| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| 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 routine application of conservation of momentum and Newton's experimental law. Part (a) is trivial impulse-momentum, part (b) applies conservation of momentum with one unknown, and part (c) requires setting up two equations (momentum conservation and the speed ratio condition) to find two possible values of e. While multi-step, all techniques are standard Further Mechanics exercises with no novel insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 3.8=5u so u=0.76(m s-1) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | (i) | 5v +3×0.6=5(0.76)or equal to 3.8 |
| A | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| before collision | A1 | 2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | KE lost = 1×5×0.762 −1×5×0.42 −1×3×0.62 | |
| 2 2 2 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| =0.504(J) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | Case 1: trucks travel in the same direction after collision: | |
| 3×0.6−4×0.2=3v+4(2v) | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 3×0.6−4×0.2=−3v+4(2v) | M1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 11 11 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 44 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | A1 | 1.1 |
Question 5:
5 | (a) | 3.8=5u so u=0.76(m s-1) | B1 | 1.1
[1]
(b) | (i) | 5v +3×0.6=5(0.76)or equal to 3.8
A | M1 | 3.3 | Conservation of linear momentum – correct
number of terms
v =0.4(m s-1) in the same direction as it was travelling
A
before collision | A1 | 2.5 | Direction may be made clear in a diagram.
[2]
(ii) | KE lost = 1×5×0.762 −1×5×0.42 −1×3×0.62
2 2 2 | M1 | 1.1 | Change in KE due to collision – correct number
of terms (with cv for v )
A
=0.504(J) | A1 | 1.1
[2]
(c) | Case 1: trucks travel in the same direction after collision:
3×0.6−4×0.2=3v+4(2v) | M1 | 3.3 | Conservation of linear momentum – correct
number of terms
Case 2: trucks travel in opposite directions after collision:
3×0.6−4×0.2=−3v+4(2v) | M1 | 3.1b | Conservation of linear momentum – correct
number of terms
v= 1 (m s-1 ) in case 1 and v=0.2(m s-1) in case 2
11 | A1 | 1.1 | For at least one correct
Case 1: e= ( 2 − 1) ÷(0.6+0.2)
11 11 | M1 | 3.3 | NEL for either case – correct number of terms and
consistent with application of CLM
= 5 (≈0.114)
44 | A1 | 1.1 | For reference: 0.11363636…
Case 2: e=(0.4+0.2)÷(0.6+0.2)= 3
4 | A1 | 1.1
[6]5 Throughout this question it may be assumed that there are no resistances to motion.\\
Model trucks A and B, with masses 5 kg and 3 kg respectively, rest on a set of straight, horizontal rails.
Truck A is given an impulse of 3.8 Ns towards B .
\begin{enumerate}[label=(\alph*)]
\item Calculate the initial speed of A.
Truck A collides directly with B. After the collision, B moves with a speed of $0.6 \mathrm {~ms} ^ { - 1 }$.
\item Determine
\begin{enumerate}[label=(\roman*)]
\item the velocity of A after the collision,
\item the kinetic energy lost due to the collision.
\end{enumerate}\item B continues to move with a speed of $0.6 \mathrm {~ms} ^ { - 1 }$ and collides with a model truck C, of mass 4 kg , which is travelling at a speed of $0.2 \mathrm {~ms} ^ { - 1 }$ towards B on the same set of rails. After the collision between B and C , the speeds of B and C are in the ratio 1 to 2 .
Determine the two possible values of the coefficient of restitution between B and C .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2020 Q5 [11]}}