Question

A 1246 kg weather rocket accelerates upward at 11.7 m/s². It explodes 1.6 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 571 m. What were the speed of the heavier fragment just after the explosion?

Answer 1

M = 1246
Let the two fragments be , m1 & m2
Given , m1 = 2 * m2
m1 + m2 = 1246 kg
2 m2 + m2 = 1246
m2 = 1246/3 = 415.33 kg
m1 = 830.7 kg

The speed of the rocket 1.6 s after take off is :

( 11.7 m/s^2)* ( 1.6 s) = 18.72 m/s

before collision : 18.72 m/s at 1246 kg

after collision : V1 at 500 kg and V2 at 1000 kg

we can determine V1 using kinematics:

the height of the rocket at explosion is:

y_i = 1/2 a ( t^2)

    = (1/2) * (11.7)* (1.6)^2

     = 15 m

the final height of the light fragment is 571m, so

(V_f)^2 - (V₁)^2 = 2ay

0 - (V₁)^2 = 2*(- 9.8) *(571 - 18.72)

      Vf = square root (2*9.8*552.28)

            104.04m/s

By conservation of momentum at the exploision :

      1246* (18.72) = 830.7*(104.04) +415.3 V₂

1246 * 18.72 = m1 * v1 + m2 * v2
1246 * 18.72 = 830.7 * v1 + 415.3 * 104.04
v1 = - 23.93 m/s ( i.e downward)

I hope help you.....!!