Answer:
20 m/s
Explanation:
If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.
The velocity of A is 60 m/s at angle 0°.
The velocity of B is 30 m/s at angle 120°.
The velocity of C is v at angle θ.
In the x direction:
Momentum before = momentum after
(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)
0 = 60m − 30m + 3m v cos θ
0 = 30m + 3m v cos θ
-30m = 3m v cos θ
-10 = v cos θ
In the y direction:
Momentum before = momentum after
(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)
0 = 0 + 30√3 m + 3m v sin θ
-30√3m = 3m v sin θ
-10√3 = v sin θ
Square the two equations and add together:
(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²
100 + 300 = v² cos² θ + v² sin² θ
400 = v² (cos² θ + sin² θ)
400 = v²
v = 20
The speed of fragment C is 20 m/s.
