Question

a uniform shell explodes into three equal parts two of the parts fly off perpendicular to each other with velocities 9 metre per second and 12 metre per second find the velocity of the third part​

Answer 1

Answer:

12 because Mr Pet Sec convert into km Per hr 5/18

Answer 2

The velocity of the third part​ is 15 $(\frac{m}{s})$.

Explanation:

1. Let

 Total mass of shell = 3 M

 Initial velocity of shell before explosion = 0$(\frac{m}{s})$.

 Mass of first, second and third parts after explosion = M (Given)

 Velocity of first parts after explosion = $9\hat{i} (\frac{m}{s})$

 Velocity of second parts after explosion = $12\hat{j} (\frac{m}{s})$

 Velocity of third parts after explosion = $\vec{V}$

2.

 Here momentum is conserved before and after explosion.

 Means

 Initial Momentum = Final Momentum    ...1)

 $3M\times 0=M\times 9\hat{i}+M\times 12\hat{j}+M\times \vec{v}$

$\vec{v}=-9\hat{i}-12\hat{j}$      ...2)

3.  Magnitude of velocity of third parts

   $\left | \vec{v} \right |=\sqrt{9^{2}+12^{2}}$

    On solving

  $\left | \vec{v} \right |=15$$(\frac{m}{s})$