| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| 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 M2 mechanics question involving coefficient of restitution and geometric series. Part (i) requires basic application of Newton's experimental law and impulse-momentum theorem. Parts (ii)-(iii) involve routine kinematics with decreasing heights forming a GP. Part (iv) requires summing an infinite GP—all standard techniques for M2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 2 \times 9.8 \times 3.136\) | M1 | Uses \(v^2 = u^2 + 2as\) or energy with \(u=0\). Signs consistent |
| \(v = 7.84\) | A1 | |
| Rebound speed \(= 7.84e\) | B1 FT | Ignore \(-ve\) |
| \(I = \pm0.5(7.84 + 7.84e) = \pm3.92(1+e)\) | B1 FT | AEF seen. FT on \(cv(v)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-7.84e = 7.84e - gt\) | M1 | Uses a complete method to find \(t\) |
| \(t = 1.6e\) | A1 AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (a) \(t_2 = 1.6e^2\) | B1 | |
| (b) \(t_3 = 1.6e^3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Time to first bounce is 0.8 s | B1 | |
| Identify total time is sum of a GP in \(e\) | B1 | Indication of sum to at least term in \(e^4\) |
| M1 | Equate 3.4 or 4.2 or 5 or 5.8 with attempt at use of formula for sum to infinity of a GP | |
| \(\dfrac{1.6e}{1-e} = 4.2\) | A1 | |
| \(e = 0.724\) | A1 | Allow 21/29 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 2 \times 9.8 \times 3.136$ | M1 | Uses $v^2 = u^2 + 2as$ or energy with $u=0$. Signs consistent |
| $v = 7.84$ | A1 | |
| Rebound speed $= 7.84e$ | B1 FT | Ignore $-ve$ |
| $I = \pm0.5(7.84 + 7.84e) = \pm3.92(1+e)$ | B1 FT | AEF seen. FT on $cv(v)$ |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-7.84e = 7.84e - gt$ | M1 | Uses a complete method to find $t$ |
| $t = 1.6e$ | A1 **AG** | |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| (a) $t_2 = 1.6e^2$ | B1 | |
| (b) $t_3 = 1.6e^3$ | B1 | |
---
## Question 6(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Time to first bounce is 0.8 s | B1 | |
| Identify total time is sum of a GP in $e$ | B1 | Indication of sum to at least term in $e^4$ |
| | M1 | Equate 3.4 or 4.2 or 5 or 5.8 with attempt at use of formula for sum to infinity of a GP |
| $\dfrac{1.6e}{1-e} = 4.2$ | A1 | |
| $e = 0.724$ | A1 | Allow 21/29 |
---6 A small ball of mass 0.5 kg is held at a height of 3.136 m above a horizontal floor. The ball is released from rest and rebounds from the floor. The coefficient of restitution between the ball and floor is $e$.\\
(i) Find in terms of $e$ the speed of the ball immediately after the impact with the floor and the impulse that the floor exerts on the ball.
The ball continues to bounce until it eventually comes to rest.\\
(ii) Show that the time between the first bounce and the second bounce is $1.6 e$.\\
(iii) Write down, in terms of $e$, the time between
\begin{enumerate}[label=(\alph*)]
\item the second bounce and the third bounce,
\item the third bounce and the fourth bounce.\\
(iv) Given that the time from the ball being released until it comes to rest is 5 s , find the value of $e$.
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2012 Q6 [13]}}