| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with speed relationships |
| Difficulty | Moderate -0.3 This is a standard M2 collision problem followed by projectile motion. Part (a) uses routine conservation of momentum and Newton's restitution formula with given numerical values. Parts (b) and (c) are straightforward projectile motion calculations using SUVAT equations. All techniques are standard textbook exercises requiring no novel insight, though the multi-part nature and combination of topics makes it slightly more substantial than the most basic questions. |
| Spec | 3.02h Motion under gravity: vector form6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| (a) cons. of mom: \(2M(U) + 0 = 2M(V) + 5M(4)\) | M1 | |
| \(U = V + 10\) | A1 | |
| \(\frac{4-V}{U-0} = \frac{3}{4} \quad \therefore 4 - V = \frac{3}{4}U\) | M1 A1 | |
| solve simul. giving \(U = 8\) | M1 A1 | |
| (b) \(s_v = -\frac{1}{2}gt^2 = 19.6\); \(t^2 = 4 \quad \therefore t = 2\) | M2 A1 | |
| (c) \(v_x = 4\), \(v_y = 0 - gt = -19.6\) | M1 A1 | |
| req'd angle \(= \tan^{-1}\frac{19.6}{4} = 78.5°\) (3sf) below horizontal | M1 A1 | (13) |
**(a)** cons. of mom: $2M(U) + 0 = 2M(V) + 5M(4)$ | M1 |
$U = V + 10$ | A1 |
$\frac{4-V}{U-0} = \frac{3}{4} \quad \therefore 4 - V = \frac{3}{4}U$ | M1 A1 |
solve simul. giving $U = 8$ | M1 A1 |
**(b)** $s_v = -\frac{1}{2}gt^2 = 19.6$; $t^2 = 4 \quad \therefore t = 2$ | M2 A1 |
**(c)** $v_x = 4$, $v_y = 0 - gt = -19.6$ | M1 A1 |
req'd angle $= \tan^{-1}\frac{19.6}{4} = 78.5°$ (3sf) below horizontal | M1 A1 | (13)6. Particle $S$ of mass $2 M$ is moving with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a smooth horizontal plane when it collides directly with a particle $T$ of mass $5 M$ which is lying at rest on the plane. The coefficient of restitution between $S$ and $T$ is $\frac { 3 } { 4 }$.
Given that the speed of $T$ after the collision is $4 \mathrm {~ms} ^ { - 1 }$,
\begin{enumerate}[label=(\alph*)]
\item find $U$.
As a result of the collision, $T$ is projected horizontally from the top of a building of height 19.6 m and falls freely under gravity. $T$ strikes the ground at the point $X$ as shown in Figure 3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ef2dd10c-5a3c-4868-af00-aaf7f2937d7e-4_663_928_740_523}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\item Find the time taken for $T$ to reach $X$.
\item Show that the angle between the horizontal and the direction of motion of $T$, just before it strikes the ground at $X$, is $78.5 ^ { \circ }$ correct to 3 significant figures.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [13]}}