| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Time to reach midpoint or specific position |
| Difficulty | Moderate -0.3 This is a straightforward M1 SUVAT question requiring standard application of kinematic equations and Newton's second law. Part (a) uses s = (u+v)t/2 with given values, part (b) applies standard equations to find time at midpoint, and part (c) uses F=ma after finding acceleration. All steps are routine with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A\) to \(B\): \(s=\left(\frac{u+v}{2}\right)t\) : \(400=\left(\frac{u+28}{2}\right)20\) | M1 | Complete method to find value of \(u\) (may use two equations, eliminate \(a\) and solve for \(u\)) |
| Other possible equations: \(28=u+20a\); \(400=20u+\frac{1}{2}a\times20^2\); \(28^2=u^2+2\times400a\); \(400=(28\times20)-\frac{1}{2}a\times20^2\) | ||
| \(u = 12\)* | A1* cso | Correctly reaches given answer. If two equations used, \(a\) must be eliminated and \(u\) found correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v=u+at\): \(28=12+20a\) (leads to \(a=0.8\)) | M1 A1 | Relevant suvat equation to find acceleration (may be found in (a) or (c)) |
| \(A\) to midpoint: \(200=12t+\frac{1}{2}(0.8)t^2\) | M1 A1 | Complete method for equation in \(t\) only |
| \(t = 12\) (s) or better (\(11.9258...\)), \(5\sqrt{29}-15\) | A1 | Negative value \(t=-41.9258...\) must be rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(D - 260 = 1200(0.8)\) | M1 A1ft | Use of \(F=ma\). Correct no. of terms, dimensionally correct. M0 if \(a=g\). A1ft on their \(a\) |
| \(D = 1220\) (N) | A1 | Accept 1200 (N) |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A$ to $B$: $s=\left(\frac{u+v}{2}\right)t$ : $400=\left(\frac{u+28}{2}\right)20$ | M1 | Complete method to find value of $u$ (may use two equations, eliminate $a$ and solve for $u$) |
| Other possible equations: $28=u+20a$; $400=20u+\frac{1}{2}a\times20^2$; $28^2=u^2+2\times400a$; $400=(28\times20)-\frac{1}{2}a\times20^2$ | | |
| $u = 12$* | A1* cso | Correctly reaches given answer. If two equations used, $a$ must be eliminated and $u$ found correctly |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v=u+at$: $28=12+20a$ (leads to $a=0.8$) | M1 A1 | Relevant suvat equation to find acceleration (may be found in (a) or (c)) |
| $A$ to midpoint: $200=12t+\frac{1}{2}(0.8)t^2$ | M1 A1 | Complete method for equation in $t$ only |
| $t = 12$ (s) or better ($11.9258...$), $5\sqrt{29}-15$ | A1 | Negative value $t=-41.9258...$ must be rejected |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $D - 260 = 1200(0.8)$ | M1 A1ft | Use of $F=ma$. Correct no. of terms, dimensionally correct. M0 if $a=g$. A1ft on their $a$ |
| $D = 1220$ (N) | A1 | Accept 1200 (N) |
---\begin{enumerate}
\item A van travels with constant acceleration along a straight horizontal road.
\end{enumerate}
The van passes a point $A$ with speed $u \mathrm {~ms} ^ { - 1 }$ and 20 seconds later passes a point $B$ with speed $28 \mathrm {~ms} ^ { - 1 }$
The distance $A B$ is 400 m .\\
(a) Show that $u = 12$\\
(b) Find the time taken for the van to travel from $A$ to the midpoint of $A B$.
The van has mass 1200 kg .\\
During its motion the van experiences a constant resistive force of magnitude 260 N\\
(c) Find the magnitude of the driving force exerted by the engine of the van as it travels from $A$ to $B$.
\hfill \mbox{\textit{Edexcel M1 2024 Q3 [10]}}