| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Topic | Motion on a slope |
| Type | Collision on slope |
| Difficulty | Standard +0.3 This is a standard A-level mechanics question involving forces on an inclined plane, constant acceleration, collision with coefficient of restitution, and motion after collision. While it has multiple parts and requires several techniques (resolving forces, Newton's laws, conservation of momentum, Newton's law of restitution), each step follows routine procedures without requiring novel insight. The calculations are straightforward once the correct approach is identified, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| All forces shown: Applied, weight and reaction | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Net force up the slope \(20 - 20\sin 30 = 10\text{(N)}\) | B1 | |
| Use 'Force = mass × acceleration' \(\Rightarrow a = 5\text{ms}^{-2}\) | B1 | |
| Applying 'suva' with \(u = 0\) and \(a = 5\) | M1 | |
| \(v = 5t\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(U\) and \(V\) (F > U) be the speeds of the particles up the slope after the collision. | ||
| An attempt at both of | ||
| COM: \(2 \times 15 - 1 \times 5 = 2 \times U + 1 \times V\) | M1 A1 ∨ | |
| NEL: \(0.2 \times (15 - (-5)) = V - U\) | M1 A1 ∨ | |
| Obtain \(U = 7\text{ms}^{-1}\) | A1 | |
| 'suva' gives \(v = 7 - 5T\), where \(T\) is time after impact | A1 ∨ | [6] |
**Part (i)**
All forces shown: Applied, weight and reaction | B1 | [1]
**Part (ii)**
Net force up the slope $20 - 20\sin 30 = 10\text{(N)}$ | B1
Use 'Force = mass × acceleration' $\Rightarrow a = 5\text{ms}^{-2}$ | B1
Applying 'suva' with $u = 0$ and $a = 5$ | M1
$v = 5t$ | A1 | [4]
**Part (iii)**
Let $U$ and $V$ (F > U) be the speeds of the particles up the slope after the collision. |
An attempt at both of |
COM: $2 \times 15 - 1 \times 5 = 2 \times U + 1 \times V$ | M1 A1 ∨
NEL: $0.2 \times (15 - (-5)) = V - U$ | M1 A1 ∨
Obtain $U = 7\text{ms}^{-1}$ | A1
'suva' gives $v = 7 - 5T$, where $T$ is time after impact | A1 ∨ | [6]\includegraphics{figure_12}
A particle $P$ of mass 2 kg can move along a line of greatest slope on a smooth plane, inclined at $30°$ to the horizontal. $P$ is initially at rest at a point on the plane, and a force of constant magnitude 20 N is applied to $P$ parallel to and up the slope (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Copy and complete the diagram, showing all forces acting on $P$. [1]
\item Find the velocity of $P$ in terms of time $t$ seconds, whilst the force of 20 N is applied. [4]
\end{enumerate}
After 3 seconds the force is removed at the instant that $P$ collides with a particle of mass 1 kg moving down the slope with speed 5 m s$^{-1}$. The coefficient of restitution between the particles is 0.2.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{2}
\item Express the velocity of $P$ as a function of time after the collision. [6]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q12 [11]}}