| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Rebound from wall or barrier |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question testing standard impulse-momentum theorem and SUVAT equations. Part (a) requires simple application of impulse = change in momentum with careful attention to sign convention. Parts (b) and (c) are routine kinematics calculations using v² = u² + 2as and s = ut + ½at². No problem-solving insight or novel approach is needed—just methodical application of standard formulas. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I = 0.2(7 - -10)\) | M1 | Impulse momentum equation. Dimensionally correct. Must be using \(\pm(mv - mu)\) |
| \(= 3.4\text{ N s}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 7^2 - 2gH\) | M1 | Complete method to find max height. Must be using 7 (\(u=10\) is M0) |
| \(H = 2.5\text{ m}\) | A1 | Must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 = 7t - 4.9t^2\) | M1 | Complete method to form equation in \(t\) (using 7) |
| \(4.9t^2 - 7t + 1 = 0\) | A1 | Or equivalent |
| \(t = \dfrac{7 \pm \sqrt{49 - 19.6}}{9.8}\) | dM1 | Solve for \(t\) (sight of either root \(\Rightarrow\) M1). Dependent on previous M1 |
| \(= 0.16\text{ s}\) or \(0.161\text{ s}\) | A1 | Final answer (do not ISW). Max 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 = 49 - 2g\) | M1 | Find speed when 1 m up and use of suvat to find \(t\) |
| \(v = \sqrt{\dfrac{147}{5}} = 7 - gt\) | A1 | Or equivalent |
| Solve for \(t\). Dependent on previous M1 | dM1 | |
| \(t = 0.16\text{ s}\) or \(0.161\text{ s}\) | A1 | Final answer (do not ISW). Max 3 s.f. |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = 0.2(7 - -10)$ | M1 | Impulse momentum equation. Dimensionally correct. Must be using $\pm(mv - mu)$ |
| $= 3.4\text{ N s}$ | A1 | |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 7^2 - 2gH$ | M1 | Complete method to find max height. Must be using 7 ($u=10$ is M0) |
| $H = 2.5\text{ m}$ | A1 | Must be positive |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 = 7t - 4.9t^2$ | M1 | Complete method to form equation in $t$ (using 7) |
| $4.9t^2 - 7t + 1 = 0$ | A1 | Or equivalent |
| $t = \dfrac{7 \pm \sqrt{49 - 19.6}}{9.8}$ | dM1 | Solve for $t$ (sight of either root $\Rightarrow$ M1). Dependent on previous M1 |
| $= 0.16\text{ s}$ or $0.161\text{ s}$ | A1 | Final answer (do not ISW). Max 3 s.f. |
**4(c) alt:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = 49 - 2g$ | M1 | Find speed when 1 m up and use of suvat to find $t$ |
| $v = \sqrt{\dfrac{147}{5}} = 7 - gt$ | A1 | Or equivalent |
| Solve for $t$. Dependent on previous M1 | dM1 | |
| $t = 0.16\text{ s}$ or $0.161\text{ s}$ | A1 | Final answer (do not ISW). Max 3 s.f. |
---\begin{enumerate}
\item A small ball of mass 0.2 kg is moving vertically downwards when it hits a horizontal floor. Immediately before hitting the floor the ball has speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Immediately after hitting the floor the ball rebounds vertically with speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Find the magnitude of the impulse exerted by the floor on the ball.
\end{enumerate}
By modelling the motion of the ball as that of a particle moving freely under gravity,\\
(b) find the maximum height above the floor reached by the ball after it has rebounded from the floor,\\
(c) find the time between the instant when the ball first hits the floor and the instant when the ball is first 1 m above the floor and moving upwards.\\
\hfill \mbox{\textit{Edexcel M1 2017 Q4 [8]}}