| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Coalescence or perfectly inelastic collision |
| Difficulty | Standard +0.3 This is a standard M2 momentum and collisions question covering routine applications of conservation of momentum, coefficient of restitution formula, and energy loss calculations. All parts follow textbook procedures with straightforward arithmetic, though the oblique collision in part (b) requires resolving velocities into components. Slightly above average difficulty due to multiple parts and the 2D collision, but no novel insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking right as positive: \(2(4) + 6(-2) = 8v\) | M1 | Conservation of momentum, both terms needed |
| \(8 - 12 = 8v\) | ||
| \(v = -0.5 \text{ ms}^{-1}\) | A1 | |
| Speed \(0.5 \text{ ms}^{-1}\) to the left | A1 | Accept \(-0.5\) if direction defined |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE before \(= \frac{1}{2}(2)(4^2) + \frac{1}{2}(6)(2^2) = 16 + 12 = 28\) J | M1 | Attempt at KE before and after |
| KE after \(= \frac{1}{2}(8)(0.5^2) = 1\) J | ||
| Energy lost \(= 27\) J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(v_P\), \(v_Q\) be velocities after collision | ||
| Conservation of momentum: \(2(4) + 6(-2) = 2v_P + 6v_Q\) | M1 | |
| \(-4 = 2v_P + 6v_Q\) ... (1) | A1 | |
| Newton's law of restitution: \(v_Q - v_P = \frac{2}{3}(4-(-2))\) | M1 | Correct use of NEL, correct relative velocity |
| \(v_Q - v_P = 4\) ... (2) | A1 | |
| Solving: \(v_Q = \frac{1}{2} \text{ ms}^{-1}\) (to left) | A1 | |
| \(v_P = -\frac{7}{2} = -3.5 \text{ ms}^{-1}\) (to left) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Impulse on P \(= 2(-3.5) - 2(4)\) | M1 | \(m(v-u)\) for P |
| \(= -7 - 8 = -15\) Ns | A1 | \(15\) Ns, direction stated or implied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Incoming speed \(= 26 \text{ ms}^{-1}\), angle \(= \arcsin\frac{12}{13}\) | ||
| Horizontal component: \(26\cos(\arcsin\frac{12}{13}) = 26 \times \frac{5}{13} = 10 \text{ ms}^{-1}\) | B1 | |
| Vertical component downward: \(26 \times \frac{12}{13} = 24 \text{ ms}^{-1}\) | B1 | |
| Horizontal component preserved: \(10 \text{ ms}^{-1}\) | B1 | |
| Rebound angle \(= \arcsin\frac{3}{5}\), so vertical component \(= w\sin(\arcsin\frac{3}{5}) = \frac{3w}{5}\) | ||
| Horizontal: \(w\cos(\arcsin\frac{3}{5}) = \frac{4w}{5} = 10\) | M1 | Using horizontal component preserved |
| \(w = 12.5 \text{ ms}^{-1}\) | A1 | |
| Vertical rebound speed \(= 12.5 \times \frac{3}{5} = 7.5 \text{ ms}^{-1}\) | ||
| \(e = \frac{7.5}{24} = \frac{5}{16}\) | M1 A1 |
# Question 1:
## Part (a)(i)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking right as positive: $2(4) + 6(-2) = 8v$ | M1 | Conservation of momentum, both terms needed |
| $8 - 12 = 8v$ | | |
| $v = -0.5 \text{ ms}^{-1}$ | A1 | |
| Speed $0.5 \text{ ms}^{-1}$ to the left | A1 | Accept $-0.5$ if direction defined |
## Part (a)(i)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE before $= \frac{1}{2}(2)(4^2) + \frac{1}{2}(6)(2^2) = 16 + 12 = 28$ J | M1 | Attempt at KE before and after |
| KE after $= \frac{1}{2}(8)(0.5^2) = 1$ J | | |
| Energy lost $= 27$ J | A1 | |
## Part (a)(ii)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $v_P$, $v_Q$ be velocities after collision | | |
| Conservation of momentum: $2(4) + 6(-2) = 2v_P + 6v_Q$ | M1 | |
| $-4 = 2v_P + 6v_Q$ ... (1) | A1 | |
| Newton's law of restitution: $v_Q - v_P = \frac{2}{3}(4-(-2))$ | M1 | Correct use of NEL, correct relative velocity |
| $v_Q - v_P = 4$ ... (2) | A1 | |
| Solving: $v_Q = \frac{1}{2} \text{ ms}^{-1}$ (to left) | A1 | |
| $v_P = -\frac{7}{2} = -3.5 \text{ ms}^{-1}$ (to left) | A1 | |
## Part (a)(ii)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Impulse on P $= 2(-3.5) - 2(4)$ | M1 | $m(v-u)$ for P |
| $= -7 - 8 = -15$ Ns | A1 | $15$ Ns, direction stated or implied |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Incoming speed $= 26 \text{ ms}^{-1}$, angle $= \arcsin\frac{12}{13}$ | | |
| Horizontal component: $26\cos(\arcsin\frac{12}{13}) = 26 \times \frac{5}{13} = 10 \text{ ms}^{-1}$ | B1 | |
| Vertical component downward: $26 \times \frac{12}{13} = 24 \text{ ms}^{-1}$ | B1 | |
| Horizontal component preserved: $10 \text{ ms}^{-1}$ | B1 | |
| Rebound angle $= \arcsin\frac{3}{5}$, so vertical component $= w\sin(\arcsin\frac{3}{5}) = \frac{3w}{5}$ | | |
| Horizontal: $w\cos(\arcsin\frac{3}{5}) = \frac{4w}{5} = 10$ | M1 | Using horizontal component preserved |
| $w = 12.5 \text{ ms}^{-1}$ | A1 | |
| Vertical rebound speed $= 12.5 \times \frac{3}{5} = 7.5 \text{ ms}^{-1}$ | | |
| $e = \frac{7.5}{24} = \frac{5}{16}$ | M1 A1 | |
---1
\begin{enumerate}[label=(\alph*)]
\item Two small spheres, $P$ of mass 2 kg and $Q$ of mass 6 kg , are moving in the same straight line along a smooth, horizontal plane with the velocities shown in Fig. 1.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31c38a58-e9d5-4d01-90af-3b41213a9c7d-2_252_647_404_708}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
Consider the direct collision of P and Q in the following two cases.
\begin{enumerate}[label=(\roman*)]
\item The spheres coalesce on collision.\\
(A) Calculate the common velocity of the spheres after the collision.\\
(B) Calculate the energy lost in the collision.
\item The spheres rebound with a coefficient of restitution of $\frac { 2 } { 3 }$ in the collision.\\
(A) Calculate the velocities of P and Q after the collision.\\
(B) Calculate the impulse on P in the collision.
\end{enumerate}\item A small ball bounces off a smooth, horizontal plane. The ball hits the plane with a speed of $26 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\arcsin \frac { 12 } { 13 }$ to it. The ball rebounds at an angle of $\arcsin \frac { 3 } { 5 }$ to the plane, as shown in Fig. 1.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31c38a58-e9d5-4d01-90af-3b41213a9c7d-2_238_545_1695_767}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}
Calculate the speed with which the ball rebounds from the plane.\\
Calculate also the coefficient of restitution in the impact.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2006 Q1 [19]}}