CAIE A-Level Mathematics AS 4.5.3 Forces Questions

$$\begin{aligned} 0^{2}=(40)^{2}+2 \times 0.04 \times a \quad[a=-20000] \text { or } 0.04=\frac{0+40}{2} t \text { gets } t=0.002, \text { so } 0=40+0.002 a[a=-20000] \end{aligned}$$
Use of a 'suvat' method to get an equation in a. Allow sign errors. Allow $\pm 20000$.

Do not allow 4 in place of 0.04 .

Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40 .

Attempt to use Newton's Second Law vertically.
$$\begin{aligned} {[-R+(1.2+0.004) g=(1.2+0.004) \times a]} {[-R+12.04=1.204 a]} \end{aligned}$$
Must have the correct number of relevant terms. Allow sign errors, but terms including masses must be effectively added. Do not allow any mass other than (1.2
+ 0.004).

$R=24100 \mathrm{~N}\left[24092.04=\frac{602301}{25}\right]$
WWW.

Note: use of wrong sign for g leads to answers 24067.96 which gets max M1M1A0.

Note: Missing weight term gets 24080 which gets Max M1M0A0.

3

2(b)

Alternative method for Question 2(b) using energy

[Change in PE =] $1.204 \mathrm{~g} \times 0.04[=0.4816]$
or [change in KE = ] $\frac{1}{2} \times 1.204 \times(40)^{2}[=963.2]$
Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40 .
B0 for kinetic energy, if extra kinetic energy terms
present.

$1.204 \mathrm{~g} \times 0.04+\frac{1}{2} \times 1.204 \times(40)^{2}=0.04 R$
Attempt at work energy equation. Must have correct number of relevant terms. dimensionally correct;
allow sign errors.

Do not allow 4 in place of 0.04.

Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40 .

$R=24100 \mathrm{~N}\left[24092.04=\frac{602301}{25}\right]$
WWW

Note: use of wrong sign for g leads to answers 24067.96 which gets max B1M1A0.

Note: Missing potential energy term gets 24 080, which gets maximum of B1M0A0.