| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Standard +0.8 This is a two-particle collision problem requiring simultaneous use of conservation of momentum and energy loss conditions to find two unknowns (the final velocities), followed by applying the restitution formula. While the techniques are standard M2 content, the problem requires careful algebraic manipulation of two equations with two unknowns and attention to sign conventions for opposite directions, making it moderately harder than average but still within typical M2 scope. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| \(4(8) + 3(-10) = 4v_A + 3v_B\) | M1* | Attempt at use of conservation of momentum |
| \(\frac{1}{2}(4)(8)^2 + \frac{1}{2}(3)(10)^2 - \frac{1}{2}(4)v_A^2 - \frac{1}{2}(3)v_B^2 = 121.5\) | M1* | Attempt at use of KE(before) – KE (after) = 121.5 |
| M1 dep* | Obtaining quadratic equation in either \(v_A\) or \(v_B\) (\(7v_B^2 - 4v_B - 416 = 0\), \(28v_A^2 - 16v_A - 935 = 0\)) and attempt to solve quadratic for either \(v_A\) or \(v_B\) | |
| \(v_A = -5.5\) (\(v_A = 6.0714...\)) so speed of A is 5.5 (m s\(^{-1}\)) | A1 | cao; must be positive |
| \(v_B = 8\) (\(v_B = -7.428...\)) so speed of B is 8 (m s\(^{-1}\)) | A1 | cao; must be positive |
| Both particles are moving in the reverse direction to their original motion | A1 [8] | Or an equivalent statement consistent with their \(v_A\) and \(v_B\); left and right not sufficient without a diagram; moving away from each other needs a diagram also |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_A - v_B = -e(8 - (-10))\) | M1 | Attempt at use of coefficient of restitution, right way round, \(v_A\) and \(v_B\) substituted |
| \(e = 0.75\) | A1 [2] |
## (i)
$4(8) + 3(-10) = 4v_A + 3v_B$ | M1* | Attempt at use of conservation of momentum
$\frac{1}{2}(4)(8)^2 + \frac{1}{2}(3)(10)^2 - \frac{1}{2}(4)v_A^2 - \frac{1}{2}(3)v_B^2 = 121.5$ | M1* | Attempt at use of KE(before) – KE (after) = 121.5
| M1 dep* | Obtaining quadratic equation in either $v_A$ or $v_B$ ($7v_B^2 - 4v_B - 416 = 0$, $28v_A^2 - 16v_A - 935 = 0$) and attempt to solve quadratic for either $v_A$ or $v_B$
$v_A = -5.5$ ($v_A = 6.0714...$) so speed of A is 5.5 (m s$^{-1}$) | A1 | cao; must be positive
$v_B = 8$ ($v_B = -7.428...$) so speed of B is 8 (m s$^{-1}$) | A1 | cao; must be positive
Both particles are moving in the reverse direction to their original motion | A1 [8] | Or an equivalent statement consistent with their $v_A$ and $v_B$; left and right not sufficient without a diagram; moving away from each other needs a diagram also
## (ii)
$v_A - v_B = -e(8 - (-10))$ | M1 | Attempt at use of coefficient of restitution, right way round, $v_A$ and $v_B$ substituted
$e = 0.75$ | A1 [2] |
---The masses of two particles $A$ and $B$ are 4 kg and 3 kg respectively. The particles are moving towards each other along a straight line on a smooth horizontal surface. $A$ has speed $8 \text{ m s}^{-1}$ and $B$ has speed $10 \text{ m s}^{-1}$ before they collide. The kinetic energy lost due to the collision is 121.5 J.
\begin{enumerate}[label=(\roman*)]
\item Find the speed and direction of motion of each particle after the collision. [8]
\item Find the coefficient of restitution between $A$ and $B$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2016 Q6 [10]}}