| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with given impulse |
| Difficulty | Standard +0.3 This is a straightforward two-part mechanics question requiring standard impulse calculation (change in momentum with sign consideration) and energy conservation with resistance forces. While it involves Further Mechanics content, both parts use direct application of formulas without requiring novel problem-solving insight or complex multi-step reasoning. |
| Spec | 6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = mv - mu = 0.2\times24 - 0.2\times-14\) | M1 (AO 1.1) | Use of \(I = \pm\Delta mv\) soi; if \(4.8 - 2.8\), then must be clear evidence of sign error in the second velocity e.g. \(\pm(0.2\times24 - 0.2\times14)\) |
| \(7.6\) Ns | A1 [2] (AO 1.1) | Magnitude must be \(> 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Initial (kinetic) energy \(= \frac{1}{2}\times0.2\times24^2\) | B1 (AO 1.1) | Use of \(\frac{1}{2}mv^2\) in an attempt to calculate initial KE of puck; 57.6 |
| Final (potential) energy \(= 0.2g\times15\sin10°\) | M1 (AO 1.1) | Use of \(mgh\) in an attempt to calculate final PE of puck; 5.105...; allow sin/cos confusion; NB count use of \(g = 9\)/GPE \(= 4.688\)... as a slip |
| Work done against resistance \(= R\times15\) | M1 (AO 1.1) | Use of "\(W = Fd\)" |
| \(15R + 0.2g\times15\sin10° = \frac{1}{2}\times0.2\times24^2\) | M1 (AO 3.4) | Balancing their energies (3 terms); all terms must be in correct direction and dimensionally correct |
| awrt \(3.50\) N | A1 [5] (AO 1.1) | If N2L used SC2 for 3.50 N www |
# Question 2:
## Part (a)
$I = mv - mu = 0.2\times24 - 0.2\times-14$ | M1 (AO 1.1) | Use of $I = \pm\Delta mv$ soi; if $4.8 - 2.8$, then must be clear evidence of sign error in the second velocity e.g. $\pm(0.2\times24 - 0.2\times14)$
$7.6$ Ns | A1 [2] (AO 1.1) | Magnitude must be $> 0$
## Part (b)
Initial (kinetic) energy $= \frac{1}{2}\times0.2\times24^2$ | B1 (AO 1.1) | Use of $\frac{1}{2}mv^2$ in an attempt to calculate initial KE of puck; 57.6
Final (potential) energy $= 0.2g\times15\sin10°$ | M1 (AO 1.1) | Use of $mgh$ in an attempt to calculate final PE of puck; 5.105...; allow sin/cos confusion; NB count use of $g = 9$/GPE $= 4.688$... as a slip
Work done against resistance $= R\times15$ | M1 (AO 1.1) | Use of "$W = Fd$"
$15R + 0.2g\times15\sin10° = \frac{1}{2}\times0.2\times24^2$ | M1 (AO 3.4) | Balancing their energies (3 terms); all terms must be in correct direction and dimensionally correct
awrt $3.50$ N | A1 [5] (AO 1.1) | If N2L used SC2 for 3.50 N www
---2 A hockey puck of mass 0.2 kg is sliding down a rough slope which is inclined at $10 ^ { \circ }$ to the horizontal. At the instant that its velocity is $14 \mathrm {~ms} ^ { - 1 }$ directly down the slope it is hit by a hockey stick. Immediately after it is hit its velocity is $24 \mathrm {~ms} ^ { - 1 }$ directly up the slope.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse exerted by the hockey stick on the puck.
After it has been hit, the puck first comes to instantaneous rest when it has travelled 15 m up the slope. While the puck is moving up the slope, the resistance to its motion has constant magnitude $R \mathrm {~N}$.
\item Use an energy method to determine the value of $R$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2022 Q2 [7]}}