| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: given height at two times |
| Difficulty | Standard +0.3 This is a standard M1 SUVAT question with three routine parts: (a) uses v²=u²+2as with given values, (b) solves s=ut+½at² for two times (quadratic), (c) applies F=ma then SUVAT again. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v^2 = u^2 + 2as \Rightarrow 28^2 = u^2 + 2 \times 9.8 \times 17.5\) | M1 A1 | |
| \(u = 21\) | A1 (3) | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s = ut + \frac{1}{2}at^2 \Rightarrow 19 = 21t - 4.9t^2\) | M1 A1 | |
| \(4.9t^2 - 21t + 19 = 0\) | ||
| \(t = \frac{21 \pm \sqrt{21^2 - 4 \times 4.9 \times 19}}{9.8}\) | DM1 A1 A1 | |
| \(t = 2.99\) or \(3.0\); \(t = 1.30\) or \(1.3\) | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| N2L: \(\quad 4g - 5000 = 4a \quad (a = -1240.2)\) | M1 A1 | |
| \(v^2 = u^2 + 2as \Rightarrow 0 = 28^2 - 2 \times 1240.2 \times s\) | M1 A1 | |
| \(s = 0.316\) (m) or 0.32 | (4) | |
| OR (Work-Energy): \(\frac{1}{2} \times 4 \times 28^2 + 4gs = 5000s\) | M1 A1 | |
| \(s = 0.316\) or \(0.32\) | M1 A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Complete method for finding \(u\), e.g. \(28^2 = u^2 + 2g \times 17.5\) | M1 | Condone sign errors |
| or \(28^2 = u^2 + 2(-g)\times(-17.5)\) | ||
| or \(28^2 = 2gs \Rightarrow s = 40\) then \(0^2 = u^2 + 2(-g)\times(22.5)\) | ||
| Correct equation(s) with \(g = 9.8\) | A1 | |
| \(u = 21\) | A1 | PRINTED ANSWER; verification method must state \(u = 21\) as conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Complete method producing equation in \(t\) only | M1 | Condone sign errors but not missing terms |
| Correct quadratic in \(t\) only OR two correct linear equations in \(t\) only | A1 | |
| Attempt to solve quadratic or one linear equation | DM1 | Dependent on first M1 |
| \(t = 3.0\) or \(3\) or \(2.99\) | A1 | |
| \(t = 1.3\) or \(1.30\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically with usual rules | M1 | |
| Correct equation | A1 | |
| Use of \(v^2 = u^2 + 2as\) with \(v=0\), \(u=28\) or \(u=0\), \(v=28\) and their \(a\) | M1 | Any other complete method producing equation in \(s\); M0 if \(a\) not calculated |
| \(s = 0.32\) or \(0.316\) | A1 | Must be positive since it's a distance |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $v^2 = u^2 + 2as \Rightarrow 28^2 = u^2 + 2 \times 9.8 \times 17.5$ | M1 A1 | |
| $u = 21$ | A1 **(3)** | cso |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $s = ut + \frac{1}{2}at^2 \Rightarrow 19 = 21t - 4.9t^2$ | M1 A1 | |
| $4.9t^2 - 21t + 19 = 0$ | | |
| $t = \frac{21 \pm \sqrt{21^2 - 4 \times 4.9 \times 19}}{9.8}$ | DM1 A1 A1 | |
| $t = 2.99$ or $3.0$; $t = 1.30$ or $1.3$ | **(5)** | |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| N2L: $\quad 4g - 5000 = 4a \quad (a = -1240.2)$ | M1 A1 | |
| $v^2 = u^2 + 2as \Rightarrow 0 = 28^2 - 2 \times 1240.2 \times s$ | M1 A1 | |
| $s = 0.316$ (m) or 0.32 | **(4)** | |
| **OR** (Work-Energy): $\frac{1}{2} \times 4 \times 28^2 + 4gs = 5000s$ | M1 A1 | |
| $s = 0.316$ or $0.32$ | M1 A1 **(4)** | |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Complete method for finding $u$, e.g. $28^2 = u^2 + 2g \times 17.5$ | M1 | Condone sign errors |
| or $28^2 = u^2 + 2(-g)\times(-17.5)$ | | |
| or $28^2 = 2gs \Rightarrow s = 40$ then $0^2 = u^2 + 2(-g)\times(22.5)$ | | |
| Correct equation(s) with $g = 9.8$ | A1 | |
| $u = 21$ | A1 | PRINTED ANSWER; verification method must state $u = 21$ as conclusion |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Complete method producing equation in $t$ only | M1 | Condone sign errors but not missing terms |
| Correct quadratic in $t$ only OR two correct linear equations in $t$ only | A1 | |
| Attempt to solve quadratic or one linear equation | DM1 | Dependent on first M1 |
| $t = 3.0$ or $3$ or $2.99$ | A1 | |
| $t = 1.3$ or $1.30$ | A1 | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically with usual rules | M1 | |
| Correct equation | A1 | |
| Use of $v^2 = u^2 + 2as$ with $v=0$, $u=28$ or $u=0$, $v=28$ and their $a$ | M1 | Any other complete method producing equation in $s$; M0 if $a$ not calculated |
| $s = 0.32$ or $0.316$ | A1 | Must be positive since it's a distance |
---\begin{enumerate}
\item A particle $P$ is projected vertically upwards from a point $A$ with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The point $A$ is 17.5 m above horizontal ground. The particle $P$ moves freely under gravity until it reaches the ground with speed $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Show that $u = 21$
\end{enumerate}
At time $t$ seconds after projection, $P$ is 19 m above $A$.\\
(b) Find the possible values of $t$.
The ground is soft and, after $P$ reaches the ground, $P$ sinks vertically downwards into the ground before coming to rest. The mass of $P$ is 4 kg and the ground is assumed to exert a constant resistive force of magnitude 5000 N on $P$.\\
(c) Find the vertical distance that $P$ sinks into the ground before coming to rest.
\hfill \mbox{\textit{Edexcel M1 2012 Q5 [12]}}