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**

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