Connected Particles

Resolving for Ball $1$:

$T – (0.06 times 9.8) = T – 0.588 = 0.06a$

Resolving for Ball $2$:

$(0.12 times 9.8) – T = 1.176 – T = 0.12a$

From here, we have

$0.12a = 2T – 1.176 = 1.176 – T$

which can be rearranged to

$3T = 2.352$, or $T = 0.784text{ N}$

Substituting this back into one of our equations, we have $1.176 – 0.784 = 0.12a$, meaning that Ball $2$ has a downward acceleration of $a = 3.27text{  ms}^{-2}$. This means that Ball $1$ accelerates towards the peg at $3.27text{  ms}^{-2} text{ (to }2text{ dp)}$.

Using SUVAT, $s = ut + dfrac{1}{2}at^2$ gives

$0.3 = dfrac{1}{2} times 3.27 times t^2$, so $t = 0.43text{ seconds}$

Weight $text{A}$ (vertical):

$R = mg = 4 times 9.8 = 39.2text{ N}$

Weight $text{A}$ (horizontal):

$T – mu R = ma$, giving $T – (0.2 times 39.2) = T – 7.84 = 4a$

Weight $text{B}$ (vertical):

$mg – T = ma$, giving $39.2 – T = 4a$

Therefore,

$4a = 39.2 – T = T – 7.84$

which can be rearranged to

$2T = 47.04$, or $T = 23.52text{ N}$

Substituting back into one of our equations, we have $23.52 – 7.84 = 4a$

giving

$a = dfrac{23.52 – 7.84}{4} = 3.92text{  ms}^{-2}$

Weight $text{A}$ (perpendicular to motion):

$R = mgcos 45° = (9 times 9.8)cos 45° = dfrac{88.2}{sqrt{2}}$

Weight $text{A}$ (parallel to motion, down slope):

$mgsin 45° – T – F = mgsin 45° – T – mu R = ma$

giving

$(9 times 9.8)sin 45° – T – left( 0.1 times dfrac{88.2}{sqrt{2}}right)$

$= dfrac{88.2}{sqrt{2}} – T – dfrac{8.82}{sqrt{2}} = 9a$.

Weight $text{B}$ (upwards):

$T – mg = ma$, giving $T – (3 times 9.8) = 3a$

or

$T – 29.4 = 3a$

Equating the two equations:

$9a = 3T – 88.2 = dfrac{79.38}{sqrt{2}} – T$

So

$4T = dfrac{79.38}{sqrt{2}} + 88.2 = 144.33$

giving

$T = 36.08text{ N(to }2text{ dp)}$

Therefore, we have

$9a = (3 times 36.08) – 88.2$

so

$a = 2.23text{  ms}^{-2}$