| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Multiple sequential collisions |
| Difficulty | Moderate -0.3 This is a standard two-collision momentum conservation problem with straightforward calculations. Part (i) applies conservation of momentum to find one unknown velocity. Part (ii) requires careful attention to directions and signs, but the mathematics is routine. The question tests basic mechanics principles without requiring novel insight or complex problem-solving, making it slightly easier than average for M1. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Momentum before collision \(= +/-(0.8 \times 4 - 0.6 \times 2)\) Momentum after collision \(= +/-0.8v_L + 0.6 \times 2\) Speed is 1 m s\(^{-1}\) | B1 B1 M1 A1 | Or momentum change L \(0.8x4 +/- 0.8v_L\) Accept inclusion of g in both terms Momentum change N \(0.6x2 + 0.6x2\) Accept inclusion of g in both terms For using the principle of conservation of momentum even if g is included throughout Accept -1 from correct work (g not used). |
| (ii)(a) \(0.6x2 - 0.7x0.5\) Total is 0.85kgms\(^{-1}\) Total momentum +ve after the collision. If N continues in its original direction, both particles have a negative momentum. N must reverse its direction. | M1 A1 DM 1 A1 | Must be a difference. SR 0.6x1 - 0.7x0.5 M1 Must be positive Or 0.6v + 0.7v is positive, confirming that the momentum is shared between two particles. No reference need be made to the physically impossible scenario where M and N both might continue in their original directions. |
| (ii)(b) \(0.6x2 - 0.7x0.5 (= 0.85) = 0.7v\) Speed is 1.21m s\(^{-1}\) | A1ft A1 | ft cv (0.85). Award M1 if not given in ii(a). Positive. Accept (a,r,t) 1.2 from correct work |
(i) Momentum before collision $= +/-(0.8 \times 4 - 0.6 \times 2)$ Momentum after collision $= +/-0.8v_L + 0.6 \times 2$ Speed is 1 m s$^{-1}$ | B1 B1 M1 A1 | Or momentum change L $0.8x4 +/- 0.8v_L$ Accept inclusion of g in both terms Momentum change N $0.6x2 + 0.6x2$ Accept inclusion of g in both terms For using the principle of conservation of momentum even if g is included throughout Accept -1 from correct work (g not used).
(ii)(a) $0.6x2 - 0.7x0.5$ Total is 0.85kgms$^{-1}$ Total momentum +ve after the collision. If N continues in its original direction, both particles have a negative momentum. N must reverse its direction. | M1 A1 DM 1 A1 | Must be a difference. SR 0.6x1 - 0.7x0.5 M1 Must be positive Or 0.6v + 0.7v is positive, confirming that the momentum is shared between two particles. No reference need be made to the physically impossible scenario where M and N both might continue in their original directions.
(ii)(b) $0.6x2 - 0.7x0.5 (= 0.85) = 0.7v$ Speed is 1.21m s$^{-1}$ | A1ft A1 | ft cv (0.85). Award M1 if not given in ii(a). Positive. Accept (a,r,t) 1.2 from correct work
---\includegraphics{figure_4}
Three uniform spheres $L$, $M$ and $N$ have masses 0.8 kg, 0.6 kg and 0.7 kg respectively. The spheres are moving in a straight line on a smooth horizontal table, with $M$ between $L$ and $N$. The sphere $L$ is moving towards $M$ with speed $4 \text{ m s}^{-1}$ and the spheres $M$ and $N$ are moving towards $L$ with speeds $2 \text{ m s}^{-1}$ and $0.5 \text{ m s}^{-1}$ respectively (see diagram).
\begin{enumerate}[label=(\roman*)]
\item $L$ collides with $M$. As a result of this collision the direction of motion of $M$ is reversed, and its speed remains $2 \text{ m s}^{-1}$. Find the speed of $L$ after the collision. [4]
\item $M$ then collides with $N$.
\begin{enumerate}[label=(\alph*)]
\item Find the total momentum of $M$ and $N$ in the direction of $M$'s motion before this collision takes place, and deduce that the direction of motion of $N$ is reversed as a result of this collision. [4]
\item Given that $M$ is at rest immediately after this collision, find the speed of $N$ immediately after this collision. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2007 Q4 [10]}}