| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Rebound from wall or barrier |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring application of standard equations of motion with friction (part a) and impulse-momentum theorem (part b). The steps are routine: calculate deceleration from friction, apply v² = u² + 2as, then use impulse = change in momentum. No novel insight or complex problem-solving required, just methodical application of familiar formulas. |
| Spec | 3.03r Friction: concept and vector form6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | −0.1(2g)=2a |
| Answer | Marks |
|---|---|
| v = 2.8 (ms- 1) | M1* |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Apply N2L with correct number of terms (must use |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | M1 | Apply work-energy principle – correct number of |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | A1 | Where F is the friction acting on the particle |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | Using F =0.1(2g)in their work-energy principle |
| v = 2.8 (ms- 1) | A1 |
| Answer | Marks |
|---|---|
| (b) | ( w−(−2.8)) |
| Answer | Marks |
|---|---|
| w = 2.1 (ms- 1) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | Use of Impulse = change in momentum – with their |
Question 3:
3 | (a) | −0.1(2g)=2a
a=−0.98
v2 =4.22 +2(−0.98)(5)
v = 2.8 (ms- 1) | M1*
A1
M1dep*
A1
[4] | 3.3
1.1
3.4
1.1 | Apply N2L with correct number of terms (must use
mass and not weight on RHS) – condone sign error
−0.1g
Use of v2 =u2 +2as(oe complete method to find v)
with their a and u=4.2,s=5
Alternative method (energy)
M1 | M1 | Apply work-energy principle – correct number of
terms – condone sign errors
1 1
(2)(4.2)2 − (2)v2 =5F
2 2 | A1 | Where F is the friction acting on the particle | Where F is the friction acting on the particle
1 1
(2)(4.2)2 − (2)v2 =5(0.1)(2g)
2 2 | M1 | Using F =0.1(2g)in their work-energy principle
v = 2.8 (ms- 1) | A1
[4]
(b) | ( w−(−2.8))
9.8=2
w = 2.1 (ms- 1) | M1
A1
[2] | 3.3
1.1 | Use of Impulse = change in momentum – with their
v≠4.2and correct values for the mass and impulse.
Condone sign error, w is the speed after impact
oe e.g. −9.8=2(w−2.8)
Allow –2.1 being changed to 2.1 for both marksA particle, of mass 2 kg, is placed at a point A on a rough horizontal surface. There is a straight vertical wall on the surface and the point on the wall nearest to A is B. The distance AB is 5 m. The particle is projected with speed 4.2 m s$^{-1}$ along the surface from A towards B. The particle hits the wall directly and rebounds. The coefficient of friction between the particle and the surface is 0.1.
\begin{enumerate}[label=(\alph*)]
\item Determine the speed of the particle immediately before impact with the wall. [4]
\end{enumerate}
The magnitude of the impulse that the wall exerts on the particle is 9.8 N s.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the speed of the particle immediately after impact with the wall. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q3 [6]}}