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 (-47 m/s) and a second piece, also of mass m, moves with velocity (-47 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

Answer:

a) 22.06m/s

b)45°

Explanation:

Let 'm' be  mass of 1st and 2nd pieces

mass of 3rd piece=3m

$v_{1}$=velocity of 1st piece = -47iˆ  m/s

$v_{2}$=velocity of 2nd piece = -47jˆ  m/s

$v_{3}$=velocity of 3rd piece=?

By considering conservation of linear momentum, we have

m$v_{1}$ + m$v_{2}$ + 3m$v_{3}$=0

$v_{1}$ + $v_{2}$ + 3$v_{3}$=0

3$v_{3}$= - ($v_{1}$ + $v_{2}$ )

$v_{3}$ = -$\frac{1}{3}$ ($v_{1}$ + $v_{2}$ )

Substituting the values of $v_{1}$ and $v_{2}$ in above equation

$v_{3}$ = -$\frac{1}{3}$ (-47iˆ -47jˆ ) => 15.6iˆ + 15.6jˆ

(a)Magnitude of the velocity of  the third piece is given by

|$v_{3}$| =√15.6²+15.6² => 22.06m/s

(b) its direction (as an angle relative to the +x axis)  'θ'

θ= $tan^{-1} (\frac{15.6}{15.6} )$ => $tan^{-1} (1 )$ =>45°