Moderate -0.5 This is a standard SUVAT kinematics problem requiring reading a velocity-time graph and applying basic formulas for displacement (area under graph) and acceleration (gradient). It's slightly easier than average as it involves straightforward graph interpretation and routine calculations with no complex problem-solving or novel insights required.
3
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\({ } ^ { P A } { } _ { P A } ^ { P B }\)
\(P\)
\(P\)
\begin{verbatim}
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\end{verbatim}
\(21/\sin 104.3 = T_B/\sin 143.1\) (or \(T_A/\sin 112.6\)) or \(T_B/\sin 143.1 = 20/\sin 112.6\) or \(T_A/\sin 112.6 = 13/\sin 143.1\)
B1
Solve for \(T_A\) and \(T_B\)
M1
For using the equations to find \(T_A\) and \(T_B\)
Tension in AP is \(20\text{ N}\) and tension in BP is \(13\text{ N}\)
A1 [5]
Both \(T_A = 20\) and \(T_B = 13\)
# Question 3:
## Main Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces acting on P horizontally or vertically |
| $T_A \times (40/50) + T_B \times (40/104) = 21$ or $T_A \times (30/50) = T_B \times (96/104)$ | A1 | |
| $T_A \times (30/50) = T_B \times (96/104)$ or $T_A \times (40/50) + T_B \times (40/104) = 21$ | B1 | |
| Solve for $T_A$ and $T_B$ | M1 | Solving for both |
| Tension in AP is $20\text{ N}$ and tension in BP is $13\text{ N}$ | A1 [5] | Both $T_A = 20$ and $T_B = 13$ |
## First Alternative Marking Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using the sine rule in the triangle of forces |
| $21/\sin 75.75$ (or $75.7$ or $75.8$) $= T_A/\sin 67.4$ (or $T_B/\sin 36.9$) | A1 | |
| $21/\sin 75.75 = T_B/\sin 36.9$ (or $T_A/\sin 67.4$) or $T_B/\sin 36.9 = 20/\sin 67.4$ | B1 | |
| Solve for $T_A$ and $T_B$ | M1 | Solving for both |
| Tension in AP is $20\text{ N}$ and tension in BP is $13\text{ N}$ | A1 [5] | Both $T_A = 20$ and $T_B = 13$ |
## Second Alternative Marking Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using Lami's Rule |
| $21/\sin 104.3 = T_A/\sin 112.6$ (or $T_B/\sin 143.1$) | A1 | |
| $21/\sin 104.3 = T_B/\sin 143.1$ (or $T_A/\sin 112.6$) or $T_B/\sin 143.1 = 20/\sin 112.6$ or $T_A/\sin 112.6 = 13/\sin 143.1$ | B1 | |
| Solve for $T_A$ and $T_B$ | M1 | For using the equations to find $T_A$ and $T_B$ |
| Tension in AP is $20\text{ N}$ and tension in BP is $13\text{ N}$ | A1 [5] | Both $T_A = 20$ and $T_B = 13$ |
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