| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Standard +0.8 This is a multi-part oblique collision problem requiring systematic application of momentum conservation (parallel and perpendicular to line of centres), coefficient of restitution, and energy considerations. While the individual techniques are standard M3 content, the problem requires careful geometric decomposition, tracking multiple velocity components across four parts, and synthesizing several mechanics principles. The 12-mark allocation and multi-stage reasoning place it moderately above average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \([\sin\gamma = 0.96 \div 1.2]\) \(\sin\gamma = 0.8\) | M1, A1 [2] | For using \(v_B\sin\gamma = u_B\sin\beta\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((m)2 - (m)u_B\cos\beta = (m)v_B\cos\gamma\) | M1, A1 | For using the principle of conservation of momentum. Allow sign error and/or \(u_A\cos\alpha\) (instead of 2) for M1. allow \(u_A\cos\alpha\) (instead of 2) for A1 |
| \(2 = v_B(0.6 + 0.28 \div 1.2)\) \(v_B = 2.4, u_B = 2\) | M1, A1, A1 [5] | For eliminating \(u_B\) or \(v_B\). Allow with \(\cos\) Or \(2 = 0.28u_B + 0.72u_B\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([(2 + u_B\cos\beta)e = v_B\cos\gamma]\) | M1 | For applying Newton's exp'tal law. Allow sign error and/or \(u_A\cos\alpha\) (instead of 2) for M1. |
| \((2 + 2 \times 0.28)e = 2.4 \times 0.6\) \(e = \frac{9}{16}\) or \(0.5625\) | A1ft, A1 [3] | ft \(u_B\) and \(v_B\) only |
| Answer | Marks | Guidance |
|---|---|---|
| \([(\text{y-component})^2 = 13 - 4]\) \(v_A = (\text{y-component})_{\text{before}} = 3\) | M1, A1 [2] | For using \(\frac{1}{2}(m)v^2 = 6.5(m)\) and \((y\text{-component})^2 = v^2 - 2^2\). Allow 1 slip. |
**Part (i):**
$[\sin\gamma = 0.96 \div 1.2]$ $\sin\gamma = 0.8$ | M1, A1 [2] | For using $v_B\sin\gamma = u_B\sin\beta$ |
**Part (ii):**
$(m)2 - (m)u_B\cos\beta = (m)v_B\cos\gamma$ | M1, A1 | For using the principle of conservation of momentum. Allow sign error and/or $u_A\cos\alpha$ (instead of 2) for M1. allow $u_A\cos\alpha$ (instead of 2) for A1 |
$2 = v_B(0.6 + 0.28 \div 1.2)$ $v_B = 2.4, u_B = 2$ | M1, A1, A1 [5] | For eliminating $u_B$ or $v_B$. Allow with $\cos$ Or $2 = 0.28u_B + 0.72u_B$ |
**Part (iii):**
$[(2 + u_B\cos\beta)e = v_B\cos\gamma]$ | M1 | For applying Newton's exp'tal law. Allow sign error and/or $u_A\cos\alpha$ (instead of 2) for M1. |
$(2 + 2 \times 0.28)e = 2.4 \times 0.6$ $e = \frac{9}{16}$ or $0.5625$ | A1ft, A1 [3] | ft $u_B$ and $v_B$ only |
**Part (iv):**
$[(\text{y-component})^2 = 13 - 4]$ $v_A = (\text{y-component})_{\text{before}} = 3$ | M1, A1 [2] | For using $\frac{1}{2}(m)v^2 = 6.5(m)$ and $(y\text{-component})^2 = v^2 - 2^2$. Allow 1 slip. |\includegraphics{figure_5}
Two uniform smooth identical spheres $A$ and $B$ are moving towards each other on a horizontal surface when they collide. Immediately before the collision $A$ and $B$ are moving with speeds $u_A$ m s$^{-1}$ and $u_B$ m s$^{-1}$ respectively, at acute angles $\alpha$ and $\beta$, respectively, to the line of centres. Immediately after the collision $A$ and $B$ are moving with speeds $v_A$ m s$^{-1}$ and $v_B$ m s$^{-1}$ respectively, at right angles and at acute angle $\gamma$, respectively, to the line of centres (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Given that $\sin \beta = 0.96$ and $\frac{v_B}{u_B} = 1.2$, find the value of $\sin \gamma$. [2]
\item Given also that, before the collision, the component of $A$'s velocity parallel to the line of centres is $2$ m s$^{-1}$, find the values of $u_B$ and $v_B$. [5]
\item Find the coefficient of restitution between the spheres. [3]
\item Given that the kinetic energy of $A$ immediately before the collision is $6.5m$ J, where $m$ kg is the mass of $A$, find the value of $v_A$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2011 Q5 [12]}}