Question

Review I Constants Periodic Table A 1600 kg weather rocket accelerates upward at 10 m/s2. It explodes 1.9 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 Part A of 520 m What was the speed of the heavier fragment just after the explosion? Express your answer using two significant figures. nα ΑΣφ ? m/s Request Answer Submit Part B What was the direction of the heavier fragment just after the explosion? O upward O downward

Answer 1

speed of rocket at t = 1.9 s

= v = u + at = 0 + 10(1.9) = 19 m/s

height of impact = s = vₐᵥₑt = [0 + 19](1.9)/2 = 18.05 m

let the fragments be m and 2m (m+2m = 1600 = 3m)

conservation of momentum(mass x velocity):

=> momentum before impact = mom. after

lighter fragment reached a maximum height(from impact point) of 520 - 18.05 = 501.95 m

=> 501.95 = [initial vertical velocity]²/(2g) = u² / 19.62

=> u = √[501.95 x 19.62] = 99.24 m/s

CoM => 3m(19) = m(99.24) + 2mu'

=> 2u' = 57 - 99.24 = - 42.24

or u' = -21.12 m/s (downwards)

Part(a) v = 21 m/s....Answer.

Part(b) Downward.... Answer.


hope this helps.