| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with speed relationships |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law. The algebraic manipulation in part (a) is straightforward, and parts (b) and (c) involve routine substitution and percentage calculation. While it requires multiple steps and two key principles, the structure is typical and the mathematics is not demanding. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| (a) cons. of mom: \(7u_1 + 4u_2 = 7\left(\frac{u_1}{2}\right) + 4v_2\) | M1 | |
| \(8v_2 = 7u_1 + 8u_2\) | A1 | |
| \(\dfrac{v_2 - \frac{1}{2}u_1}{u_1 - u_2} = e\) \(\therefore\) \(v_2 = eu_1 - eu_2 + \frac{1}{2}u_1\) | M1 A1 | |
| eliminate \(v_2\) giving \(7u_1 + 8u_2 = 8eu_1 - 8eu_2 + 4u_1\) | M1 A1 | |
| \(8u_2 + 8eu_2 = 8eu_1 - 3u_1\) \(\therefore\) \(8u_2(e+1) = u_1(8e-3)\) | A1 | |
| (b) sub. in for \(u_1\) and \(u_2\): \(24(e+1) = 14(8e-3)\) | M1 | |
| \(24e + 24 = 112e - 42\) giving \(e = \dfrac{3}{4}\) | M1 A1 | |
| (c) speeds of \(A\), \(B\) after impact are \(v_1\) and \(v_2\) resp. | ||
| \(v_1 = 7\) ms\(^{-1}\), \(v_2 = \left(\frac{7}{8}\right)14 + 3 = 15.25\) ms\(^{-1}\) | A1 | |
| original KE \(= \frac{1}{2} \times 7 \times 14^2 + \frac{1}{2} \times 4 \times 3^2 = 704\) J | M1 A1 | |
| final KE \(= \frac{1}{2} \times 7 \times 7^2 + \frac{1}{2} \times 4 \times 15.25^2 = 636.625\) J | M1 A1 | |
| % KE lost \(= \dfrac{704 - 636.625}{704} \times 100 = 9.6\%\) (2sf) | M1 A1 | (17) |
## Question 7:
| Answer/Working | Marks | Notes |
|---|---|---|
| **(a)** cons. of mom: $7u_1 + 4u_2 = 7\left(\frac{u_1}{2}\right) + 4v_2$ | M1 | |
| $8v_2 = 7u_1 + 8u_2$ | A1 | |
| $\dfrac{v_2 - \frac{1}{2}u_1}{u_1 - u_2} = e$ $\therefore$ $v_2 = eu_1 - eu_2 + \frac{1}{2}u_1$ | M1 A1 | |
| eliminate $v_2$ giving $7u_1 + 8u_2 = 8eu_1 - 8eu_2 + 4u_1$ | M1 A1 | |
| $8u_2 + 8eu_2 = 8eu_1 - 3u_1$ $\therefore$ $8u_2(e+1) = u_1(8e-3)$ | A1 | |
| **(b)** sub. in for $u_1$ and $u_2$: $24(e+1) = 14(8e-3)$ | M1 | |
| $24e + 24 = 112e - 42$ giving $e = \dfrac{3}{4}$ | M1 A1 | |
| **(c)** speeds of $A$, $B$ after impact are $v_1$ and $v_2$ resp. | | |
| $v_1 = 7$ ms$^{-1}$, $v_2 = \left(\frac{7}{8}\right)14 + 3 = 15.25$ ms$^{-1}$ | A1 | |
| original KE $= \frac{1}{2} \times 7 \times 14^2 + \frac{1}{2} \times 4 \times 3^2 = 704$ J | M1 A1 | |
| final KE $= \frac{1}{2} \times 7 \times 7^2 + \frac{1}{2} \times 4 \times 15.25^2 = 636.625$ J | M1 A1 | |
| % KE lost $= \dfrac{704 - 636.625}{704} \times 100 = 9.6\%$ (2sf) | M1 A1 | **(17)** |
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**Total: (75)**7. Particle $A$ of mass 7 kg is moving with speed $u _ { 1 }$ on a smooth horizontal surface when it collides directly with particle $B$ of mass 4 kg moving in the same direction as $A$ with speed $u _ { 2 }$.
After the impact, $A$ continues to move in the same direction but its speed has been halved. Given that the coefficient of restitution between the particles is $e$,
\begin{enumerate}[label=(\alph*)]
\item show that $8 u _ { 2 } ( e + 1 ) = u _ { 1 } ( 8 e - 3 )$.
Given also that $u _ { 1 } = 14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $u _ { 2 } = 3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item find $e$,
\item show that the percentage of the kinetic energy of the particles lost as a result of the impact is $9.6 \%$, correct to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [17]}}