| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| 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 parallel and perpendicular to the line of centres, application of conservation of momentum and Newton's restitution law, then reconstruction of final velocities. While it involves multiple steps and careful component work, it follows a well-established procedure taught explicitly in M3 with no novel insight required. The given cos θ = 0.6 simplifies calculations. Slightly above average difficulty due to the multi-step nature and potential for sign errors, but remains a textbook exercise. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Conservation of momentum | *M1 | Must have 4 terms; allow sign errors, \(\cos\theta\) omitted |
| \(0.1 \times 3 + 0.2 \times 1 \times \cos\theta = 0.1 \times a + 0.2 \times b\) | A1 | \(a\) and \(b\) are vel components of \(A\) and \(B\) to right, respectively, after collision |
| Newton's experimental law | *M1 | Must have 4 terms and 0.8; allow sign errors, \(\cos\theta\) omitted |
| \(b - a = -0.8(1 \times \cos\theta - 3)\) | A1 | |
| Attempt to solve their 2 sim eqns | M1* | Dep both previous M marks; allow 1 slip |
| \(0.12\) in same direction as before | A1 | Direction may be implied by working; withhold if direction stated to left |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(b = 2.04\) | B1 | Must be seen/used in (ii) |
| vel of \(B\) perp to line of centres \(= 0.8\) | B1 | \((1 \times \sin\theta)\) |
| Direction of \(B\) after collision makes angle \(21.4°\) with line of centres | M1 | \(\tan\varphi = 0.8/2.04\); Allow with their 0.8 and 2.04 (\(b\) from (i)); allow \(\tan\varphi = 2.04/0.8\), if angle clear, leading to \(68.4°\) for A1 |
| or \(0.374\) rads | A1 | |
| Angle turned through by \(B\) is \(31.7°\) | A1ft | or \(0.554\) rads; allow \(+/-\); \(53.1(3) - \varphi\), \(0.927 - 0.374\) rads |
| [5] |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Conservation of momentum | *M1 | Must have 4 terms; allow sign errors, $\cos\theta$ omitted |
| $0.1 \times 3 + 0.2 \times 1 \times \cos\theta = 0.1 \times a + 0.2 \times b$ | A1 | $a$ and $b$ are vel components of $A$ and $B$ to right, respectively, after collision |
| Newton's experimental law | *M1 | Must have 4 terms and 0.8; allow sign errors, $\cos\theta$ omitted |
| $b - a = -0.8(1 \times \cos\theta - 3)$ | A1 | |
| Attempt to solve their 2 sim eqns | M1* | Dep both previous M marks; allow 1 slip |
| $0.12$ in same direction as before | A1 | Direction may be implied by working; withhold if direction stated to left |
| **[6]** | | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $b = 2.04$ | B1 | Must be seen/used in (ii) |
| vel of $B$ perp to line of centres $= 0.8$ | B1 | $(1 \times \sin\theta)$ |
| Direction of $B$ after collision makes angle $21.4°$ with line of centres | M1 | $\tan\varphi = 0.8/2.04$; Allow with their 0.8 and 2.04 ($b$ from (i)); allow $\tan\varphi = 2.04/0.8$, if angle clear, leading to $68.4°$ for A1 |
| or $0.374$ rads | A1 | |
| Angle turned through by $B$ is $31.7°$ | A1ft | or $0.554$ rads; allow $+/-$; $53.1(3) - \varphi$, $0.927 - 0.374$ rads |
| **[5]** | | |4 Two uniform smooth spheres $A$ and $B$ of equal radius are moving on a horizontal surface when they collide. $A$ has mass 0.1 kg and $B$ has mass 0.2 kg . Immediately before the collision $A$ is moving with speed $3 \mathrm {~ms} ^ { - 1 }$ along the line of centres, and $B$ is moving away from $A$ with speed $1 \mathrm {~ms} ^ { - 1 }$ at an acute angle $\theta$ to the line of centres, where $\cos \theta = 0.6$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{3e8248ca-74f1-443f-a5db-d7da532d2815-3_422_844_431_612}
The coefficient of restitution between the spheres is 0.8 . Find\\
(i) the velocity of $A$ immediately after the collision,\\
(ii) the angle turned through by the direction of motion of $B$ as a result of the collision.
\hfill \mbox{\textit{OCR M3 2013 Q4 [11]}}