| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Standard +0.3 This is a standard Further Mechanics collision problem requiring conservation of momentum and Newton's restitution law. Part (a) involves routine algebraic manipulation with two equations, part (b) is a straightforward 'show that' verification, and parts (c)-(d) require understanding that e ranges from 0 to 1 with corresponding physical interpretations. While it's a multi-part question requiring several techniques, all steps follow standard textbook procedures with no novel insight needed. Slightly above average difficulty due to the algebraic manipulation and being Further Mechanics content. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | PCLM: 0.1𝑢 = 0.1𝑎+0.2𝑏 |
| Answer | Marks |
|---|---|
| dir as A before. | Candidates may define |
| Answer | Marks | Guidance |
|---|---|---|
| NEL: 𝑏−𝑎 = −𝑒(0−𝑢) | M1 | 3.3 |
| PCLM and NEL equations both correct | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | A1 | 2.2a |
| Answer | Marks |
|---|---|
| (b) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 3 | M1 | 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| 30 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | When e = 1 | B1 |
| KE loss is 0 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 | E1 | 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | When e = 0 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 30 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| same direction as A initially moved. | E1 | 3.2a |
Question 4:
4 | (a) | PCLM: 0.1𝑢 = 0.1𝑎+0.2𝑏 | M1 | 3.3 | oe: a is speed of A after, and b
is speed of B after, both in same
dir as A before. | Candidates may define
speeds differently
NEL: 𝑏−𝑎 = −𝑒(0−𝑢) | M1 | 3.3
PCLM and NEL equations both correct | A1 | 1.1
1
𝑎 = 𝑢(1−2𝑒)
3 | M1 | 1.1 | Attempt to solve – can be implied
by one (nearly) correct answer
1
𝑏 = 𝑢(1+𝑒)
3 | A1 | 2.2a | cao for both
[5]
(b) | 2
1 1 1
0.1𝑢2 − 0.1( 𝑢(1−2𝑒))
2 2 3
2
1 1
− 0.2( 𝑢(1+𝑒))
2 3 | M1 | 1.2
1
𝑢2(1−𝑒2)
30 | A1 | 1.1 | AG correctly obtained
[2]
(c) | When e = 1 | B1 | 2.2a
KE loss is 0 | B1 | 1.1
A and B move away from each other, A with
1 2
speed 𝑢 and B with speed 𝑢
3 3 | E1 | 3.2a
[3]
(d) | When e = 0 | B1 | 2.2a
KE loss is 1 𝑢2
30 | B1 | 1.1
1
A and B coalesce and move with speed 𝑢 in
3
same direction as A initially moved. | E1 | 3.2a
[3]4 Two model railway trucks, A of mass 0.1 kg and B of mass 0.2 kg , are constrained to move on a smooth straight level track.\\
Initially B is stationary and A is moving towards B with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ before they collide. The coefficient of restitution between A and B is $e$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of A and the speed of B after the collision, giving your answers in terms of $e$ and $u$.
\item Show that the loss of kinetic energy in the collision is $\frac { 1 } { 30 } u ^ { 2 } \left( 1 - e ^ { 2 } \right)$.
\item For the case in which the loss of kinetic energy is least
\begin{itemize}
\item state the value of $e$
\item state the loss in kinetic energy
\item describe the subsequent motion of the trucks.
\item For the case in which the loss of kinetic energy is greatest
\item state the value of $e$
\item state the loss in kinetic energy
\item describe the subsequent motion of the trucks.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2019 Q4 [13]}}