| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.2 This is a standard M3 oblique collision problem requiring resolution of velocities, application of conservation of momentum and Newton's law of restitution along the line of centres, followed by analysis of a subsequent wall collision. While it involves multiple steps and careful component resolution, it follows a well-established procedure taught in M3 with no novel insights required. The second part requires logical reasoning about relative velocities but is straightforward once the first part is complete. Slightly above average difficulty due to the multi-stage nature and algebraic manipulation involved. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| C of M \(2m \times 5 \cos α - 3m \times 3\frac{1}{4}\cos β = 2ma + 3mb\) | M1* | allow sign and number slips. |
| Answer | Marks | Guidance |
|---|---|---|
| \(b - a = -\frac{2}{3}(1.25 - 4)\) | A1, M1*, A1 | Or equivalent; allow sign slips. \((2a + 3b = 4.25)\) |
| Attempt to solve simultaneous equations | *M1, A1 | CAO |
| \(b = 2.25\) (ms⁻¹) | A1 7 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\) and \(B\) both have same component of velocity perp to l o c | B1, B1 | May be implied. Do not allow ≥ |
| Answer | Marks | Guidance |
|---|---|---|
| Coeff of restitution > 5/9 (accept > 0.5 recurring, or 0.556) | B1 3 |
## (i)
C of M $2m \times 5 \cos α - 3m \times 3\frac{1}{4}\cos β = 2ma + 3mb$ | M1* | allow sign and number slips. | $a$ and $b$ are vels of $A$ and $B$ to right
$2mx4 - 3mx1.25 = 2ma + 3mb$
Newton's experimental law
$b - a = -\frac{2}{3}(1.25 - 4)$ | A1, M1*, A1 | Or equivalent; allow sign slips. $(2a + 3b = 4.25)$
Attempt to solve simultaneous equations | *M1, A1 | CAO
$b = 2.25$ (ms⁻¹) | A1 7 | CAO | Consistent. $-2.25$ if $b$ defined to left
($a = -1.25$ so speed of $A = 3.25$ (ms⁻¹))
## (ii)
$A$ and $B$ both have same component of velocity perp to l o c | B1, B1 | May be implied. Do not allow ≥
After collision with wall $B$ must move faster than $A$
Coeff of restitution > 5/9 (accept > 0.5 recurring, or 0.556) | B1 3 | | Ignore $e ≤ 1$, etc
---\includegraphics{figure_3}
Two uniform smooth spheres $A$ and $B$, of equal radius, have masses $2m$ kg and $3m$ kg respectively. The spheres are approaching each other on a horizontal surface when they collide. Before the collision $A$ is moving with speed $5$ m s$^{-1}$ in a direction making an angle $\alpha$ with the line of centres, where $\cos \alpha = \frac{4}{5}$, and $B$ is moving with speed $3\frac{1}{4}$ m s$^{-1}$ in a direction making an angle $\beta$ with the line of centres, where $\cos \beta = \frac{5}{13}$. A straight vertical wall is situated to the right of $B$, perpendicular to the line of centres (see diagram). The coefficient of restitution between $A$ and $B$ is $\frac{2}{5}$.
\begin{enumerate}[label=(\roman*)]
\item Find the speed of $A$ after the collision. Find also the component of the velocity of $B$ along the line of centres after the collision. [7]
\end{enumerate}
$B$ subsequently hits the wall.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Explain why $A$ and $B$ will have a second collision if the coefficient of restitution between $B$ and the wall is sufficiently large. Find the set of values of the coefficient of restitution for which this second collision will occur. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2016 Q3 [10]}}