## Question 11(a):

For $P$: $40 - T - 8 = 3a$ | **A1** | —

For $Q$: $T - 2g = 2a$ | **A1** | —

$32 - 2g = 5a \Rightarrow a = \ldots$ | **Dep\*M1** | Attempt to solve simultaneous equations; M1\* attempt N2L for $P$ or $Q$ (must include correct number of terms — use of weight for mass is **M0**)

$a = 2.48 \text{ m s}^{-2}$ | **A1** | AG — must show sufficient working to justify the given answer; M1 A2 for $40 - 8 - 2g = 5a$ for M1 must have correct number of terms and mass must be $5$ not $5g$

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## Question 11(b):

$T - 2g = 2(2.48)$ | **M1** | Substitute given value of $a$ into either equation; must include correct number of terms — use of weight for mass is **M0**

$T = 24.56 \text{ N}$ | **A1** | cao; allow 24.6

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## Question 11(c):

$v = 2.48(0.5)$ | **B1** | Speed after 0.5 seconds $= 1.24$

$s = 0.5(2.48)(0.5)^2$ | **B1** | Distance travelled in this time $= 0.31$

$-(2 + 0.31) = 1.24t - 0.5(9.8)t^2$ | **M1** | Applying $s = ut + 0.5at^2$ correctly — allow sign errors; **M0** if not using relevant displacement

$t = 0.825 \text{ s}$ | **A1** | BC; $0.8246986\ldots$

**ALT:** $0 = 1.24^2 - 2(9.8)s \Rightarrow s = 0.0784$; $2 + 0.31 + 0.0784 = 0.5(9.8)t_1^2$; $t_1 = 0.698$; $t = 0.825$ | **M1, A1, A1** | Complete method to calculate time down; correct value for time down

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## Question 11(d):

$v^2 = 1.24^2 + 2(-9.8)(-2.31)$ | **M1** | Applying $v^2 = u^2 + 2as$ correctly with their 1.24 and 2.31 or any other complete method; **M0** if not using total time or relevant displacement

$v = 6.84 \text{ m s}^{-1}$ | **A1** | Allow 6.85; $6.8420464\ldots$

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## Question 11(e):

$19.6 \text{ N}$ | **B1** | Accept $2g$

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## Question 11(f):

e.g. include a more accurate value for $g$; e.g. include a variable resistance in the model rather than a constant; e.g. include the dimension of the pulley in the model so that the string is not parallel to the table; e.g. include a frictional force at the pulley | **B1** | AO 3.5c