Question

A bomb at rest, with mass m, explodes into three unequal pieces. 1/3 of the mass moves north with a speed of 48.0 m/s. ½ the mass moves 58° west of north with a speed of 27.0 m/s. Find the speed and direction of the last piece.

Answer 1

Total mass of the bomb is,

m = m1m+m2+m3

Hence, the mass of the third piece is,

m3 = m - m1 -m2 = m -(1/3)m - (1/2)m = (1/6)m

In this case, the momentum is conserved. Since the bomb is at rest initially,

the initial momentum is is zero. Pix = Piy = 0

X-direction:

pfx = m1(0) + m₂v₂cos 58^o + m₃v₃

But, Pfx = Pix

-m₂v₂sin 58^o i^ + m₃v_3x = 0

-(1/2)m*27*sin58^o i^+ (1/6)mv3x =0

-(1/2)*27*sin58^o i^+ (1/6)v_3x =0

Hence, the velocity of the thirs piece is,

   v_3x = [(1/2)*27*sin58^o]* 6 i^ = 68.69 i^

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In y-direction:

   pfy = m1v1 j^ m2v2cos58 j^ + m3v3

   = (1/3)m*48 j^ + 1/2*m*27*cos58 j^ +1/6 mv3y

But Pfy = Piy

(1/3)m*48 j^ + 1/2*m*27*cos58 j^ +1/6 mv3y = 0

(1/3)*48 j^ + 1/2*27*cos58 j^ +1/6 v_3y = 0

Hence, the velocity of third piece in y-direction is,

v_3y = -[ (1/3)*48 j^ + 1/2*27*cos58 j^]*6

= -138.9 m/s j^

Hence, the velocity of the third piece is,

v =  68.69 i^ -138.9 m/s j^

Magnitude: v3 = 154.97 m/s = 155 m/s

Direction: theta = tan-1[-138.9 / 68.69] = -63.68^o = 26.3^o