$ to the horizontal.\\
(v) Find the stopping distance when Robin is driving at $20 \mathrm {~ms} ^ { - 1 }$ down this hill.\\
(vi) By what percentage is the stopping distance increased by the fact that the car is going down the hill? Give your answer to the nearest 1\%.
\hfill \mbox{\textit{OCR MEI M1 Q2 [18]}}