A1300 kg weather rocket accelerates upward at 10m/s2. it explodes 1.5 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 500 m . part a what was the speed of the heavier fragment just after the explosion?

To solve this problem it is necessary to apply the conservation equations of the Momentum. The equation that represents such momentum conservation is given by,

Final momentum = Initial momentum

Where,

=mass light

= mass heavy

= velocity of mass light

= velocity of mass heavy

= final velocity

One of the masses is heavier than the other previously detached so we will assume that mass is 1/3 of the total, and the heavy mass is 2/3 of the total.

We start finding the initial velocity, then

The distance traveled is then calculated by the kinematic equations of motion,

The speed of the lighter piece is given also for the kinematic equation of movement,

Replacing at the conservation of momentum equation we have,

Therefore the speed of the heavier fragment just after the explosion is 18.52m/s against the direction of movement.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day