# Question 8:

## Part (a)
The velocity of incoming chemical is directed into the pipe or there is no work done on the liquid as it enters the pipe | B1 [1] (AO 3.3) | There is no change in KE is insufficient; comments relating to energy changes as the liquid enters the tube; do not accept trivial statements such as constant velocity; ignore "other resistances"; ignore any comments relating to changes of energy within the tube, or changes in density/compressibility

## Part (b)
In one hour, increase in KE $= \frac{1}{2}\times1500\times(14.3^2 - 6.2^2)$ | B1 (AO 3.4) | Change in KE soi; could be divided by 3600 $(42.6... - 8.00... = 34.59...)$ for 1 second (or $= 83.025$ for 1 kg); $153367.5 - 28830 = 124537.5$; NB may be seen in part c); could also see reference to 2.354 kg in 5.65s to go through the tube
In one hour, increase in PE $= 1500\times9.8\times35\sin26°$ | M1 (AO 3.4) | Allow $\cos26°$ but not $1500g\times35$; could be divided by $3600$ $(= 62.65...)$ (or $150.36...$ for 1kg); $225541.955...$; NB may be seen in part c)
Rate at which work is done against resistance in the tube is $40\times6.2$ | B1 (AO 1.1) | Or work done against resistance $= 40\times(6.2\times3600) = 892800$ J; allow $40\times35$ if divided by 5.65s; 248W; do not allow if the resistance is treated as a driving force or used to find a driving force of 40N
Power at which the pump is working is $\dfrac{\Delta KE + \Delta PE}{3600} +$ "$40(6.2)$" | M1 (AO 3.1b) | oe: could have total energy $\div 3600$; allow if $40\times35$ used and added to the total energy; must be dimensionally correct
$345$ W | A1 [5] (AO 1.1) | Accept any valid units for power; do not allow use of suvat

**Alternative Method:**
(At the start): PE $= 0$ and KE $= \frac{1}{2}\times1500\times6.2^2$ and final KE $= \frac{1}{2}\times1500\times14.3^2$ | B1 | oe, e.g. initial and final KE seen in a balanced equation; could be expressed per second or for 1kg or for 35m (5.65s)
(At the pump end): PE $= 1500\times9.8\times35\sin26°$ | M1 |
Work done against resistance $= 40\times6.2\times3600$ | B1 | Or rate $= 40\times6.2$ $(= 248)$
$\frac{1}{2}\times1500\times6.2^2 + 3600P + 40\times6.2\times3600 = 1500\times9.8\times35\sin26° + \frac{1}{2}\times1500\times(14.3^2)$ oe | M1 | Must be dimensionally correct
$P = 345$ W | A1 [5] |

## Question (c)(i):

**Calculation of useful energy output**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $450 \times 3600 - (124537.5 + 225541.955...)$ | M1 | Must not include resistance, e.g. $345 \times 3600$ accounted for (see alternative method) |
| $1620000 - 350079... = 1269921...$ NB $1620000 - 351479...$ is M0 | | |
| $= 1270 \text{ kJ}$ to 3sf | A1 | A0 for $1268520...$ |
| **Alternative Method:** $(450 - 345 + 40 \times 6.2) \times 3600 = 1270 \text{ kJ}$ | M1, A1 | Use of excess power output $\times 3600$ |

**[2 marks total]**

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## Question (c)(ii):

**Explanation of modelling assumption limitation**

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. work must be done against other resistance forces (e.g. at the nozzle) or a blockage; or the pump would heat up the air/chemical/tube(s); which requires energy e.g. due to friction between fluid and pump blades (exclude internal resistance); or total resistance to motion may be more than 40N so more energy is required; or model ignores other resistances to motion; or fluid not all moving at same velocity (turbulent flow) or may be viscous so there would be internal resistance to overcome; or some energy required to change direction of velocity of liquid at intake; so pump needs to provide more energy to get intake liquid to velocity of $6.2 \text{ ms}^{-1}$ up the tube | B1 | An explanation which looks at one of the modelling assumptions and shows that higher power output or more energy may be required if it does not hold. Ignore anything referring to internal losses in the pump. Candidates need to give a valid reason or example not covered by the question text. B0 for considering internal resistance of motor/electrical energy. Exclude: "energy will not always be constant in the system"; "velocity is not always constant"; "the model only considers mechanical energy, not electrical energy"; "the resistance to motion is not constant"; "there would be more resistance"; "power output of motor is not constant"; "It doesn't consider resistance inside the pump"; "the fluid cannot be modelled as a particle"; "energy loss due to inefficiency in delivery of power"; "there will be friction"; "no liquid escapes the tube"; "no thermal or sound energy escapes"; "The flow of liquid is laminar". References to heat or noise, unless clearly associated with movement of fluid, excluded. |

**[1 mark total]**