The forces acting on the body are: Normal reaction (N$=m\mathrm{g)}$(upward) and Friction force f$\le \mu mg$(opposing applied force)

The resultant force is:

$$F = \sqrt{N^2 + f^2} = \sqrt{mg^2 + (\mu mg)^2} = mg \sqrt{1 + \mu^2}$$

$F \leq mg(1 + \mu^2)^{1/2}$

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.

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:

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

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 \(s = \frac{P^2}{2\mu m^2g}\).

At the highest point, velocity is horizontal: \(v_x = u\cos(37^\circ) = 100 \times 0.8 = 80\text{ m/s}\). By conservation of linear momentum along the horizontal: \(m v_x = m_1 v_1 + m_2 v_2\). Here \(m_1 = m/4\) (comes to rest, \(v_1 = 0\)) and \(m_2 = 3m/4\). Thus, \(m(80) = \frac{3m}{4} v_2\), which gives \(v_2 = \frac{320}{3}\text{ m/s}\).

Impulse is defined as the product of the average force and the short time interval over which it acts, i.e., \(\text{Impulse} = f \times t\).

The maximum limiting frictional force is \(f_{max} = \mu m g = 0.2 \times 20 \times 10 = 40\text{ N}\) (using \(g = 10\text{ m/s}^2\)). Since the applied force of \(40\text{ N}\) is equal to the limiting friction, the net horizontal force on the block is zero, resulting in zero acceleration.

The velocity of the ball just before striking the ground is \(u = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 14.14\text{ m/s}\). Since it rebounds to the same height, its velocity just after is \(v = 14.14\text{ m/s}\) upwards. The change in momentum is \(Delta p = m(v - (-u)) = 2mu = 2 \times 0.15 \times 14.14 \approx 4.24\text{ kg m/s}\).