June 2024 Paper 3 Q10
10
A block of mass \(m\) kg is on smooth horizontal ground with one end of a light inextensible rope attached to its upper surface. The other end of the rope is attached to an object of mass 5 kg. The rope passes over a small smooth pulley, and the object hangs vertically below the pulley. The part of the rope between the block and the pulley makes an angle of 50° with the horizontal. A force of magnitude \(X\) N acts on the block at an angle of 20° above the horizontal in the vertical plane containing the rope (see diagram).
You are given that the block is in equilibrium.
You are also given that the magnitude of the contact force exerted by the ground on the block is 147 N.
| Scheme | Marks | AO |
|---|---|---|
| \(R(\uparrow \text{object}):\quad T = 5g\) | B1 | 1.1 |
| \(R(\rightarrow \text{block}):\quad T\cos 50 = X\cos 20\) | M1 | 3.3 |
| \(X = 33.5\) | A1 | 1.1 |
| [3] |
Notes
B1: Possibly implied by later working
Allow \(T - 5g = 0\)
M1: Resolving horizontally to form an equation – correct number of terms (possibly with \(T\) replaced with \(5g\)) – allow sin/cos mix but must be using correct angles
Must be components of both \(T\) and \(X\)
A1: awrt 33.5 – condone including of ‘N’
33.51797…
| Scheme | Marks | AO |
|---|---|---|
| \(X\sin 20 + 147 + T\sin 50 = mg\) | M1 | 3.3 |
| \((\text{‘33.5…’})\sin 20 + 147 + 5g\sin 50 = mg\) | A1FT | 1.1 |
| \(m = 20\) | A1 | 1.1 |
| [3] |
Notes
M1: Resolving vertically to form an equation for the block with the correct number of relevant terms – allow sign errors and sin/cos mix but must be using correct angles
Allow \(X\), \(T\) or their value(s) for \(X\) and \(T\) – relevant means both dimensionally correct (so must be using \(mg\)) and that we must see components of \(T\) and \(X\) but no component of the weight or the 147
A1FT: Correct equation for \(m\) FT their value of \(X\) only (so not just \(X\))
A1: Accept either 20 or awrt 20.0; condone inclusion of ‘kg’
www