| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Standard +0.3 This is a standard two-part momentum conservation problem requiring students to (i) prove direction change using momentum principles and (ii) solve a system with two scenarios (same/opposite directions). While it involves multiple steps and careful sign consideration, the techniques are routine for M1 and the question provides clear guidance on the approach. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Calculation for both "before" momentum (magnitudes) | M1 | Must not include \(g\) |
| Compares both terms without arithmetic error | A1* | |
| Shows direction of after total momentum conflicts with the before velocity/momentum of Q | D*A1 | Vector nature of momentum by word or sign \((+/-)\); Explicit reference to after momentum or conservation of momentum essential |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{TMB} = +/-(0.2 \times 4 + 0.3 \times (-2))\) | B1 | Accept inclusion of \(g\); LHS must be difference for both M1 marks |
| \(0.8 - 0.6 = 0.2v + 0.3v\) | M1 | Allow if \(g\) included in all terms |
| \(v = 0.4 \text{ m s}^{-1}\) | A1 | Not awarded if \(g\) included |
| \(0.8 - 0.6 = -0.2v + 0.3v\) | M1 | Allow if \(g\) included in all terms; SC \(0.8 - 0.6 = 0.2v - 0.3v\) M1 |
| \(v = 2 \text{ m s}^{-1}\) | A1 | Not awarded if \(g\) included; Speed \(= 2\) and the direction of motion of Q is reversed A1 |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculation for both "before" momentum (magnitudes) | M1 | Must not include $g$ |
| Compares both terms without arithmetic error | A1* | |
| Shows direction of after total momentum conflicts with the before velocity/momentum of Q | D*A1 | Vector nature of momentum by word or sign $(+/-)$; Explicit reference to after momentum or conservation of momentum essential |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{TMB} = +/-(0.2 \times 4 + 0.3 \times (-2))$ | B1 | Accept inclusion of $g$; LHS must be difference for both M1 marks |
| $0.8 - 0.6 = 0.2v + 0.3v$ | M1 | Allow if $g$ included in all terms |
| $v = 0.4 \text{ m s}^{-1}$ | A1 | Not awarded if $g$ included |
| $0.8 - 0.6 = -0.2v + 0.3v$ | M1 | Allow if $g$ included in all terms; SC $0.8 - 0.6 = 0.2v - 0.3v$ M1 |
| $v = 2 \text{ m s}^{-1}$ | A1 | Not awarded if $g$ included; Speed $= 2$ **and** the direction of motion of Q is reversed A1 |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{66eb8290-3a80-40bf-be40-a936ed7d5a1b-3_136_824_260_623}
Particles $P$ and $Q$ are moving towards each other with constant speeds $4 \mathrm {~ms} ^ { - 1 }$ and $2 \mathrm {~ms} ^ { - 1 }$ along the same straight line on a smooth horizontal surface (see diagram). $P$ has mass 0.2 kg and $Q$ has mass 0.3 kg . The two particles collide.\\
(i) Show that $Q$ must change its direction of motion in the collision.\\
(ii) Given that $P$ and $Q$ move with equal speed after the collision, calculate both possible values for their speed after they collide.
\hfill \mbox{\textit{OCR M1 2014 Q4 [8]}}