| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.3 This is a standard Further Mechanics collision problem with straightforward application of conservation of momentum, coefficient of restitution formula, and impulse calculations. All parts follow routine procedures with no novel insights required, though it involves multiple steps across three collisions. Slightly easier than average A-level due to the mechanical nature of the calculations. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5m + (-3)m = (-2)m + mv_B\) | M1 | Conservation of momentum. \(u_A\) must be \(> u_B\) |
| \(v_B = 4\) | A1 | |
| in the direction of motion of \(A\) before the collision | A1 | Must be clearly stated or shown. Direction of B is reversed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e = (4-(-2))/(5-(-3))\) | M1 | Attempt at restitution - condone sign error as long as consistent. \(0 \leq (\pm)e \leq 1\). Must have sufficient detail, e.g. 6/8 on its own is M0. |
| \(= 6/8 = \frac{3}{4}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial \(\text{KE} = \frac{1}{2} \times 4 \times 5^2 + \frac{1}{2} \times 4 \times (-3)^2 = 68\) J | M1 | Attempt to calculate total initial or final KE. Both values must be positive. Or KE loss for \(A = \frac{1}{2}\times4\times5^2 - \frac{1}{2}\times4\times(-2)^2 = 42\) J or KE gain for \(B = \frac{1}{2}\times4\times4^2 - \frac{1}{2}\times4\times3^2 = 14\) J |
| Final \(\text{KE} = \frac{1}{2} \times 4 \times (-2)^2 + \frac{1}{2} \times 4 \times 4^2 = 40\) J | ||
| so loss is \(68 - 40 = 28\) J | A1 | \(42\text{ J} - 14\text{ J} = 28\text{ J}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e = 4/4 = 1\) | B1FT | FT their \(v_B\) provided that \(0 < e \leq 1\). Allow \(e = 1\) without working, provided (a)(i) is correct |
| The collision is perfectly elastic. | B1FT | FT their \(e\) provided that \(0 < e \leq 1\). Do not accept phrases such as "completely elastic" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((-2)m + (-4)m = mV_A + mV_B\) | M1* | Conservation of momentum with consistent signs. \(2V_A + 2V_B = -12\). If "positive" direction reversed: \(2m + 4m = mV_A + mV_B\). M0 if approach speed \(< 0\) |
| \(\frac{3}{4} = (V_B - V_A)/((-2)-(-4))\) | M1* | Restitution with consistent signs. \(V_B - 2V_A = 3\). Allow use of e (e.g. \(V_B - V_A = \pm 2e\)) |
| \(2V_A + 2V_B = -12\), \(2V_B - 2V_A = 3\) | M1dep p | Attempt to solve both equations simultaneously. Allow use of e (e.g. \(V_A = \pm(3+e)\) or \(V_B = \pm(3-e)\)) |
| \(V_B = -2.25\) or \(V_A = -3.75\) | A1 | A numerical value is required here (may be implied by a correct final answer). |
| Impulse on \(A\) = change in \(A\)'s momentum \(= 4(-3.75-(-2)) = -7 \Rightarrow 7\) Ns | A1 | Ignore wrong units. ISW e.g. any statements regarding direction of travel |
# Question 5:
## Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5m + (-3)m = (-2)m + mv_B$ | M1 | Conservation of momentum. $u_A$ must be $> u_B$ |
| $v_B = 4$ | A1 | |
| in the direction of motion of $A$ before the collision | A1 | Must be clearly stated or shown. Direction of B is reversed |
## Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e = (4-(-2))/(5-(-3))$ | M1 | Attempt at restitution - condone sign error as long as consistent. $0 \leq (\pm)e \leq 1$. Must have sufficient detail, e.g. 6/8 on its own is M0. |
| $= 6/8 = \frac{3}{4}$ | A1 | AG |
## Part (a)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial $\text{KE} = \frac{1}{2} \times 4 \times 5^2 + \frac{1}{2} \times 4 \times (-3)^2 = 68$ J | M1 | Attempt to calculate total initial or final KE. Both values must be positive. Or KE loss for $A = \frac{1}{2}\times4\times5^2 - \frac{1}{2}\times4\times(-2)^2 = 42$ J or KE gain for $B = \frac{1}{2}\times4\times4^2 - \frac{1}{2}\times4\times3^2 = 14$ J |
| Final $\text{KE} = \frac{1}{2} \times 4 \times (-2)^2 + \frac{1}{2} \times 4 \times 4^2 = 40$ J | | |
| so loss is $68 - 40 = 28$ J | A1 | $42\text{ J} - 14\text{ J} = 28\text{ J}$ |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e = 4/4 = 1$ | B1FT | FT their $v_B$ provided that $0 < e \leq 1$. Allow $e = 1$ without working, provided (a)(i) is correct |
| The collision is perfectly elastic. | B1FT | FT their $e$ provided that $0 < e \leq 1$. Do not accept phrases such as "completely elastic" |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(-2)m + (-4)m = mV_A + mV_B$ | M1* | Conservation of momentum with consistent signs. $2V_A + 2V_B = -12$. If "positive" direction reversed: $2m + 4m = mV_A + mV_B$. M0 if approach speed $< 0$ |
| $\frac{3}{4} = (V_B - V_A)/((-2)-(-4))$ | M1* | Restitution with consistent signs. $V_B - 2V_A = 3$. Allow use of e (e.g. $V_B - V_A = \pm 2e$) |
| $2V_A + 2V_B = -12$, $2V_B - 2V_A = 3$ | M1dep p | Attempt to solve both equations simultaneously. Allow use of e (e.g. $V_A = \pm(3+e)$ or $V_B = \pm(3-e)$) |
| $V_B = -2.25$ or $V_A = -3.75$ | A1 | A numerical value is required here (may be implied by a correct final answer). |
| Impulse on $A$ = change in $A$'s momentum $= 4(-3.75-(-2)) = -7 \Rightarrow 7$ Ns | A1 | Ignore wrong units. ISW e.g. any statements regarding direction of travel |
---5 Two identical spheres, $A$ and $B$, each of mass 4 kg , are moving directly towards each other along the same straight line on a smooth horizontal surface until they collide. Before they collide, the speeds of $A$ and $B$ are $5 \mathrm {~ms} ^ { - 1 }$ and $3 \mathrm {~ms} ^ { - 1 }$ respectively. Immediately after they collide, the speed of $A$ is $2 \mathrm {~ms} ^ { - 1 }$ and its direction of motion has been reversed.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the velocity of $B$ immediately after $A$ and $B$ collide.
\item Show that the coefficient of restitution between $A$ and $B$ is $\frac { 3 } { 4 }$.
\item Calculate the total loss of kinetic energy due to this collision.
Sphere $B$ goes on to strike a fixed wall directly. As a result of this collision $B$ moves along the same straight line with a speed of $4 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}\item Find the coefficient of restitution between $B$ and the wall, stating whether the collision between $B$ and the wall is perfectly elastic.
\item Determine the magnitude of the impulse that $B$ exerts on $A$ the next time that they collide.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2023 Q5 [14]}}