B.

486 J

The linear momentum of the exploding part will remain conserved.

Applying coservation of linear momentum,

m₁ u₁ =m₂ u₂

Here, m₁ = 18kg, m₂ = 12kg

u₁ = 6ms^-1, u₂ = ?

${\mathrm{u}}_2 = \frac{18\times 6}{12} 9 {\mathrm{ms}}^{-1}$

Thus kinetic energy of 12 kg mass

$$\begin{gathered} {\mathrm{K}}_2 =\frac{1}{2}{\mathrm{m}}_1{\mathrm{u}}_2^2 \\ = \frac{1}{2}\times 12\times {\left(9\right)}^2 \\ = 6\times 81 \\ {\mathrm{K}}_2 = 486 \mathrm{J} \end{gathered}$$

B.

63.90^o

A body that travels an equal distance in equal amounts of time along a circular path has a constant speed but not constant velocity. This is because velocity is a vector and thus it has magnitude as well as direction.

               v = ( Rg tanθ )^1/2

               $\tan \mathrm{\theta } = \frac{{\mathrm{v}}^2}{\mathrm{r} \mathrm{g}}$

               = $\frac{400}{20 \times 900}$

         tan θ = 63.70^o 

         tanθ ≈ 63.90^o

A.

200 m/s

Linear momentum is a product of the mass (m) of an object and the velocity (v) of the object.

According to conservation of linear momentum

          m₁ v₁ = m₁ v + m₂ v₂

where v₁ is the velocity of the bullet before the collision, v  is the velocity of bullet after the collision and v₂ is the velocity of the block.

∴     0.02 × 600 = 0.02 v + 4 v₂

We have

       ${\mathrm{v}}_2 = \sqrt{2\mathrm{gh}}$

           = $\sqrt{2 \times 10 \times 0.2}$

       v₂ = 2 m/s

∴      0.02 × 600 = 0.02 v + 4 × 2

⇒             0.02 v = 12 - 8

⇒                    v = $\frac{4}{0.02}$

⇒                      v = 200 m/s

A.

11.2m/s

   $$\begin{gathered} {\mathrm{V}}_{\max }=\sqrt{\mathrm{\mu rg}}=\sqrt{0.64\times 20\times 9.8} \\ \mathrm{where},\mathrm{\mu }-\mathrm{coefficient} \mathrm{of} \mathrm{friction} \\ {\mathrm{V}}_{\max }=11.2 \mathrm{m}/\mathrm{s} \end{gathered}$$

A.

22.4 m/s

A car moves in a circular track so it perform circular motion. 

According to second law the force providing this acceleration is

           f_c = $\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

But according to static friction

         f_s ≤ μ_s N

     f = $\frac{\mathrm{m} {\mathrm{v}}^2}{\mathrm{R}} ⩽ {\mathrm{\mu }}_{\mathrm{s}} \mathrm{N}$

     ${\mathrm{v}}_{\max } = \sqrt{\mathrm{\mu } \mathrm{R} \mathrm{g}}$

Given:-  μ = 1.28 

where μ is  the coefficient of friction 

  R = 40 m 

The maximum velocity

          v_max = $\sqrt{\mathrm{\mu } \mathrm{R} \mathrm{g}}$ 

                   = $\sqrt{1.28 \times 40 \times 9.8}$

             v_max = 22.4 m/s