| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find final speed |
| Difficulty | Moderate -0.8 This is a straightforward application of conservation of momentum with clearly defined initial conditions and one unknown. Part (i) requires simple substitution into Δp = m(v - u), and part (ii) uses conservation of momentum with basic algebra. No conceptual challenges or problem-solving insight needed—purely routine mechanics calculation below average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\Delta Mom P = 0.5(2.4 + 0.2)\) \(\Delta Mom P = +/-1.3 \text{ kgms}^{-1}\) | M1 A1 [2] | \(+/- 0.5(2.4 \pm 0.2)\) |
| \(\text{Momentum before} = 0.5 \times 2.4 - 0.8 \times 1.5\) \(0.5 \times 2.4 +/- 0.8 \times 1.5 = +/-(0.5 \times 0.2 +/- 0.8v)\) \(\text{Speed} = 0.125 \text{ ms}^{-1}\) OR \(\Delta Mom Q = +/- (+/- 0.8v - 0.8 \times 1.5)\) \(1.3 = +/-(0.8v - 0.8 \times 1.5)\) \(\text{Speed} = 0.125 \text{ ms}^{-1}\) | B1 M1 A1ft A1 [4] | Uses mom before = mom after CV(Expression for before momentum) 1/8, +ve (not 0.13) Uses \(\Delta Mom P = \Delta Mom Q\) CV(ans(i)) = +/-(+/- 0.8v - 0.8 \times 1.5)$ 1/8, +ve (not 0.13) |
| Cont MR \(0.5 \times 2.4 - 0.8 \times 1.5 = +/-(0.8 \times 2 +/- 0.5v)\) \(0.32\) B1 M1A1A1 ft |
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $\Delta Mom P = 0.5(2.4 + 0.2)$ $\Delta Mom P = +/-1.3 \text{ kgms}^{-1}$ | M1 A1 [2] | $+/- 0.5(2.4 \pm 0.2)$ |
| $\text{Momentum before} = 0.5 \times 2.4 - 0.8 \times 1.5$ $0.5 \times 2.4 +/- 0.8 \times 1.5 = +/-(0.5 \times 0.2 +/- 0.8v)$ $\text{Speed} = 0.125 \text{ ms}^{-1}$ OR $\Delta Mom Q = +/- (+/- 0.8v - 0.8 \times 1.5)$ $1.3 = +/-(0.8v - 0.8 \times 1.5)$ $\text{Speed} = 0.125 \text{ ms}^{-1}$ | B1 M1 A1ft A1 [4] | Uses mom before = mom after CV(Expression for before momentum) 1/8, +ve (not 0.13) Uses $\Delta Mom P = \Delta Mom Q$ CV(ans(i)) = +/-(+/- 0.8v - 0.8 \times 1.5)$ 1/8, +ve (not 0.13) |
| | | **Cont MR** $0.5 \times 2.4 - 0.8 \times 1.5 = +/-(0.8 \times 2 +/- 0.5v)$ $0.32$ B1 M1A1A1 ft |1 Two particles $P$ and $Q$ are projected directly towards each other on a smooth horizontal surface. $P$ has mass 0.5 kg and initial speed $2.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and $Q$ has mass 0.8 kg and initial speed $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. After a collision between $P$ and $Q$, the speed of $P$ is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the direction of its motion is reversed. Calculate\\
(i) the change in the momentum of $P$,\\
(ii) the speed of $Q$ after the collision.
\hfill \mbox{\textit{OCR M1 2011 Q1 [6]}}