| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Standard +0.3 This is a standard M3 mechanics question involving straightforward applications of SUVAT equations, coefficient of restitution, and geometric series concepts. Part (a) is routine kinematics, part (b)(i) is a 'show that' requiring basic algebra with given answer, parts (b)(ii) and (c) follow the same pattern with algebraic manipulation, and part (d) tests understanding of modeling assumptions. While multi-part, each step uses well-practiced techniques without requiring novel insight or complex problem-solving. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum3.02h Motion under gravity: vector form6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 2 \times 9.8 \times 2.5\) | M1 | Use of \(v^2 = u^2 + 2as\) with \(u=0\), \(a=9.8\), \(s=2.5\) |
| \(v = 7 \text{ m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Speed after first bounce \(= 7e\) | B1 | |
| Time to reach max height: \(0 = 7e - 9.8t\), so \(t = \frac{7e}{9.8} = \frac{5e}{7}\) | M1 | |
| Total time between first and second contact \(= 2 \times \frac{5e}{7} = \frac{10e}{7}\) | A1 | Shown correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Time between 2nd and 3rd contact \(= \frac{10e^2}{7}\) | B1 | ft from (b)(i), multiply by \(e\) |
| Answer | Marks | Guidance |
|---|---|---|
| Distance fallen initially \(= 2.5\) | B1 | |
| Height after 1st bounce \(= \frac{(7e)^2}{2 \times 9.8} = \frac{49e^2}{19.6} = 2.5e^2\) | M1 A1 | |
| Height after 2nd bounce \(= 2.5e^4\) | A1 | |
| Total distance \(= 2.5 + 2(2.5e^2) + 2(2.5e^4) = 2.5 + 5e^2 + 5e^4\) | M1 A1 | Sum of initial drop plus two bounces up and down |
| Answer | Marks | Guidance |
|---|---|---|
| The floor/surface is smooth or no air resistance or the floor is perfectly horizontal | B1 | Accept any valid modelling assumption other than ball is a particle |
## Question 5:
### Part (a):
| $v^2 = 2 \times 9.8 \times 2.5$ | M1 | Use of $v^2 = u^2 + 2as$ with $u=0$, $a=9.8$, $s=2.5$ |
| $v = 7 \text{ m s}^{-1}$ | A1 | |
### Part (b)(i):
| Speed after first bounce $= 7e$ | B1 | |
| Time to reach max height: $0 = 7e - 9.8t$, so $t = \frac{7e}{9.8} = \frac{5e}{7}$ | M1 | |
| Total time between first and second contact $= 2 \times \frac{5e}{7} = \frac{10e}{7}$ | A1 | Shown correctly |
### Part (b)(ii):
| Time between 2nd and 3rd contact $= \frac{10e^2}{7}$ | B1 | ft from (b)(i), multiply by $e$ |
### Part (c):
| Distance fallen initially $= 2.5$ | B1 | |
| Height after 1st bounce $= \frac{(7e)^2}{2 \times 9.8} = \frac{49e^2}{19.6} = 2.5e^2$ | M1 A1 | |
| Height after 2nd bounce $= 2.5e^4$ | A1 | |
| Total distance $= 2.5 + 2(2.5e^2) + 2(2.5e^4) = 2.5 + 5e^2 + 5e^4$ | M1 A1 | Sum of initial drop plus two bounces up and down |
### Part (d):
| The floor/surface is smooth **or** no air resistance **or** the floor is perfectly horizontal | B1 | Accept any valid modelling assumption other than ball is a particle |
---5 A ball is dropped from a height of 2.5 m above a horizontal floor. The ball bounces repeatedly on the floor.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of the ball when it first hits the floor.
\item The coefficient of restitution between the ball and the floor is $e$.
\begin{enumerate}[label=(\roman*)]
\item Show that the time taken between the first contact of the ball with the floor and the second contact of the ball with the floor is $\frac { 10 e } { 7 }$ seconds.
\item Find, in terms of $e$, the time taken between the second contact and the third contact of the ball with the floor.
\end{enumerate}\item Find, in terms of $e$, the total vertical distance travelled by the ball from when it is dropped until its third contact with the floor.
\item State a modelling assumption for answering this question, other than the ball being a particle.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2011 Q5 [12]}}