| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2022 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Direct collision, find velocities |
| Difficulty | Standard +0.8 This is a substantial Further Mechanics question requiring multiple collision/explosion principles across three parts: finding collision point using kinematics, applying conservation of momentum and restitution coefficient, then solving an explosion problem with both momentum and energy conservation leading to simultaneous equations. While the individual techniques are standard for Further Maths, the multi-stage nature, algebraic manipulation with mass ratios, and the 72% energy increase calculation make this moderately challenging. |
| Spec | 3.02h Motion under gravity: vector form6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Let the particles collide T seconds after projection. |
| 1 7 .5 T − 4 .9 T 2 + 4 .9 T 2 = 2 0 | M1 | 3.4 |
| T = 87 ( = 1 .1 4 2 8 ) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| collision. | A1 | 1.1 |
| Answer | Marks |
|---|---|
| 2 1 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at COLM. |
| Answer | Marks | Guidance |
|---|---|---|
| 1 1 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Immediately before explosion, KE of system = 3 m g 1 0 . | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | Speed of particle before explosion = 0 2 + 2 g 1 0 = 1 4 ms-1 | |
| . | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 1 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 1 1 1 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| at greater downward speed than lower fragment | B1 | 3.2b |
Question 5:
5 | (a) | Let the particles collide T seconds after projection.
1 7 .5 T − 4 .9 T 2 + 4 .9 T 2 = 2 0 | M1 | 3.4
T = 87 ( = 1 .1 4 2 8 ) | A1 | 1.1 | Award if clearly seen used later on.
So P travels at 1 7 .5 − 9 .8 87 = 6 .3 ms-1 upwards and Q travels
at 9 .8 87 = 1 1 .2 ms-1 downwards immediately before
collision. | A1 | 1.1 | Award if clearly seen used later on.
Let P and Q travel respectively at u and u ms-1 upwards
1 2
immediately after collision.
4 m ( 6 .3 ) − m ( 1 1 .2 ) = 4 m u + m u 4 u + u = 1 4
1 2 1 2
u − u = 0 .6 ( 1 7 .5 ) = 1 0 .5
2 1 | M1
M1
A1 | 3.3
3.3
1.1 | Attempt at COLM.
Attempt at NEL.
Both correct.
5 u = 3 .5 u = 0 .7 as required.
1 1 | A1 | 1.1 | AG
u = 1 1 .2
2 | A1 | 1.1 | Directions must be made clear at
some point in the working.
[8]
(b) | Immediately before explosion, KE of system = 3 m g 1 0 . | B1 | 1.1 | Seen or implied. Can also be
obtained by SUVAT.
12 2 m v 2 + 12 m v 2 = 1 .7 2 3 0 m g
1 2 | M1 | 3.4
2 v 2 + v 2 = 1 0 1 1 .3 6
1 2 | A1 | 2.2a
[3]
(c) | Speed of particle before explosion = 0 2 + 2 g 1 0 = 1 4 ms-1
. | B1 | 3.4 | Can be seen or implied.
Can also be obtained using energy.
3 m 1 4 = 2 m v − m v 2 v − v = 4 2 .
1 2 1 2 | M1 | 3.3 | Attempt at COLM
2 v 2 + ( 2 v − 4 2 ) 2 = 1 0 1 1 .3 6 v 2 − 2 8 v + 1 2 5 .4 4 = 0
1 1 1 1 | M1 | 1.1 | Forming a three-term quadratic in
either v or v
1 2
v = 2 2 .4 ( o r 5 .6 )
1 | A1 | 1.1
v =2.8 ( or −30.8 )
2 | A1 | 1.1
Second solution not possible as upper fragment cannot travel
at greater downward speed than lower fragment | B1 | 3.2b
[6]5 Point A lies 20 m vertically below a point B . A particle P of mass 4 m kg is projected upwards from A , at a speed of $17.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At the same time, a particle Q of mass $m \mathrm {~kg}$ is released from rest at point B . The particles collide directly, and it is given that the coefficient of restitution in the collision between P and Q is 0.6 .
\begin{enumerate}[label=(\alph*)]
\item Show that, immediately after the collision, P continues to travel upwards at $0.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and determine, at this time, the corresponding velocity of Q .
In another situation, a particle of mass $3 m \mathrm {~kg}$ is released from rest and falls vertically. After it has fallen 10 m , it explodes into two fragments. Immediately after the explosion, the lower fragment, of mass $2 m \mathrm {~kg}$, moves vertically downwards with speed $v _ { 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and the upper fragment, of mass $m \mathrm {~kg}$, moves vertically upwards with speed $v _ { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Given that, in the explosion, the kinetic energy of the system increases by $72 \%$, show that $2 v _ { 1 } ^ { 2 } + v _ { 2 } ^ { 2 } = 1011.36$.
\item By finding another equation connecting $v _ { 1 }$ and $v _ { 2 }$, determine the speeds of the fragments immediately after the explosion.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2022 Q5 [17]}}