Question

1460 kg weather rocket accelerates upward at 12 m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

Answer #1

Let the two fragments be , m1 & m2

Given , m1 = 2 * m2

m1 + m2 = 1460 kg

2 m2 + m2 = 1460

m2 = 1460/3 = 486.7 kg

m1 = 973.4 kg

vf = vi + a*t

vf = 0 + 12 * 2.0

vf = 24.0 m/s

h = 1/2*12*2.0^2

h = 24 m

vf^2 = vi^2 + 2*a*h

0 = vi^2 - 2*9.8*(530 - 24)

vi = 99.58 m/s

Now,Using momentum conservation,

1460 * 24 = m1 * v1 + m2 * v2

1460 * 24 = 973.4 * v1 + 486.7 * 99.58

v1 = - 13.8 m/s

So,

**Speed of the heavier fragment = 13.8 m/s
Direction = Downwards !!**

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