| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions with wall rebound |
| Difficulty | Challenging +1.2 This is a standard two-stage collision problem requiring conservation of momentum and Newton's restitution law applied twice, with an inequality condition to prevent a second collision. Part (a) is routine algebraic manipulation (showing a given result), while part (b) requires setting up and solving an inequality involving the coefficient of restitution—a common Further Mechanics exercise but requiring careful reasoning about relative velocities after both collisions. |
| Spec | 6.03b 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 |
|---|---|---|
| Mom\(^n\): \(1\times1.79 + 2.74\times-0.08 = v_P + 2.74v_Q\) | M1 (AO 3.3) | Attempt at equating momentum before and after collision between \(P\) and \(Q\) with 4 terms; 1.5708; allow 1 incorrect mass and one sign slip |
| Rest\(^n\): \(e = -(v_P - v_Q)/(1.79 - -0.08)\) | M1 (AO 3.3) | Attempt at using NEL; accept global sign error; allow sign error in \(u_Q\) provided that this is shown clearly; \(v_Q - v_P = 1.87e\) |
| A1 (AO 1.1) | Both equations correct | |
| \(v_P = v_Q - 1.87e\) and \(v_P = 1.5708 - 2.74v_Q\) | M1 (AO 1.1) | Attempt to solve simultaneously e.g. by substituting for \(v_P\) |
| \(\rightarrow v_Q - 1.87e = 1.5708 - 2.74v_Q\) | Or using elimination: \(1.5708 = v_P + 2.74v_Q\) and \(1.87e = v_Q - v_P\); \(1.5708 + 1.87e = 3.74v_Q\) | |
| \(v_Q = (1.5708 + 1.87e)/3.74 = 0.42 + 0.5e\) AG | A1 [5] (AO 1.1) | AG; intermediate working must be shown; final value must be positive; so \(v_Q = 0.42 + 0.5e\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_P = v_Q - 1.87e = 0.42 - 1.37e\) | M1 (AO 1.1) | Deriving \(v_P\) from the equations and/or answer in (a); or \(1.5708 = v_P + 2.74(0.42 + 0.5e)\) |
| After \(Q\) hits wall: \(w_Q = \pm e(0.42 + 0.5e)\) | M1 (AO 3.1b) | \((-)e \times\) their \(v_Q\) |
| No \(2^{\text{nd}}\) collision so their \(v_P \leq v_Q\) soi | B1 (AO 2.2a) | Condition on velocities given no \(2^{\text{nd}}\) collision occurs; allow strict inequality for this mark; if using left hand reference then \(v_P \geq v_Q\) |
| \(0.42 - 1.37e \leq \pm e(0.42 + 0.5e)\) | M1* (AO 3.1b) | Condone any inequality or equality sign; must be derived from an attempt at \(v_P\) and \(w_Q\) in terms of \(e\) |
| \(e^2 - 1.9e + 0.84 \leq 0\) | M1dep* (AO 1.1) | Rearranging to 3-term inequality; must see zero on one side of the inequality |
| Critical values for \(e\): \(1.2, 0.7\) | A1FT (AO 1.1) | BC (correct CVs for their inequality, which must be a 3-term quadratic); NB if \(w_Q = 0.42e + 0.5e^2\) then expect to see \(-3.80\) and \(0.221\); at least one value must be positive |
| \(0.7 \leq e \leq 1.2\) and \(0 \leq e \leq 1 \Rightarrow 0.7 \leq e \leq 1\) | A1 [7] (AO 2.3) | cao; do not allow strict inequality; if derived from equality, then inequality must be fully justified |
# Question 7:
## Part (a)
Mom$^n$: $1\times1.79 + 2.74\times-0.08 = v_P + 2.74v_Q$ | M1 (AO 3.3) | Attempt at equating momentum before and after collision between $P$ and $Q$ with 4 terms; 1.5708; allow 1 incorrect mass and one sign slip
Rest$^n$: $e = -(v_P - v_Q)/(1.79 - -0.08)$ | M1 (AO 3.3) | Attempt at using NEL; accept global sign error; allow sign error in $u_Q$ provided that this is shown clearly; $v_Q - v_P = 1.87e$
| A1 (AO 1.1) | Both equations correct
$v_P = v_Q - 1.87e$ and $v_P = 1.5708 - 2.74v_Q$ | M1 (AO 1.1) | Attempt to solve simultaneously e.g. by substituting for $v_P$
$\rightarrow v_Q - 1.87e = 1.5708 - 2.74v_Q$ | | Or using elimination: $1.5708 = v_P + 2.74v_Q$ and $1.87e = v_Q - v_P$; $1.5708 + 1.87e = 3.74v_Q$
$v_Q = (1.5708 + 1.87e)/3.74 = 0.42 + 0.5e$ **AG** | A1 [5] (AO 1.1) | AG; intermediate working must be shown; final value must be positive; so $v_Q = 0.42 + 0.5e$
## Part (b)
$v_P = v_Q - 1.87e = 0.42 - 1.37e$ | M1 (AO 1.1) | Deriving $v_P$ from the equations and/or answer in **(a)**; or $1.5708 = v_P + 2.74(0.42 + 0.5e)$
After $Q$ hits wall: $w_Q = \pm e(0.42 + 0.5e)$ | M1 (AO 3.1b) | $(-)e \times$ their $v_Q$
No $2^{\text{nd}}$ collision so their $v_P \leq v_Q$ soi | B1 (AO 2.2a) | Condition on velocities given no $2^{\text{nd}}$ collision occurs; allow strict inequality for this mark; if using left hand reference then $v_P \geq v_Q$
$0.42 - 1.37e \leq \pm e(0.42 + 0.5e)$ | M1* (AO 3.1b) | Condone any inequality or equality sign; must be derived from an attempt at $v_P$ and $w_Q$ in terms of $e$
$e^2 - 1.9e + 0.84 \leq 0$ | M1dep* (AO 1.1) | Rearranging to 3-term inequality; must see zero on one side of the inequality
Critical values for $e$: $1.2, 0.7$ | A1FT (AO 1.1) | BC (correct CVs for their inequality, which must be a 3-term quadratic); NB if $w_Q = 0.42e + 0.5e^2$ then expect to see $-3.80$ and $0.221$; at least one value must be positive
$0.7 \leq e \leq 1.2$ and $0 \leq e \leq 1 \Rightarrow 0.7 \leq e \leq 1$ | A1 [7] (AO 2.3) | cao; do not allow strict inequality; if derived from equality, then inequality must be fully justified
---7 Two particles, $P$ and $Q$, are on a smooth horizontal floor. $P$, of mass 1 kg , is moving with speed $1.79 \mathrm {~ms} ^ { - 1 }$ directly towards a vertical wall. $Q$, of mass 2.74 kg , is between $P$ and the wall, moving directly towards $P$ with speed $0.08 \mathrm {~ms} ^ { - 1 }$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{98053e88-1aec-4b0d-ae5f-ece4ad340266-4_232_830_1370_246}\\
$P$ and $Q$ collide directly and the coefficient of restitution for this collision is denoted by $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that after this collision the speed of $Q$ is given by $0.42 + 0.5 e \mathrm {~ms} ^ { - 1 }$.
After this collision, $Q$ then goes on to collide directly with the wall. The coefficient of restitution for the collision between $Q$ and the wall is also $e$. There is then no subsequent collision between $P$ and $Q$.
\item Determine the range of possible values of $e$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2022 Q7 [12]}}