| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Moderate -0.3 This is a straightforward application of conservation of momentum and Newton's experimental law (coefficient of restitution) with clearly defined masses and velocities. All parts follow standard procedures: (a) uses momentum conservation, (b) applies the restitution formula, (c) calculates KE before and after, (d) is a simple interpretation. The question requires careful arithmetic but no problem-solving insight or novel approaches, making it slightly easier than a typical A-level question. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(50 \times 2.1 + 70 \times -0.8 = 50 \times 0.35 + 70 \times v_B\) | M1 (AO 1.1) | Conservation of momentum with correct masses and velocities substituted; allow one sign error |
| \(v_B = 0.45\) | A1 [2] (AO 1.1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\pm e = ("0.45" - 0.35)/(2.1 - -0.8)\) (oe) | M1 (AO 1.1) | NEL with correct velocities substituted; allow one sign error; NB "0.45" + 0.35 is M0 unless clearly going in opposite directions |
| \(1/29\) or awrt \(0.0345\) | A1 [2] (AO 1.1) |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. initial KE for A \(= \frac{1}{2} \times 50 \times 2.1^2\) | M1 (AO 1.1) | Correct calculation of any initial or final KE (using their values); values: 110.25, 3.0625, 22.4, 7.0875 |
| so KE loss \(= \frac{1}{2}\times50\times2.1^2 + \frac{1}{2}\times70\times0.8^2 - (\frac{1}{2}\times50\times0.35^2 + \frac{1}{2}\times70\times0.45^2)\) oe | M1 (AO 1.1) | Attempt to find difference between total final KE and total initial KE; NB \(132.65 - 10.15\); if evaluating each object separately, then \(107.875 + 15.3125\) (must be sum) |
| \(122.5\) J | A1 [3] (AO 1.1) | Must be positive; or 123J |
| Answer | Marks | Guidance |
|---|---|---|
| Not perfectly elastic since (kinetic) energy is lost | B1 [1] (AO 2.4) | or since \(e < 1\) or \(e \neq 1\); prefer to be specific but accept "energy" only |
# Question 1:
## Part (a)
$50 \times 2.1 + 70 \times -0.8 = 50 \times 0.35 + 70 \times v_B$ | M1 (AO 1.1) | Conservation of momentum with correct masses and velocities substituted; allow one sign error
$v_B = 0.45$ | A1 [2] (AO 1.1) |
## Part (b)
$\pm e = ("0.45" - 0.35)/(2.1 - -0.8)$ (oe) | M1 (AO 1.1) | NEL with correct velocities substituted; allow one sign error; NB "0.45" + 0.35 is M0 unless clearly going in opposite directions
$1/29$ or awrt $0.0345$ | A1 [2] (AO 1.1) |
## Part (c)
e.g. initial KE for A $= \frac{1}{2} \times 50 \times 2.1^2$ | M1 (AO 1.1) | Correct calculation of any initial or final KE (using their values); values: 110.25, 3.0625, 22.4, 7.0875
so KE loss $= \frac{1}{2}\times50\times2.1^2 + \frac{1}{2}\times70\times0.8^2 - (\frac{1}{2}\times50\times0.35^2 + \frac{1}{2}\times70\times0.45^2)$ oe | M1 (AO 1.1) | Attempt to find difference between total final KE and total initial KE; NB $132.65 - 10.15$; if evaluating each object separately, then $107.875 + 15.3125$ (must be sum)
$122.5$ J | A1 [3] (AO 1.1) | Must be positive; or 123J
## Part (d)
Not perfectly elastic since (kinetic) energy is lost | B1 [1] (AO 2.4) | or since $e < 1$ or $e \neq 1$; prefer to be specific but accept "energy" only
---1 Two stones, A and B , are sliding along the same straight line on a horizontal sheet of ice. Stone A, of mass 50 kg , is moving with a constant velocity of $2.1 \mathrm {~ms} ^ { - 1 }$ towards stone B. Stone B, of mass 70 kg , is moving with a constant velocity of $0.8 \mathrm {~ms} ^ { - 1 }$ towards stone A.
A and B collide directly. Immediately after their collision stone A's velocity is $0.35 \mathrm {~ms} ^ { - 1 }$ in the same direction as its velocity before the collision.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of stone B immediately after the collision.
\item Find the coefficient of restitution for the collision.
\item Find the total loss of kinetic energy caused by the collision.
\item Explain whether the collision was perfectly elastic.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2022 Q1 [8]}}