Foundation November 2022 Paper 1 Q24
24. A glass of water is placed on a small table. The table stands on horizontal ground.
Diagram not drawn to scale
The total mass of the table and the glass of water is 9·6 kg.
You are given the ratios,
mass of table : mass of glass of water = 11 : 1,
mass of empty glass : mass of water = 3 : 5.
What is the mass of the empty glass? [3]
| Use: Pressure = \(\dfrac{\text{Force (N)}}{\text{Area (cm}^2\text{)}}\) |
Diagram not drawn to scale
The base of the table has an area of \(1600\,\text{cm}^2\).
Some books are also placed on the table.
The books, glass of water and table exert a pressure of \(0{\cdot}1\,\text{N/cm}^2\) on the ground.
Calculate the force exerted on the ground by the books, glass of water and table.
Assume that the whole of the base of the table is in contact with the ground. [2]
In fact, the assumption made in part (b) is incorrect. Part of the base of the table is not in contact with the ground.
Describe how this changes your answer to part (b). [1]
| Scheme | Marks | Comments |
|---|---|---|
| \(9.6 \div 12\) (= 0.8) | M1 | |
| \(0.8 \div 8 \times 3\) | M1 | FT ‘their 0.8’ including place value error from conversion of kg to g |
| 0.3 (kg) or 300g | A1 | CAO |
Alternative
| Scheme | Marks | Comments |
|---|---|---|
| 88 : 3 : 5 | B1 | |
| \(\dfrac{3}{88 + 3 + 5} \times 9.6\) | M1 | |
| 0.3 (kg) or 300g | A1 | CAO |
| Scheme | Marks | Comments |
|---|---|---|
| (Total force = ) \(1600 \times 0.1\) | M1 | |
| 160 (N) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| Valid impact e.g. ‘The force would be less’ | E1 | Ignore any extraneous comments e.g. ‘The pressure would increase, and the force will be lower’. |
| (6) |