| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical motion: energy loss on impact |
| Difficulty | Moderate -0.8 This is a straightforward multi-part SUVAT question with standard applications: using v²=u²+2as for speeds before/after impact, impulse = change in momentum, and basic kinematics for time calculations. All parts follow routine procedures with no problem-solving insight required, making it easier than average but not trivial due to the multiple steps and careful sign management needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(v^2 = u^2 + 2as\): \(v^2 = 4g\), \(v = \sqrt{4g}\) or \(6.3\) or \(6.26\ (\text{m s}^{-1})\) | M1, A1 | M1 for complete method; A1 must be positive; allow \(0 = u^2 - 4g\) or \(v^2 = 4g\) but not \(0 = u^2 + 4g\) or \(v^2 = -4g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Rebounds to \(1.5\) m: \(0 = u^2 - 3g\), \(u = \sqrt{3g}\), \(5.4\) or \(5.42\ (\text{m s}^{-1})\) | M1A1 | Allow \(0 = u^2 - 3g\) or \(v^2 = 3g\) but not \(0 = u^2 + 3g\) or \(v^2 = -3g\); must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impulse \(= 0.3(6.3 + 5.4) = 3.5\) (Ns) | M1A1 | M1 for \(\pm 0.3(\text{their } (b) \pm \text{their } (a))\); unless definitely adding momenta i.e. using \(I = m(v+u)\) which is M0; extra \(g\) is M0; A1 for 3.5(Ns) or 3.50(Ns), must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| First line (from origin to \(v\), \(v\) marked on axis) | B1 | Straight line from origin; \(v\) must be marked on axis |
| Second line (parallel, correctly positioned) | B1 | Parallel straight line correctly positioned; if continuous vertical lines clearly included as part of graph then B0 |
| \(-u\), \(u\) correctly marked | B1 | Provided second line is correctly positioned; NB: reflection of graph in \(t\)-axis (upwards +ve) is also acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of suvat to find \(t_1\) or \(t_2\): \(\sqrt{4g} = gt_1\), \(t_1 = \sqrt{\frac{4}{g}} = 0.64\) s | M1A1 | M1 for use of suvat or area under \(v\)-\(t\) graph to find either \(t_1\) or \(t_2\) or \(2t_2\); A1 for correct value of either \(t_1\) or \(t_2\) (can be in terms of \(g\) or surds or unsimplified e.g. \(6.3/9.8\)) |
| \(\sqrt{3g} = gt_2\), \(t_2 = \sqrt{\frac{3}{g}} = 0.55\) s | ||
| Total time \(= t_1 + 2t_2 = 1.7\) s or \(1.75\) s | DM1A1 | Second M1 dependent on first M1 for \(t_1 + 2t_2\); A1 for 1.7(s) or 1.75(s) |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $v^2 = u^2 + 2as$: $v^2 = 4g$, $v = \sqrt{4g}$ or $6.3$ or $6.26\ (\text{m s}^{-1})$ | M1, A1 | M1 for complete method; A1 must be positive; allow $0 = u^2 - 4g$ or $v^2 = 4g$ but not $0 = u^2 + 4g$ or $v^2 = -4g$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Rebounds to $1.5$ m: $0 = u^2 - 3g$, $u = \sqrt{3g}$, $5.4$ or $5.42\ (\text{m s}^{-1})$ | M1A1 | Allow $0 = u^2 - 3g$ or $v^2 = 3g$ but not $0 = u^2 + 3g$ or $v^2 = -3g$; must be positive |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse $= 0.3(6.3 + 5.4) = 3.5$ (Ns) | M1A1 | M1 for $\pm 0.3(\text{their } (b) \pm \text{their } (a))$; unless definitely adding momenta i.e. using $I = m(v+u)$ which is M0; **extra $g$ is M0**; A1 for 3.5(Ns) or 3.50(Ns), must be positive |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| First line (from origin to $v$, $v$ marked on axis) | B1 | Straight line from origin; $v$ must be marked on axis |
| Second line (parallel, correctly positioned) | B1 | Parallel straight line correctly positioned; if continuous vertical lines clearly included as part of graph then B0 |
| $-u$, $u$ correctly marked | B1 | Provided second line is correctly positioned; NB: reflection of graph in $t$-axis (upwards +ve) is also acceptable |
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of suvat to find $t_1$ or $t_2$: $\sqrt{4g} = gt_1$, $t_1 = \sqrt{\frac{4}{g}} = 0.64$ s | M1A1 | M1 for use of suvat or area under $v$-$t$ graph to find either $t_1$ or $t_2$ or $2t_2$; A1 for correct value of either $t_1$ or $t_2$ (can be in terms of $g$ or surds or unsimplified e.g. $6.3/9.8$) |
| $\sqrt{3g} = gt_2$, $t_2 = \sqrt{\frac{3}{g}} = 0.55$ s | | |
| Total time $= t_1 + 2t_2 = 1.7$ s or $1.75$ s | DM1A1 | Second M1 **dependent on first M1** for $t_1 + 2t_2$; A1 for 1.7(s) or 1.75(s) |
---\begin{enumerate}
\item A ball of mass 0.3 kg is released from rest at a point which is 2 m above horizontal ground. The ball moves freely under gravity. After striking the ground, the ball rebounds vertically and rises to a maximum height of 1.5 m above the ground, before falling to the ground again. The ball is modelled as a particle.\\
(a) Find the speed of the ball at the instant before it strikes the ground for the first time.\\
(b) Find the speed of the ball at the instant after it rebounds from the ground for the first time.\\
(c) Find the magnitude of the impulse on the ball in the first impact with the ground.\\
(d) Sketch, in the space provided, a velocity-time graph for the motion of the ball from the instant when it is released until the instant when it strikes the ground for the second time.\\
(e) Find the time between the instant when the ball is released and the instant when it strikes the ground for the second time.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q3 [13]}}