| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Rebound from wall or barrier |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring standard application of impulse-momentum theorem and SUVAT equations. Part (a) uses impulse = change in momentum with clear sign conventions, part (b) is routine kinematics, and part (c) is a basic sketch. The multi-part structure and need to track signs makes it slightly above average, but all techniques are standard M1 content with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(V^2 = U^2 + 2g \times 2.5\) | M1A1 | First M1 for complete method using suvat to find equation in \(U\) and \(V\) only |
| Eliminate \(V\) and solve for \(U\) | A1 (DM1) | Treat as third DM1, dependent on other two M's |
| \(7 = 0.2(10 - {-V})\) | M1A1 | Second M1 for Impulse = Change in Momentum (must have \(0.2\) in both terms, use \(10\) as one velocity); condone sign errors |
| \(U = 24\) | A1 (6) | Must appear here |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(1 = 10t - 4.9t^2\) OR e.g. \(v^2 = 10^2 - 2\times9.8\times1\) and \(v = 10 - 9.8t\) | M1 A1 | First M1 for complete method using suvat to produce equation in \(t\) only, using \(s = 1\) or \(-1\) |
| \(t = \frac{10 \pm \sqrt{100 - 19.6}}{9.8}\) so \(t = \frac{10 - \sqrt{10^2 - 2\times9.8\times1}}{9.8}\) | DM1 | Dependent on first M1, for solving equation |
| \(t = 0.11\) s or \(0.105\) s | A1 (4) | Must be only ONE answer |
| Answer | Marks | Guidance |
|---|---|---|
| Feature | Marks | Guidance |
| Straight line, positive gradient, starting at their \(U\) value on positive \(v\)-axis | B1ft | First line |
| Parallel (approx.) line placed correctly — starts where \(t\)-coordinate equals end of first line, \(v\)-coordinate negative | B1 | B0 if continuous vertical line included |
| Second line starting on \(v = -10\) | B1 (3) | SC: if second line extends to or beyond \(t\)-axis, lose second B1 but can score third B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Use suvat formulae with \(s = 1\) or \(s = -1\) to produce equation in \(t\) only | M1 | Complete method |
| Correct equation in \(t\) only | A1 | First A1 |
| Solve equation | DM1 | Dependent on first M1 |
| \(t = 0.040\) s or \(0.0403\) s | A1 | Second A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Straight line, positive gradient, starting on positive \(v\)-axis | B2 | |
| Starting at their \(U\) value (or just at \(U\)) | B1ft | Follow through on their \(U\) value |
# Question 4:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $V^2 = U^2 + 2g \times 2.5$ | M1A1 | First M1 for complete method using suvat to find equation in $U$ and $V$ only |
| Eliminate $V$ and solve for $U$ | A1 (DM1) | Treat as third DM1, dependent on other two M's |
| $7 = 0.2(10 - {-V})$ | M1A1 | Second M1 for Impulse = Change in Momentum (must have $0.2$ in both terms, use $10$ as one velocity); condone sign errors |
| $U = 24$ | A1 (6) | Must appear here |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $1 = 10t - 4.9t^2$ **OR** e.g. $v^2 = 10^2 - 2\times9.8\times1$ and $v = 10 - 9.8t$ | M1 A1 | First M1 for complete method using suvat to produce equation in $t$ only, using $s = 1$ or $-1$ |
| $t = \frac{10 \pm \sqrt{100 - 19.6}}{9.8}$ so $t = \frac{10 - \sqrt{10^2 - 2\times9.8\times1}}{9.8}$ | DM1 | Dependent on first M1, for solving equation |
| $t = 0.11$ s or $0.105$ s | A1 (4) | Must be only ONE answer |
## Part (c):
| Feature | Marks | Guidance |
|---------|-------|----------|
| Straight line, positive gradient, starting at their $U$ value on positive $v$-axis | B1ft | First line |
| Parallel (approx.) line placed correctly — starts where $t$-coordinate equals end of first line, $v$-coordinate negative | B1 | B0 if continuous vertical line included |
| Second line starting on $v = -10$ | B1 (3) | SC: if second line extends to or beyond $t$-axis, lose second B1 but can score third B1 |
**Total: (13)**
# Question 4 (continued):
## Part 4(b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Use suvat formulae with $s = 1$ or $s = -1$ to produce equation in $t$ only | M1 | Complete method |
| Correct equation in $t$ only | A1 | First A1 |
| Solve equation | DM1 | Dependent on first M1 |
| $t = 0.040$ s or $0.0403$ s | A1 | Second A1 |
## Part 4(c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Straight line, positive gradient, starting on positive $v$-axis | B2 | |
| Starting at their $U$ value (or just at $U$) | B1ft | Follow through on their $U$ value |
---4. A ball of mass 0.2 kg is projected vertically downwards with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point A which is 2.5 m above horizontal ground. The ball hits the ground. Immediately after hitting the ground, the ball rebounds vertically with a speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The ball receives an impulse of magnitude 7 Ns in its impact with the ground. By modelling the ball as a particle and ignoring air resistance, find
\begin{enumerate}[label=(\alph*)]
\item the value of $U$.
After hitting the ground, the ball moves vertically upwards and passes through a point $B$ which is 1 m above the ground.
\item Find the time between the instant when the ball hits the ground and the instant when the ball first passes through $B$.
\item Sketch a velocity-time graph for the motion of the ball from when it was projected from $A$ to when it first passes through $B$. (You need not make any further calculations to draw this sketch.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q4 [13]}}