For the block to remain stationary relative to the wedge,

the component of  pseudo force up the incline balances the gravitational component along down the incline:

$$ma \cos \theta = mg \sin \theta$$

$$\mathrm{\Rightarrow \; a}=g\tan \theta$$

The total mass of the system:

$$M = 4 + 2 + 1 = 7 \text{ kg}$$

The acceleration of the system:

$$a = \frac{\text{Total Force}}{\text{Total Mass}} = \frac{14}{7} = 2 \text{ m/s}^2$$

Now, considering block B and C together (mass =

$2 + 1 = 3$

kg), the force required to accelerate them:

$$F_{AB} = (3 \times 2) = 6 \text{ N}$$

Thus, the contact force between A and B is 6 N.

In the movable pulley system, tension in the string connecting m2 and m3 is:

$$T = \frac{2 m_2 m_3 g}{m_2 + m_3}$$

Since this tension acts twice to balance

$m_1$

, we equate:

$$2T = m_1 g \Rightarrow \frac{4 m_2 m_3 g}{m_2 + m_3} = m_1 g$$

Cancelling

$g$

and rearranging gives:

$$\boxed{ \frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}} }$$

For ideal pulleys, the product of tension and acceleration remains constant:

$$T_1 a = T_2 a_2$$

Since

$T_2 = 2T_1$

, substitute to get:

$$T_1 a = 2T_1 a_2 \Rightarrow a_2 = \frac{a}{2}$$

$\boxed{\text{So, } M_2 \text{ accelerates upward with } \frac{a}{2}}$

🔹 Step-by-step:

• Block of mass M is on a frictionless incline of angle
  $\theta$ 
• Force
  $F$ 

  is applied via a pulley system, split into two components:

  • One acts up along the incline: F
  • One acts horizontally, which when resolved along the incline becomes: F
    $F \cos \theta$ 

Total upward force along incline =

$F + F \cos \theta$

Downward component of weight =

$Mg \sin \theta$

🔹 For the block to just start moving upward:

$$F+F\cos \theta =Mg\sin \theta$$ $$F(1+\cos \theta )=Mg\sin \theta$$ $$F = \frac{Mg \sin \theta}{1 + \cos \theta}$$

Now use the identity:

$$\frac{\sin \theta}{1 + \cos \theta} = \cot\left(\frac{\theta}{2}\right)$$

 Final Answer:

$$\boxed{F = Mg \cot\left(\frac{\theta}{2}\right)}$$

The angle with the normal is \(90^\circ - 60^\circ = 30^\circ\). The change in momentum perpendicular to the wall is \(\Delta p = 2mv\cos(30^\circ) = 2(1)(1)\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}\text{ kg m s}^{-1}\). Average force \(F = \frac{\Delta p}{\Delta t} = \frac{\sqrt{3}}{0.1} = 10\sqrt{3}\text{ N}\).

Differentiating position equations twice yields accelerations: \(a_x = 48\text{ m/s}^2\) and \(a_y = 18t\text{ m/s}^2\). At \(t = 2\text{ s}\), \(a_y = 36\text{ m/s}^2\). Total acceleration \(a = \sqrt{a_x^2 + a_y^2} = 60\text{ m/s}^2\). The net force is \(F = ma = 3.0 \times 60 = 180\text{ N}\).

For a body in equilibrium, the net external force acting on it is zero. According to Newton's second law, \(F_{\\text{net}} = ma = 0\), which means the acceleration of the body must be zero.

The stopping distance is given by \(s = \frac{v^2}{2a}\), where the frictional retardation is \(a = \mu g\). Substituting the relation for momentum \(v = \frac{P}{m}\) into the formula yields \(s = \frac{P^2}{2\mu m^2g}\).