In: Physics
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?
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:
yi = 1/2 a ( t^2)
= (1/2) * (11.7)* (1.6)^2
= 15 m
the final height of the light fragment is 571m, so
(Vf)^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.....!!