Question

A bomb explodes into 3 pieces. First piece m1=1kg, moves at v1=2m/s at 30 degree to the horizontal m2=2kg moves at v2=1.5m/s to the right and m3=3kg moves at v3=4m/s at 45 degree below the horizontal. What was the bomb's initial velocity(magnitude and direction)?

Answer 1

let v be the speed of bomb before explosion

momentum before explosion Pi = (m1+m2+m3)*V

after explosion

v1 = 2*cos30i + 2*sin3o j

v2 = 1.5 i

v3 = 4*cos45i - 4*sin45 j

momentum after explosion Pf = m1*v1 + m2*v2 + m3*v3

from momentum conservation

pi = Pf

(m1+m2+m3)*V = m1*v1 + m2*v2 + m3*v3

(1 + 2 + 3)*V = (1*(2*cos30i + 2*sin3o j)) + (2*1.5i) + (3*(4*cos45i - 4*sin45 j))

6*V = (2*cos30 + 3 + 12*cos45)i + (2*sin30 - 12*sin45)j

V = 2.2 i - 1.25 j

magnitde = sqrt(2.2^2+1.25^2) = 2.53 m/s

direction = tan^-1(-1.25/2.2) = -29.6

direction is 29.6 degrees below the horizantal