| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with speed relationships |
| Difficulty | Standard +0.3 This is a standard Further Mechanics collision problem requiring conservation of momentum, coefficient of restitution, and energy calculations. While it has multiple parts (6 total), each part follows routine procedures: (a) uses momentum conservation with the given constraint, (b) applies the standard restitution formula, (c-e) are bookwork/interpretation, and (f) uses impulse-momentum theorem. The given relationship v_B = 2v_A simplifies the algebra significantly. This is slightly easier than average for Further Mechanics AS, as it's a straightforward application of standard collision formulas without requiring novel insight or complex problem-solving. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.5 \times 3.15 = 0.5v_A + 0.8 \times 2v_A\) | M1 | Or \(0.5 \times 3.15 = 0.5 \times \frac{1}{2}v_B + 0.8 \times v_B\) |
| \(v_A = 0.75\) | A1 | \(v_B = 1.5\) |
| So \(v_B = 2v_A = 1.5\) | A1 | \(v_A = \frac{1}{2}v_B = 0.75\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e = (\pm)\dfrac{\text{"1.5"} - \text{"0.75"}}{3.15 - 0}\) | M1 | Speed of separation over speed of approach; using their values from 3(a) provided c.o.m. used |
| \(\dfrac{5}{21}\) or awrt \(0.238\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Because \(e\) is the ratio of two speeds (in ms\(^{-1}\)) the units cancel and so it is a dimensionless quantity. | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Initial KE \(= \frac{1}{2} \times 0.5 \times 3.15^2\) | M1 | \(\frac{3969}{1600} = 2.48\)... Correct KE calc; or change/gain of KE of \(B = 0.8 \times\) "1.5"\(^2\) |
| Final KE \(= \frac{1}{2} \times 0.5 \times\) "0.75"\(^2 + \frac{1}{2} \times 0.8 \times\) "1.5"\(^2\) | M1 | \(\frac{333}{320} = 1.04\)... KE calculation with correct \(m\) and their \(u\) and \(2u\); change/loss of KE of \(A = \pm\frac{1}{2} \times 0.5 \times\) "0.75"\(^2 \mp \frac{1}{2} \times 0.5 \times 3.15^2\) |
| KE Loss \(= 2.48... - 1.04... = 1.44\) J | A1 | FT their speeds if positive; \(\frac{36}{25} = 1.44\); must be positive value for the amount lost |
| Answer | Marks | Guidance |
|---|---|---|
| Not perfectly elastic since KE is lost oe | B1 | eg \(e \neq 1\) oe (but just \(e = 0.238\)... is insufficient) |
| Answer | Marks | Guidance |
|---|---|---|
| Change in \(B\)'s momentum \(= 0.8 \times\) "1.5" | M1 | Using impulse = change in momentum (condone sign error); or by finding change in \(A\)'s momentum: \(0.5 \times 0.75 - 0.5 \times 3.15 = (\pm)1.2\) Ns |
| \((\pm)1.2\) Ns or kgms\(^{-1}\) | A1 | Impulse on \(B\) (hence impulse \(B\) exerts on \(A\) is \((\pm)1.2\) Ns) |
| in the opposite direction to \(A\)'s original direction of motion | A1 | This statement oe needed for full marks; in the opposite direction to \(A\)'s original motion |
# Question 3:
## Part (a)
Conservation of Momentum:
$0.5 \times 3.15 = 0.5v_A + 0.8 \times 2v_A$ | **M1** | Or $0.5 \times 3.15 = 0.5 \times \frac{1}{2}v_B + 0.8 \times v_B$
$v_A = 0.75$ | **A1** | $v_B = 1.5$
So $v_B = 2v_A = 1.5$ | **A1** | $v_A = \frac{1}{2}v_B = 0.75$
## Part (b)
$e = (\pm)\dfrac{\text{"1.5"} - \text{"0.75"}}{3.15 - 0}$ | **M1** | Speed of separation over speed of approach; using their values from 3(a) provided c.o.m. used
$\dfrac{5}{21}$ or awrt $0.238$ | **A1** |
## Part (c)
Because $e$ is the ratio of two speeds (in ms$^{-1}$) the units cancel and so it is a dimensionless quantity. | **B1** | oe
## Part (d)
Initial KE $= \frac{1}{2} \times 0.5 \times 3.15^2$ | **M1** | $\frac{3969}{1600} = 2.48$... Correct KE calc; or change/gain of KE of $B = 0.8 \times$ "1.5"$^2$
Final KE $= \frac{1}{2} \times 0.5 \times$ "0.75"$^2 + \frac{1}{2} \times 0.8 \times$ "1.5"$^2$ | **M1** | $\frac{333}{320} = 1.04$... KE calculation with correct $m$ and their $u$ and $2u$; change/loss of KE of $A = \pm\frac{1}{2} \times 0.5 \times$ "0.75"$^2 \mp \frac{1}{2} \times 0.5 \times 3.15^2$
KE Loss $= 2.48... - 1.04... = 1.44$ J | **A1** | FT their speeds if positive; $\frac{36}{25} = 1.44$; must be positive value for the amount lost
## Part (e)
Not perfectly elastic since KE is lost oe | **B1** | eg $e \neq 1$ oe (but just $e = 0.238$... is insufficient)
## Part (f)
Change in $B$'s momentum $= 0.8 \times$ "1.5" | **M1** | Using impulse = change in momentum (condone sign error); or by finding change in $A$'s momentum: $0.5 \times 0.75 - 0.5 \times 3.15 = (\pm)1.2$ Ns
$(\pm)1.2$ Ns or kgms$^{-1}$ | **A1** | Impulse on $B$ (hence impulse $B$ exerts on $A$ is $(\pm)1.2$ Ns)
in the opposite direction to $A$'s original direction of motion | **A1** | This statement oe needed for full marks; in the opposite direction to $A$'s original motion3 A particle $A$ of mass 0.5 kg is moving with a speed of $3.15 \mathrm {~ms} ^ { - 1 }$ on a smooth horizontal surface when it collides directly with a particle $B$ of mass 0.8 kg which is at rest on the surface. The velocities of $A$ and $B$ immediately after the collision are denoted by $\mathrm { v } _ { \mathrm { A } } \mathrm { ms } ^ { - 1 }$ and $\mathrm { v } _ { \mathrm { B } } \mathrm { ms } ^ { - 1 }$ respectively. You are given that $\mathrm { v } _ { \mathrm { B } } = 2 \mathrm { v } _ { \mathrm { A } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $\mathrm { V } _ { \mathrm { A } }$ and $\mathrm { V } _ { \mathrm { B } }$.
\item Find the coefficient of restitution between $A$ and $B$.
\item Explain why the coefficient of restitution is a dimensionless quantity.
\item Calculate the total loss of kinetic energy as a result of the collision.
\item State, giving a reason, whether or not the collision is perfectly elastic.
\item Calculate the impulse that $B$ exerts on $A$ in the collision.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2021 Q3 [13]}}