Question

A 1300 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?

Answer 1

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

$m_1v_2+m_2v_2 = (m_1+m_2)v$

Where,

$m_1$=mass light

$m_2$ = mass heavy

$v_1$= velocity of mass light

$v_2$ = velocity of mass heavy

$v_f$ = 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

$V=a*t$

$V=10m/s^2*2s = 20 m/s$

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

$X = V_0 t+\frac{1}{2}at^2$

$X=0+\frac{1}{2}*10*4$

$X=20m$

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

$V_f = V_i-2ax$

$0 = V_i - 2*9.81*(500-20)$

$v_i=9417.6$

$v_i=97.04 m/s$

Replacing at the conservation of momentum equation we have,

$m_1v_2+m_2v_2 = (m_1+m_2)v$

$(1300*\frac{1}{3})*97.04 + (1300*\frac{2}{3})*V_2= 1300*20$

$26000 = 42050.66 + 866.66V_2$

$V_2= -18.52 m/s$

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