Li a 1500 kg weather rocket accelerates upward at io mis 2. it explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. photos reveal that the lighter frag- ment traveled straight up and reached a maximum height of 530 m. what were the speed and direction of the heavier frag- ment just after the explosion?

speed is  -20.0155 m/s

and direction is downward

Explanation:

given data

mass of rocket = 1500 kg

rocket accelerates upward = 10 m/s²

time t = 2 s

maximum height = 530 m

to find out

speed and direction of the heavier fragment just after the explosion

solution

we use here principle of conservation of momentum  that is

Mt Vt=M1 V1 + M2 V2       1

so here

first we find the initial velocity  that is express as

v = a × t     2

v = 10 × 2

v = 20 m/s

and distance S

S =u × t + 0.5 × a × t²      3

S = 0 + 0.5 × 10 ×   2²

S = 20 m

and now we find speed of the lighter piece  that is

v² = v1² + 2 ×  a × S        4

0 = v1² + 2 ×  -9.81 × (530-20)

v1² = 10006.2

v1 = 100.031 m/s

so now put all these in equation 1 we get  

Mt Vt = M1 V1 + M2 V2

1500 × 20 = (1500 × 1/3 ) × 100.031 + (1500 × 2/3) + V2

30000 = 50015.5 + 1000 V2

V2 = -20.0155 m/s

so speed is  -20.0155 m/s

that mean vertical component of second fragment will be move downward

0.55x745.7j and a calculator

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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