| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Collision on slope |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question covering routine topics (friction on slopes, collisions with restitution, impulse). While multi-part with several steps, each component uses straightforward textbook methods: resolving forces, applying conservation of momentum, using coefficient of restitution formula, and calculating motion with friction. The 'show that' parts guide students through the solution. Slightly easier than average due to its structured, methodical nature. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| \(16.66 > 14.7\) so at rest | B1, M1, B1, F1, B1, E1 | Accept any form and implied. Use of \(F_{\max} = \mu R\). Expression for \(R\); may be implied. FT their \(R\). FT if their \(F\) and weight component show given result. If \(g\) omitted, allow B1M1B0F1B0E1, so 4/6. Award as follows for use of \(\tan \alpha < \mu\): B1 \(\tan \alpha = \frac{3}{4}\); E1 \(\tan \alpha < \mu\) shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_B = 2\) so 2 m s\(^{-1}\) down plane | M1, A1, M1, A1, E1, F1 | PCLM. Any form. NEL. Allow sign errors. Condone direction not clear if +8.4 seen. Condone direction not clear if +2 seen. SCI if 2 equations obtained and 8.4 substituted into one to obtain answer 2 (instead of E1F1) |
| Answer | Marks | Guidance |
|---|---|---|
| so 21 Ns up the plane | M1, A1, A1 | Use of \(m(v - u)\) impulse on \(B\). ± 21 N s. Direction explicitly commented on |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 10.7142 \ldots\) 10.7 s (3 s.f.) | M1, B1, A1, A1, M1, A1, M1, A1, M1, A1, A1 | Using Impulse-momentum (must use 8.4). Sufficient to consider one term on LHS. Either side correct. Allow only sign errors. cao. Using N2L; sufficient to consider one force term. Allow sign errors. Using appropriate suvat must use \(a\) or-a found by use of N2L and \(u = 8.4\) cao. Use energy with 8.4, sufficient to consider one non-KE term. Using appropriate suvat. cao |
## Part (i)
Let normal reaction be $R$
$\sin \alpha = \sqrt{1 - 0.8^2} = 0.6$
$R = 2.5 \times 9.8 \times 0.8$
$F_{\max} = 0.85 \times R = 16.66$
Wt cpt down slope is $2.5 \times 9.8 \times 0.6 = 14.7$
$16.66 > 14.7$ so at rest | B1, M1, B1, F1, B1, E1 | Accept any form and implied. Use of $F_{\max} = \mu R$. Expression for $R$; may be implied. FT their $R$. FT if their $F$ and weight component show given result. If $g$ omitted, allow B1M1B0F1B0E1, so 4/6. Award as follows for use of $\tan \alpha < \mu$: B1 $\tan \alpha = \frac{3}{4}$; E1 $\tan \alpha < \mu$ shown | 6 marks
## Part (ii)
Let the seeds down the plane be $v_A$ and $v_B$.
PCLM down the plane
$1.5 \times 16 = 2.5v_A + 1.5v_B$
so $5v_A + 3v_B = 48$
NEL +ve down the plane
$\frac{v_A - v_B}{0 - 16} = -0.4$
$v_A - v_B = 6.4$
$v_A = 8.4$ so 8.4 m s$^{-1}$ down plane
$v_B = 2$ so 2 m s$^{-1}$ down plane | M1, A1, M1, A1, E1, F1 | PCLM. Any form. NEL. Allow sign errors. Condone direction not clear if +8.4 seen. Condone direction not clear if +2 seen. SCI if 2 equations obtained and 8.4 substituted into one to obtain answer 2 (instead of E1F1) | 6 marks
## Part (iii)
$1.5 \times (2 - 16)$ down plane
$= - 21$ N s down the plane
so 21 Ns up the plane | M1, A1, A1 | Use of $m(v - u)$ impulse on $B$. ± 21 N s. Direction explicitly commented on | 3 marks
## Part (iv)
**either**
$(2.5 \times 9.8 \times 0.6 - F_{\max}) \times t = 2.5(0 - 8.4)$
so $t = 10.7142 \ldots$ 10.7 s (3 s.f.)
**or**
Using N2L down the plane
$a = -0.784$
using $v = u + at$, $t = 10.7142 \ldots$ 10.7 s (3 s.f.)
**or**
$0.5 \times 2.5 \times 8.4^2 + (14.7 - 16.66) \times x = 0$
$x = 45$
$T = 10.7142 \ldots$ 10.7 s (3 s.f.) | M1, B1, A1, A1, M1, A1, M1, A1, M1, A1, A1 | Using Impulse-momentum (must use 8.4). Sufficient to consider one term on LHS. Either side correct. Allow only sign errors. cao. Using N2L; sufficient to consider one force term. Allow sign errors. Using appropriate suvat must use $a$ or-a found by use of N2L and $u = 8.4$ cao. Use energy with 8.4, sufficient to consider one non-KE term. Using appropriate suvat. cao | 4 marks
---Fig. 1.1 shows block A of mass 2.5 kg which has been placed on a long, uniformly rough slope inclined at an angle $\alpha$ to the horizontal, where $\cos \alpha = 0.8$. The coefficient of friction between A and the slope is 0.85.
\includegraphics{figure_1}
\begin{enumerate}[label=(\roman*)]
\item Calculate the maximum possible frictional force between A and the slope.
Show that A will remain at rest. [6]
\end{enumerate}
With A still at rest, block B of mass 1.5 kg is projected down the slope, as shown in Fig. 1.2. B has a speed of 16 m s$^{-1}$ when it collides with A. In this collision the coefficient of restitution is 0.4, the impulses are parallel to the slope and linear momentum parallel to the slope is conserved.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the velocity of A immediately after the collision is 8.4 m s$^{-1}$ down the slope.
Find the velocity of B immediately after the collision. [6]
\item Calculate the impulse on B in the collision. [3]
\end{enumerate}
The blocks do not collide again.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item For what length of time after the collision does A slide before it comes to rest? [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2011 Q1 [19]}}