| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 12 m 6 2 + m g 6 .5 − 4 2 = 12 m 1 2 2 |
| Answer | Marks |
|---|---|
| 9.7𝑚 = 42, so m = 4 .3 3 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Any two of 1𝑚(62), 1𝑚(122), 𝑚𝑔(6.5) |
| Answer | Marks |
|---|---|
| (b) | 1 𝑚(32)+(50cos𝜃)(5)−58 = 1 𝑚(62)+𝑚𝑔(5sin30°) |
| Answer | Marks |
|---|---|
| 𝜃 = 27.1° | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Work done by F is (50cos𝜃)(5) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 2×5 | B1 | B1 |
| 50cos𝜃−11.6−4.33×9.8sin30° = 4.33×2.7 | M1 | N2L Allow sign errors and one missing term |
| 𝜃 = 27.1° | A1 | |
| (c) | 5 0 c o s 6 | |
| =267(W) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | Use of P = D v where D 5 0 c o s = or D=50sin with |
Question 5:
5 | (a) | 12 m 6 2 + m g 6 .5 − 4 2 = 12 m 1 2 2
18𝑚+63.7𝑚−42 = 72𝑚
9.7𝑚 = 42, so m = 4 .3 3 | B1
M1
A1
[3] | 1.1
3.3
1.1 | Any two of 1𝑚(62), 1𝑚(122), 𝑚𝑔(6.5)
2 2
Attempt at WEP on stage BC; Allow sign errors and
one missing term (but must contain 42)
AG Fully correct working. At least one step required
between WEP equation and the given answer
Allow verification
(b) | 1 𝑚(32)+(50cos𝜃)(5)−58 = 1 𝑚(62)+𝑚𝑔(5sin30°)
2 2
19.5+250cos𝜃−58 = 77.9+106.1
38𝑚+58 222.5
cos𝜃 = = = 0.8901…
250 250
𝜃 = 27.1° | B1
B1
M1
A1
[4] | 1.1
1.1
3.3
1.1 | Work done by F is (50cos𝜃)(5)
Change in PE is 𝑚𝑔(5sin30°)
Attempt at WEP on stage AB; Allow sign errors and
one missing term (but must contain WD by F term)
Assuming constant resistance and acceleration (Max 3/4)
58 62−32
Resistance is (= 11.6), Acceleration is (= 2.7)
5 2×5 | B1 | B1 | For both | For both
50cos𝜃−11.6−4.33×9.8sin30° = 4.33×2.7 | M1 | N2L Allow sign errors and one missing term
𝜃 = 27.1° | A1
(c) | 5 0 c o s 6
=267(W) | M1
A1 FT
[2] | 3.4
1.1 | Use of P = D v where D 5 0 c o s = or D=50sin with
their value of from part (b), and 𝑣 = 6
FT is 300cos(their 𝜃)5 In the diagram below, points $\mathrm { A } , \mathrm { B }$ and C lie in the same vertical plane. The slope AB is inclined at an angle of $30 ^ { \circ }$ to the horizontal and $\mathrm { AB } = 5 \mathrm {~m}$. The point B is a vertical distance of 6.5 m above horizontal ground. The point C lies on the horizontal ground.\\
\includegraphics[max width=\textwidth, alt={}, center]{a96a0ebe-8f4f-4d79-9d11-9d348ef72314-6_601_1285_395_244}
Starting at A , a particle P , of mass $m \mathrm {~kg}$, moves along the slope towards B , under the action of a constant force $\mathbf { F }$. The force $\mathbf { F }$ has a magnitude of 50 N and acts at an angle of $\theta ^ { \circ }$ to AB in the same vertical plane as A and B . When P reaches $\mathrm { B } , \mathbf { F }$ is removed, and P moves under gravity landing at C .
It is given that
\begin{itemize}
\item the speed of P at A is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item the speed of P at B is $6 \mathrm {~ms} ^ { - 1 }$,
\item the speed of P at C is $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item 58 J of work is done against non-gravitational resistances as P moves from A to B ,
\item 42 J of work is done against non-gravitational resistances as P moves from B to C .
\begin{enumerate}[label=(\alph*)]
\item By considering the motion from B to C, show that $m = 4.33$ correct to 3 significant figures.
\item By considering the motion from A to B , determine the value of $\theta$.
\item Calculate the power of $\mathbf { F }$ at the instant that P reaches B .
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2024 Q5 [9]}}