OCR FM1 AS 2018 March — Question 2 5 marks

Exam BoardOCR
ModuleFM1 AS (Further Mechanics 1 AS)
Year2018
SessionMarch
Marks5
TopicWork done and energy
TypeAverage power over journey
DifficultyStandard +0.3 This is a straightforward work-energy problem requiring calculation of power from two components: gravitational PE gain (mgh) and KE of the jet (½mv²). The setup is clear, the physics is standard A-level mechanics, and it's a direct application of formulas with no conceptual subtlety. Slightly above average difficulty only because it's Further Maths and requires combining two energy terms correctly.
Spec6.01a Dimensions: M, L, T notation6.02a Work done: concept and definition6.02k Power: rate of doing work6.02l Power and velocity: P = Fv

2 A pump is pumping still water from the base of a well at a constant rate of 300 kg per minute. The well is 4.5 m deep and water is released from the pump at ground level in a horizontal jet with a speed of \(6.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Ignoring any energy losses due to resistance, calculate the power generated by the pump.

AnswerMarks Guidance
AnswerMarks Guidance
\(5 \text{ kg per second}\)B1 [5] Could be gained at the end if, for example, a minute is considered or for their time period
Gain in PE per second \(= 5 \times g \times 4.5\)M1 or for their considered time period
Gain in KE per second \(= \frac{1}{2} \times 5 \times 6.2^2\)M1
awrt 96 or awrt 221A1 Could be implied by correct answer and could be multiplied by their time period
awrt 320 W or 320 J \(\text{s}^{-1}\)A1
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5 \text{ kg per second}$ | B1 [5] | Could be gained at the end if, for example, a minute is considered or for their time period |
| Gain in PE per second $= 5 \times g \times 4.5$ | M1 | or for their considered time period |
| Gain in KE per second $= \frac{1}{2} \times 5 \times 6.2^2$ | M1 | |
| awrt 96 or awrt 221 | A1 | Could be implied by correct answer and could be multiplied by their time period |
| awrt 320 W or 320 J $\text{s}^{-1}$ | A1 | |
2 A pump is pumping still water from the base of a well at a constant rate of 300 kg per minute. The well is 4.5 m deep and water is released from the pump at ground level in a horizontal jet with a speed of $6.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Ignoring any energy losses due to resistance, calculate the power generated by the pump.

\hfill \mbox{\textit{OCR FM1 AS 2018 Q2 [5]}}