| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.8 This is a multi-stage collision problem requiring conservation of momentum and Newton's restitution law applied three times (A-B collision, B-wall collision, second A-B collision), with the constraint that B comes to rest. The algebraic manipulation across multiple collisions and the need to track velocities through three distinct events elevates this above standard single-collision questions, though the techniques themselves are M2-standard. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) cons. of mom: \(4m(u) + 0 = 4mv_1 + 5mv_2\) | M1 | |
| \(4u = 4v_1 + 5v_2\) | A1 | |
| \(\frac{v_2 - v_1}{u - 0} = \frac{1}{2} \therefore u = 2v_2 - 2v_1\) | M1 A1 | |
| solve sim. eqns. to get \(v_1 = \frac{1}{6}u, v_2 = \frac{5}{3}u\) | M1 A1 | |
| \(\therefore v_2 = \frac{4}{6}u = 4 \times v_1\) | A1 | |
| (b) | ||
| speed of B after collision with wall = \(\frac{2}{3}ue\) | M1 A1 | |
| cons. of mom: \(4m(\frac{1}{6}u) - 5m(\frac{2}{3}ue) = 4mv_1 + 0\) | M1 A1 | |
| \(\frac{2}{3}u - \frac{10}{3}ue = 4w_1 \therefore 12w_1 = 2u - 10ue\) | A1 | |
| \(\frac{(u-w_1)}{w_1+ue} = -\frac{1}{2} \therefore w_1 = \frac{1}{12}u + ue\) giving \(-12w_1 = u + 4ue\) | M1 A1 | |
| eliminating \(w_1\) gives \(u + 4ue + 2u - 10ue = 0\) | M1 | |
| \(3u = 6ue \therefore e = \frac{1}{2}\) | A1 | (16) |
| Total | (75) |
**(a)** cons. of mom: $4m(u) + 0 = 4mv_1 + 5mv_2$ | M1 |
$4u = 4v_1 + 5v_2$ | A1 |
$\frac{v_2 - v_1}{u - 0} = \frac{1}{2} \therefore u = 2v_2 - 2v_1$ | M1 A1 |
solve sim. eqns. to get $v_1 = \frac{1}{6}u, v_2 = \frac{5}{3}u$ | M1 A1 |
$\therefore v_2 = \frac{4}{6}u = 4 \times v_1$ | A1 |
**(b)** | | |
speed of B after collision with wall = $\frac{2}{3}ue$ | M1 A1 |
| | |
cons. of mom: $4m(\frac{1}{6}u) - 5m(\frac{2}{3}ue) = 4mv_1 + 0$ | M1 A1 |
$\frac{2}{3}u - \frac{10}{3}ue = 4w_1 \therefore 12w_1 = 2u - 10ue$ | A1 |
$\frac{(u-w_1)}{w_1+ue} = -\frac{1}{2} \therefore w_1 = \frac{1}{12}u + ue$ giving $-12w_1 = u + 4ue$ | M1 A1 |
eliminating $w_1$ gives $u + 4ue + 2u - 10ue = 0$ | M1 |
$3u = 6ue \therefore e = \frac{1}{2}$ | A1 | (16)
**Total** | (75)7. A smooth sphere $A$ of mass $4 m$ is moving on a smooth horizontal plane with speed $u$. It collides directly with a stationary smooth sphere $B$ of mass $5 m$ and with the same radius as $A$.
The coefficient of restitution between $A$ and $B$ is $\frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that after the collision the speed of $B$ is 4 times greater than the speed of $A$.\\
(7 marks)\\
Sphere $B$ subsequently hits a smooth vertical wall at right angles. After rebounding from the wall, $B$ collides with $A$ again and as a result of this collision, $B$ comes to rest.
Given that the coefficient of restitution between $B$ and the wall is $e$,
\item find $e$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [16]}}