| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against air resistance - vertical motion |
| Difficulty | Standard +0.8 This is a substantial multi-part mechanics question requiring energy methods, work against resistance, coefficient of restitution, and impulse-momentum. While each individual part uses standard A-level techniques, the question requires sustained problem-solving across five parts with different concepts, and part (d) requires careful tracking of energy losses through multiple stages of motion. This is more demanding than a typical single-concept mechanics question but doesn't require novel insights beyond syllabus material. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.03i Coefficient of restitution: e6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑚𝑚×7.2 =𝑚𝑚𝑚𝑚ℎ | A1 | 1.1 |
| 2 | [2] | |
| (b) | Let the amount of work done per metre against air resistance |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 3.1b | Work-energy principle. All three |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (c) | Let v be the speed of the ball just before impact with ground |
| Answer | Marks |
|---|---|
| 2 | M1 |
| B1 | 3.3 |
| 1.1 | Work-energy principle: all three |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | Let V be the speed of the ball just after impact with ground. | |
| M1 | 3.3 | Work-energy principle - all three |
| Answer | Marks | Guidance |
|---|---|---|
| 1 V =7 2 .6197 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8.7007 | A1ft | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| (e) | −mv+12=mV | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| v+V | A1 | 1.1 |
Question 4:
4 | (a) | M1 | 3.4 | Must use energy method as directed
in question.
1 h=2.6448 2 9 so maximum height is 4.24 m
𝑚𝑚×7.2 =𝑚𝑚𝑚𝑚ℎ | A1 | 1.1 | 4.2448979…
2 | [2]
(b) | Let the amount of work done per metre against air resistance
be W
M1 | 3.1b | Work-energy principle. All three
terms present. Condone sign errors.
1 W =0.568 2 m
𝑚𝑚×7.2 −2.5𝑊𝑊=2.5𝑚𝑚𝑚𝑚
2 | A1 | 1.1 | AG – sufficient working must be
shown
[2]
(c) | Let v be the speed of the ball just before impact with ground
4.1mg−4.1W = 1mv2
2 | M1
B1 | 3.3
1.1 | Work-energy principle: all three
terms present (or all four if starting
from when ball leaves the hand).
4.1W with W =0.568m
1mv2 =37.8512m⇒v=8.70ms-1
2 | A1 | 1.1 | 8.7007126…
[3]
(d) | Let V be the speed of the ball just after impact with ground.
M1 | 3.3 | Work-energy principle - all three
terms present
1 V =7 2 .6197 | A1 | 1.1
𝑚𝑚𝑉𝑉 −2.8𝑊𝑊=2.8𝑚𝑚𝑚𝑚
2Coefficient of restitution = 7.6197=0.876
8.7007 | A1ft | 3.4 | FT their answer to (c) | 0.87576318…
[3]
(e) | −mv+12=mV | M1 | 3.3 | Using impulse = change in
momentum. Correct number of
terms but allow sign errors. FT their
values for v and V
m= 12 =0.735
v+V | A1 | 1.1
[2]4 A child throws a ball of mass $m \mathrm {~kg}$ vertically upwards with a speed of $7.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The ball leaves the child's hand at a height of 1.6 m above horizontal ground.
\begin{enumerate}[label=(\alph*)]
\item Ignoring any possible air resistance, use an energy method to determine the maximum height reached by the ball above the ground.
In fact, the ball only reaches a height of 4.1 m above the ground. For the rest of this question you should assume that the air resistance may be modelled as a constant force acting in the opposite direction to the ball's motion.
\item Show that the ball does 0.568 mJ of work against air resistance per metre travelled.
\item Calculate the speed of the ball just before it hits the ground.
The ball bounces off the ground and first comes instantaneously to rest 2.8 m above the ground.
\item Determine the coefficient of restitution between the ball and the ground.
In the first impact between the ball and the ground, the magnitude of the impulse exerted on the ball by the ground is 12 Ns .
\item Determine the value of $m$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2021 Q4 [12]}}