Question

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-23 m/s) and a second piece, also of mass m, moves with velocity (-23 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the x axis) of the velocity of the third piece

Answer 1

Step-by-step explanation:

a. Given that:

`m-` mass of first & second piece,
`3m-`mass of 3rd piece,
`\bar v_1`-velocity of first piece(
`-23\dot i \ m/s`) and
`\bar v_2` as velocity of 2nd piece (
`-23\dot j \ m/s`)

Let
`\bar v_3` be velocity of 3rd piece=?

#Vessel is at rest before explosion. Considering conservation of linear momentum:

`m\bar v_1 +m\bar v_2 +3m \bar v_3=0` #Dividing both sides by
`m, m\\eq 0`

`\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-(1)/(3)(\bar v_1 +\bar v_2)`

#Plug the
`\bar v_1 ,\bar v_2` values:

`\bar v_3=-(1)/(3)(-23\dot i -23\dot j)=(1)/(3)(23\dot i +23\dot j)`

#So the magnitude of the third piece is:

`|\bar v_3|=√((23/3)^2+(23/3)^2)\\=10.84m/s`

Magnitude of the 3rd piece is 10.84 m/s

b. To find direction of the magnitude (as an angle relative to the
`x`-axis), we find
`\angle \theta`. The angle is obtained by getting the tan inverse as:

`\theta=tan^-^1((23/3)/(23/3))\\=45\textdegree`

-The direction of the magnitude (angle relative to the x-axis) is 45°