Answer:
a) 942.06 m/s
b) - 220.74 m/s
c) 3.7 x 10⁶ J
Explanation:
M = mass of the body = 17 kg
v₀ = initial velocity of body before explosion = 320 i + 0 j + 0 k
m₁ = mass of one part = 7.9 kg
v₁ = velocity of one part after explosion = 0 i + 190 j + 0 k
m₂ = mass of second part = 2.3 kg
v₂ = velocity of second part after explosion = - 420 i + 0 j + 0 k
m₃ = mass of the third part = M - m₁ - m₂ = 17 - 7.9 - 2.3 = 6.8 kg
v₃ = velocity of third part after explosion = ?
Using conservation of momentum
M v₀ = m₁ v₁ + m₂ v₂ + m₃ v₃
17 (320 i + 0 j + 0 k) = (7.9) (0 i + 190 j + 0 k) + (2.3) (- 420 i + 0 j + 0 k ) + (6.8) v₃
5440 i = 1501 j - 966 i + (6.8) v₃
(6.8) v₃ = 6406 i - 1501 j
v₃ = 942.06 i - 220.74 j
x-component = 942.06 m/s
b)
y-component = - 220.74 m/s
c)
Energy released in the explosion is given as
E = (0.5) (M v₀² - m₁ v₁² - m₂ v₂² - m₃ v₃²)
E = (0.5) ((17) (320)² - (7.9) (190)² - (2.3) (420)² + (6.8) (sqrt(942.06² + (- 220.74))²))
E = 3.7 x 10⁶ J