| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +0.3 This is a standard sequential collision problem requiring conservation of momentum, coefficient of restitution, and energy considerations across multiple collisions. While it has multiple parts and involves vertical motion with energy changes, each step follows routine A-level mechanics procedures without requiring novel insight. The 'show that' parts provide target values, reducing problem-solving demand. Slightly easier than average due to its structured, guided nature. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Let the velocity of B after collision be u m s-1 in the direction |
| Answer | Marks | Guidance |
|---|---|---|
| 0 .3 8 − 0 .5 1 .6 = 0 .5 u | M1 | 3.3 |
| u = 3 .2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 .6 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Let B reach the lower section with speed v m s-1. | |
| 12 0 .5 3 .2 2 + 0 .5 9 .8 0 .4 5 = 12 0 .5 v 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒ 𝑣2 = 19.06 , 𝑣 = 4.36577… | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (c) | Let the speed of C after collision be w m s-1. |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| collide with A next. | A1 | 2.1 |
| OR | Assume C reaches the top section with speed x ms−1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | 1 ×0.7×3.11842−0.7×9.8×0.45 = 1 ×0.7×𝑥2 | |
| 2 2 | A1 | A1 |
| 𝑥2 (= 0.9044) > 0 ( x = 0.951 ) | Or C can reach a height of 0.496 m ( > 0.45 ) | |
| So C will collide with A next | A1 |
| Answer | Marks |
|---|---|
| (d) | OR |
| OR | [ If 𝑣 = 0, ] by COLM, 𝑣 > 𝑢 giving 𝑒 > 1 (or increase |
| Answer | Marks | Guidance |
|---|---|---|
| B C B B C B B | B2 | 3.5a |
Question 4:
4 | (a) | Let the velocity of B after collision be u m s-1 in the direction
AB.
0 .3 8 − 0 .5 1 .6 = 0 .5 u | M1 | 3.3 | COLM – all terms present but allow sign errors
u = 3 .2 | A1 | 1.1
= 3+ .21 = 13
So coefficient of restitution
8 .6 | A1 | 1.1 | AG Must be correctly obtained
[3]
(b) | Let B reach the lower section with speed v m s-1.
12 0 .5 3 .2 2 + 0 .5 9 .8 0 .4 5 = 12 0 .5 v 2 | M1 | 3.3 | Attempt at WEP: require two KE terms and a
GPE term.
M0 if 𝑣2 = 𝑢2+2𝑎𝑠 used
⇒ 𝑣2 = 19.06 , 𝑣 = 4.36577… | A1 | 1.1 | AG Must be correctly obtained
[2]
(c) | Let the speed of C after collision be w m s-1.
0 .7 w = 0 .5 v 5 ( ⇒ 𝑤 = 𝑣 = 3.1184 )
7 | M1 | 3.4 | COLM
KE of C is 1 ×0.7×(3.1184)2 (= 3.40357…) J
2 | A1 | 1.1
C needs to gain 0 .7 9 .8 0 .4 5 = 3 .0 8 7 J of GPE so C will
collide with A next. | A1 | 2.1 | Argument must be clear.
OR | Assume C reaches the top section with speed x ms−1
1 ×0.7×3.11842−0.7×9.8×0.45 = 1 ×0.7×𝑥2
2 2 | 1 ×0.7×3.11842−0.7×9.8×0.45 = 1 ×0.7×𝑥2
2 2 | A1 | A1 | For KE of C | For KE of C
𝑥2 (= 0.9044) > 0 ( x = 0.951 ) | Or C can reach a height of 0.496 m ( > 0.45 )
So C will collide with A next | A1
[3]
(d) | OR
OR | [ If 𝑣 = 0, ] by COLM, 𝑣 > 𝑢 giving 𝑒 > 1 (or increase
B C B
of KE) which is impossible [contradicting 𝑒 ≤ 1 ]
[ If 𝑣 = 0, since 𝑒 ≤ 1 ] 𝑣 ≤ 𝑢 and so 𝑚 𝑣 < 𝑚 𝑢
B C B C C B B
contradicting COLM
𝑚 𝑣 +𝑚 𝑣 = 𝑚 𝑢 and 𝑣 −𝑣 ≤ 𝑢
B B C C B B C B B
⟹ (𝑚 +𝑚 )𝑣 ≥ (𝑚 −𝑚 )𝑢 ⟹ 𝑣 > 0
B C B B C B B | B2 | 3.5a | Correct explanation; must mention momentum
Give B1 for an explanation which includes at least
one of the following (ignore incorrect statements)
• This requires 𝑣 > 𝑢
C B
• This requires 𝑒 > 1
• If 𝑣 = 𝑢 then 𝑚 𝑣 < 𝑚 𝑢
C B C C B B
[2]4 The diagram shows three beads, A, B and C, of masses $0.3 \mathrm {~kg} , 0.5 \mathrm {~kg}$ and 0.7 kg respectively, threaded onto a smooth wire circuit consisting of two straight and two semi-circular sections. The circuit occupies a vertical plane, with the two straight sections horizontal and the upper section 0.45 m directly above the lower section.\\
\includegraphics[max width=\textwidth, alt={}, center]{a87d62b8-406d-44cd-9ffa-384005329566-5_361_961_450_248}
Initially, the beads are at rest. A and B are each given an impulse so that they move towards each other, A with a speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and B with a speed of $1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. In the subsequent collision between A and $\mathrm { B } , \mathrm { A }$ is brought to rest.
\begin{enumerate}[label=(\alph*)]
\item Show that the coefficient of restitution between A and B is $\frac { 1 } { 3 }$.
Bead B next collides with C.
\item Show that the speed of B before this collision is $4.37 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to 3 significant figures.
In this collision between B and C , B is brought to rest.
\item Determine whether C next collides with A or with B .
\item Explain why, if B has a greater mass than C , B could not be brought to rest in their collision.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2023 Q4 [10]}}