(a) Consider the two particles as one system.

The total mass is

\[ 0.3+0.2=0.5\text{ kg}. \]

The component of weight down the plane is

\[ 0.5(10)\sin30^\circ=2.5\text{ N}. \]

The total frictional force is

\[ 0.4(0.5)(10)\cos30^\circ=\sqrt3\text{ N}. \]

Using \(F=ma\) up the plane:

\[ 12-2.5-\sqrt3=0.5a. \]

So

\[ a=15.535\ldots \]

Therefore

\[ a=15.5\text{ m s}^{-2}. \]

Now consider particle \(P\).

For \(P\), the component of weight down the plane is

\[ 0.3(10)\sin30^\circ=1.5\text{ N}. \]

The frictional force on \(P\) is

\[ 0.4(0.3)(10)\cos30^\circ=1.2\cos30^\circ. \]

Using \(F=ma\) up the plane:

\[ T-1.5-1.2\cos30^\circ=0.3a. \]

Substituting \(a=15.535\ldots\),

\[ T=7.20\text{ N}. \]

Answer: acceleration \(=15.5\text{ m s}^{-2}\), tension \(=7.20\text{ N}\).

(b) After the string breaks, \(P\) is still moving up the plane. The forces on \(P\) down the plane are the component of weight and friction.

The deceleration of \(P\) is

\[ 10\sin30^\circ+0.4(10)\cos30^\circ. \]

\[ =5+4\cos30^\circ=8.464\ldots\text{ m s}^{-2}. \]

Using \(v=u+at\), with final velocity \(0\),

\[ 0=2-8.464\ldots t. \]

So

\[ t=0.236\text{ s}. \]

Answer: \(0.236\text{ s}\).