| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Coalescence or perfectly inelastic collision |
| Difficulty | Moderate -0.8 This is a straightforward multi-part mechanics question testing standard momentum conservation and Newton's second law. Part (i) is basic F=ma kinematics, (ii) is direct application of momentum conservation for coalescence, (iii) adds coefficient of restitution which is routine M2 content, and part (b) is standard projectile motion with bouncing. All parts follow textbook procedures with no novel problem-solving required, making it easier than average A-level. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model3.03c Newton's second law: F=ma one dimension6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Using impulse-momentum: \(13 \times T = 10 \times 4.75 - 10 \times (-1.75)\) | M1 | Apply impulse = change in momentum, allow sign errors |
| \(13T = 47.5 + 17.5 = 65\) | A1 | Correct equation |
| \(T = 5\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| CLM: \(10 \times 4.75 + 15 \times (-0.5) = 25v\) | M1 | Apply conservation of momentum |
| \(47.5 - 7.5 = 25v\) | A1 | Correct equation |
| \(v = 1.6 \text{ ms}^{-1}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| CLM: \(10 \times 4.75 + 15 \times (-0.5) = 10 \times 1 + 15v_Q\) | M1 | Apply CLM |
| \(40 = 10 + 15v_Q\) | A1 | Correct equation |
| \(v_Q = 2 \text{ ms}^{-1}\) | A1 | cao |
| \(e = \frac{\text{speed of separation}}{\text{speed of approach}}\) | M1 | Correct formula for \(e\) |
| Speed of approach \(= 4.75 - (-0.5) = 5.25\) | A1 | |
| Speed of separation \(= 2 - 1 = 1\) | ||
| \(e = \frac{1}{5.25} = \frac{4}{21} \approx 0.190\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Horizontal component: \(u_x = 14\cos 30° = 7\sqrt{3}\) | B1 | Seen or implied |
| Vertical component on projection: \(u_y = -14\sin 30° = -7\) (downward) | B1 | |
| Finding speed on first hitting ground using \(v^2 = u^2 + 2as\): \(v_y^2 = 49 + 2(9.8)(3.125)\) | M1 | Correct use of kinematics vertically |
| \(v_y^2 = 49 + 61.25 = 110.25\), so \(v_y = 10.5 \text{ ms}^{-1}\) | A1 | |
| After first bounce (vertical): \(v_1 = 0.6 \times 10.5 = 6.3 \text{ ms}^{-1}\) (upward) | M1 | Apply \(e = 0.6\) to vertical component |
| After first bounce, ball rises then falls: speed hitting ground second time: \(v_2^2 = 6.3^2\), so \(v_2 = 6.3 \text{ ms}^{-1}\) | M1 | Recognise vertical speed same magnitude at ground level |
| After second bounce: vertical component \(= 0.6 \times 6.3 = 3.78 \text{ ms}^{-1}\) | A1 | |
| Horizontal component unchanged throughout: \(7\sqrt{3} \approx 12.124 \text{ ms}^{-1}\) | B1 | |
| \(\tan\theta = \frac{3.78}{7\sqrt{3}}\) | M1 | Correct method for angle |
| \(\theta = \arctan\left(\frac{3.78}{12.124}\right) \approx 17.3°\) | A1 | cao |
# Question 1:
## Part (a)(i)
| Working/Answer | Marks | Guidance |
|---|---|---|
| Using impulse-momentum: $13 \times T = 10 \times 4.75 - 10 \times (-1.75)$ | M1 | Apply impulse = change in momentum, allow sign errors |
| $13T = 47.5 + 17.5 = 65$ | A1 | Correct equation |
| $T = 5$ | A1 | cao |
## Part (a)(ii)
| Working/Answer | Marks | Guidance |
|---|---|---|
| CLM: $10 \times 4.75 + 15 \times (-0.5) = 25v$ | M1 | Apply conservation of momentum |
| $47.5 - 7.5 = 25v$ | A1 | Correct equation |
| $v = 1.6 \text{ ms}^{-1}$ | A1 | cao |
## Part (a)(iii)
| Working/Answer | Marks | Guidance |
|---|---|---|
| CLM: $10 \times 4.75 + 15 \times (-0.5) = 10 \times 1 + 15v_Q$ | M1 | Apply CLM |
| $40 = 10 + 15v_Q$ | A1 | Correct equation |
| $v_Q = 2 \text{ ms}^{-1}$ | A1 | cao |
| $e = \frac{\text{speed of separation}}{\text{speed of approach}}$ | M1 | Correct formula for $e$ |
| Speed of approach $= 4.75 - (-0.5) = 5.25$ | A1 | |
| Speed of separation $= 2 - 1 = 1$ | | |
| $e = \frac{1}{5.25} = \frac{4}{21} \approx 0.190$ | A1 | cao |
## Part (b)
| Working/Answer | Marks | Guidance |
|---|---|---|
| Horizontal component: $u_x = 14\cos 30° = 7\sqrt{3}$ | B1 | Seen or implied |
| Vertical component on projection: $u_y = -14\sin 30° = -7$ (downward) | B1 | |
| Finding speed on first hitting ground using $v^2 = u^2 + 2as$: $v_y^2 = 49 + 2(9.8)(3.125)$ | M1 | Correct use of kinematics vertically |
| $v_y^2 = 49 + 61.25 = 110.25$, so $v_y = 10.5 \text{ ms}^{-1}$ | A1 | |
| After first bounce (vertical): $v_1 = 0.6 \times 10.5 = 6.3 \text{ ms}^{-1}$ (upward) | M1 | Apply $e = 0.6$ to vertical component |
| After first bounce, ball rises then falls: speed hitting ground second time: $v_2^2 = 6.3^2$, so $v_2 = 6.3 \text{ ms}^{-1}$ | M1 | Recognise vertical speed same magnitude at ground level |
| After second bounce: vertical component $= 0.6 \times 6.3 = 3.78 \text{ ms}^{-1}$ | A1 | |
| Horizontal component unchanged throughout: $7\sqrt{3} \approx 12.124 \text{ ms}^{-1}$ | B1 | |
| $\tan\theta = \frac{3.78}{7\sqrt{3}}$ | M1 | Correct method for angle |
| $\theta = \arctan\left(\frac{3.78}{12.124}\right) \approx 17.3°$ | A1 | cao |
---1
\begin{enumerate}[label=(\alph*)]
\item Sphere P , of mass 10 kg , and sphere Q , of mass 15 kg , move with their centres on a horizontal straight line and have no resistances to their motion. $\mathrm { P } , \mathrm { Q }$ and the positive direction are shown in Fig. 1.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-2_332_803_434_712}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
Initially, P has a velocity of $- 1.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is acted on by a force of magnitude 13 N acting in the direction PQ .
After $T$ seconds, P has a velocity of $4.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and has not reached Q .
\begin{enumerate}[label=(\roman*)]
\item Calculate $T$.
The force of magnitude 13 N is removed. P is still travelling at $4.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it collides directly with Q , which has a velocity of $- 0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Suppose that P and Q coalesce in the collision to form a single object.
\item Calculate their common velocity after the collision.
Suppose instead that P and Q separate after the collision and that P has a velocity of $1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ afterwards.
\item Calculate the velocity of Q after the collision and also the coefficient of restitution in the collision.
\end{enumerate}\item Fig. 1.2 shows a small ball projected at a speed of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ below the horizontal over smooth horizontal ground.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-2_424_832_1918_699}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}
The ball is initially 3.125 m above the ground. The coefficient of restitution between the ball and the ground is 0.6 .
Calculate the angle with the horizontal of the ball's trajectory immediately after the second bounce on the ground.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2011 Q1 [19]}}